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Home » March 2014
— More properties of continuous functions —
As an application of the previous definition let us look into $ {f(x)= \sin x/x}$. It is $ {D= \mathbb{R}\setminus \{0\}}$.
Since $ {\lim_{x \rightarrow 0} \sin x/x=1}$ we can define $ {\tilde{f}}$ as
$ \displaystyle \tilde{f}(x)=\begin{cases} \sin x/x \quad x \neq 0 \\ 1 \quad x=0 \end{cases}$
As another example let us look into $ {f(x)=1/x}$ Since $ {\lim_{x\rightarrow 0^+}f(x)=+\infty}$ and $ {\lim_{x\rightarrow 0^-}f(x)=-\infty}$ we can't define $ {\tilde{f}}$ for $ {1/x}$.
Finally if we let $ {f(x)=1/x^2}$ we have $ {\lim_{x\rightarrow 0^+}f(x)=lim_{x\rightarrow 0^-}f(x)=+\infty}$. Since the limits are divergent we still can't define $ {\tilde{f}}$.
In general one can say that given $ {f: D\rightarrow \mathbb{R}}$ and $ {c \in D'\setminus D}$ $ {\tilde{f}}$ exists if and only if $ {\lim_{x \rightarrow c}f(x)}$ exists and is finite.
| Theorem 42 Let $ {D \subset \mathbb{R}}$; $ {f,g: D\rightarrow \mathbb{R}}$ and $ {c \in D}$. If $ {f}$ and $ {g}$ are continuous functions then $ {f+g}$, $ {fg}$ and (if $ {g(c)\neq 0}$)$ {f/g}$ are also continuous functions. Proof: We'll prove that $ {fg}$ is continuous and let the other cases for the reader. Let $ {x_n}$ be a sequence of points in $ {D}$ such that $ {x_n \rightarrow c}$. Then $ {f(x_n) \rightarrow f(c)}$ and $ {g(x_n) \rightarrow c)}$ (since $ {f}$ and $ {g}$ are continuous functions). Hence it follows $ {f(x_n)g(x_n) \rightarrow f(x)g(x)}$ from property $ {6}$ of Theorem 19. Which is the definition of a continuous function. $ \Box$ |
Let $ {f(x))5x^2-2x+4}$. First we note that $ {f_1(x)=5}$, $ {f_2(x)=-2}$ and $ {f_3(x)=4}$ are continuous functions. Now $ {f_4(x)=4}$ also a continuous function. $ {f_5(x)=x^2}$ is continuous since it is the product of $ {2}$ continuous functions. $ {f_6(x)=-2x}$ is continuous since it is the product of $ {2}$ continuous functions. Finally $ {f(x)=5x^2-2x+4}$ is continuous since it is the sum of continuous functions.
As an application of the previous theorem let $ {f(x)=a^x}$. Since $ {a^x=e^{\log a^x}=e^{x \log a}}$ we can write $ {a^x=e^t \circ t=x\log a}$. Since $ {f(t)=e^t}$ is a continuous function and $ {g(x)=x \log a}$ is also a continuous function it follows that $ {a^x}$ is a continuous function (it is the composition of two continuous functions).
By the same argument we can also show that with $ {\alpha \in \mathbb{R}}$, $ {x^\alpha}$ (for $ {x \in \mathbb{R}^+}$) is also a continuous function in $ {\mathbb{R}^+}$.
Find $ {\displaystyle \lim_{x \rightarrow +\infty} \sin (1/x)}$.
We can write $ {\sin (1/x)= \sin t \circ (t=1/x)}$. Since $ {\displaystyle \lim_{x \rightarrow + \infty}(1/x)=0}$ it is, from Theorem 44, $ {\displaystyle \lim_{x \rightarrow +\infty} \sin (1/x)=\lim_{t \rightarrow 0}\sin t =0}$.
In general if $ {\displaystyle \lim_{x \rightarrow c} g(x)= a \in \mathbb{R}}$ it is $ {\displaystyle \lim_{x \rightarrow c} \sin (g(x))=\lim_{t \rightarrow a} \sin t = \sin a}$. In conclusion
$ \displaystyle \lim_{x \rightarrow c}\sin (g(x))=\sin (\lim_{x \rightarrow c}g(x)) $
Suppose that $ {\displaystyle \lim_{x \rightarrow c}g(x)=0}$ and let $ {\tilde{f}}$ be the function that makes $ {\sin x/x}$ be continuous in $ {x=0}$.
It is $ {\sin x = \tilde{f}(x)x}$, hence it is $ {\sin g(x) = \tilde{f}(g(x))g(x)}$.
By definition $ {\tilde{f}}$ is continuous so by Theorem 44 $ {\displaystyle \lim_{x \rightarrow c^+}f(g(x))=\lim_{t \rightarrow 0}\tilde{f}(t)=1}$.
Thus we can conclude that when $ {\displaystyle \lim_{x \rightarrow c}g(x)=0}$ it is
$ \displaystyle \sin (g(x))\sim g(x)\quad (x \rightarrow c)$
For example $ {\sin (x^2-1) \sim (x^2-1)\quad (x \rightarrow 1)}$.
Let $ {\displaystyle \lim_{x \rightarrow c}g(x)=a \in \mathbb{R}}$. By Theorem 44 it is $ {\displaystyle \lim_{x \rightarrow c} e^{g(x)}=\lim_{t \rightarrow a}e^t=e^a}$ (with the conventions $ {e^{+\infty}=+\infty}$ and $ {e^{-\infty}=0}$). Thus $ {\displaystyle \lim_{x \rightarrow c}e^{g(x)}=e^{\lim_{x \rightarrow c g(x)}}}$.
Analogously one can show that $ {\displaystyle \lim_{x \rightarrow c} \log g(x)= \log (\lim_{x \rightarrow c}g(x))}$ (with the conventions $ {\displaystyle \lim_{x \rightarrow +\infty} \log g(x)=+\infty}$ and $ {\displaystyle \lim_{x \rightarrow 0} \log g(x)=-\infty}$).
Let $ {a>1}$. It is $ {\displaystyle \lim_{x \rightarrow +\infty}a^x =\lim_{x \rightarrow +\infty}e^{x\log a}=e^{\displaystyle\lim_{x \rightarrow +\infty} x\log a}=+\infty }$ (since $ {\log a>0}$).
On the other hand for $ {\alpha > 0}$ it also is $ {\displaystyle \lim_{x \rightarrow +\infty}x^\alpha =\lim_{x \rightarrow +\infty}e^{\alpha \log x}= e^{\displaystyle \lim_{x \rightarrow +\infty}\alpha \log x}=+\infty}$.
The question we want to answer is $ { \lim_{x \rightarrow +\infty}\dfrac{a^x}{x^\alpha} }$ since the answer to this question tell us which of the functions tends more rapidly to $ {+\infty}$.
| Theorem 45 Let $ { a<1}$ and $ {\alpha > 0}$. Then Proof: Let $ {b=a^{1/(2\alpha)}}$ ($ {b>1}$). It is $ {a=b^{2\alpha}}$. Hence $ {a^x=b^{2\alpha x}}$. Moreover $ {\dfrac{a^x}{x^\alpha}=\dfrac{b^{2\alpha x}}{x^\alpha}=\dfrac{b^{2\alpha x}}{\sqrt{x}^{2\alpha}}}$. which is Let $ {[x]}$ denote the nearest integer function and using Bernoulli's Inequality ($ {b^m\geq 1+ m(b-1)}$) it is $ {b^x\geq x^{}[x]\geq 1+[x](b-1)>[x](b-1)>(x-1)(b-1)}$. Hence $ {\dfrac{b^x}{\sqrt{x}}>\dfrac{x-1}{\sqrt{x}}(b-1)=\left( \sqrt{x}-1/\sqrt{x}\right)(b-1)}$. Since $ {\displaystyle \lim_{x \rightarrow +\infty}\left( \sqrt{x}-1/\sqrt{x}\right)(b-1)=+\infty}$ it follows from Theorem 32 that $ {\displaystyle\lim_{x \rightarrow \infty} \frac{b^x}{\sqrt{x}}=+\infty}$. Using 18 and setting $ {t=b^x/\sqrt{x}}$ it is $ {\displaystyle\lim_{x \rightarrow \infty}\frac{a^x}{x^\alpha}=\lim_{t \rightarrow +\infty}t^{2\alpha}=+\infty}$ $ \Box$ |
| Theorem 47 Let $ {a>1}$, then $ {\displaystyle \lim \frac{a^n}{n!}}$=0. Proof: First remember that $ {\log n!=n\log n -n + O(\log n)}$ which is Stirling's Approximation. Since $ {\dfrac{\log n}{n} \rightarrow 0}$ it also is $ {\dfrac{O(\log n)}{n} \rightarrow 0}$. And
$ \displaystyle \frac{a^n}{n!}=e^{\log (a^n/n!)}=e^{n\log a - \log n!}$
Thus
$ \displaystyle \lim \frac{a^n}{n!}=e^{\lim(n\log a - \log n!)}$
For the argument of the exponential function it is $ {\begin{aligned} \lim(n\log a - \log n!) &= \lim n\log a-n\log n+n-O(\log n) \\ &=\lim \left(n\left(\log a -\log n+1 -\dfrac{O(\log n)}{n}\right)\right) \\ &=+\infty\times -\infty=-\infty \end{aligned}}$ Hence $ {\displaystyle \lim \frac{a^n}{n!}=e^{-\infty}=0}$. $ \Box$ |
| Lemma 48 Proof: Omitted. $ \Box$ |
| Theorem 49 Proof: Will be proven as an exercise. $ \Box$ |
| Corollary 50 Proof: Left as exercise for the reader. Make the change of variables $ {e^x=t+1}$ and use the previous theorem. $ \Box$ |
Generalizing the previous results one can write with full generality:
- $ {\sin g(x) \sim g(x) \quad (x \rightarrow c)}$ if $ {\displaystyle \lim_{x \rightarrow c} g(x)=0}$
- $ {\log (1+g(x)) \sim g(x) \quad (x \rightarrow c)}$ if $ {\displaystyle \lim_{x \rightarrow c} g(x)=0}$
- $ {e^{g(x)}-1 \sim g(x) \quad (x \rightarrow c)}$ if $ {\displaystyle \lim_{x \rightarrow c} g(x)=0}$
The $ {\epsilon}$ $ {\delta}$ condition is somewhat hard to get into our heads as neophytes. On top of that the similarity of the $ {\epsilon}$ $ {\delta}$ definition for limit and continuity can increase the confusion and to try to counter those frequent turn of events the first part of this post will try to clarify the $ {\epsilon}$ $ {\delta}$ condition by means of examples.
— $ {\epsilon}$ $ {\delta}$ for Continuity —
First we'll start things off with something really simple.
Let $ {f(x)=\alpha}$ which is obviously continuous.
The gist of the the $ {\epsilon}$ $ {\delta}$ reasoning is that we want to show that no matter the $ {\delta}$ that is chosen at first it is always possible to find an $ {\epsilon}$ that satisfies Heine's criterion for continuity.
Getting back to our function $ {f(x)=\alpha}$ it is $ {|f(x)-f(c)| < \delta}$. Here $ {f(x)=f(c)=\alpha}$ so
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |\alpha-\alpha| &< \delta \\ |0| &< \delta \\ 0 &< \delta \end{aligned}}$
Which is trivially true since $ {\delta > 0}$ by assumption. Hence any value of $ {\epsilon}$ will satisfy Heine's criterion for continuity and $ {f(x)=\alpha}$ is continuous at $ {c}$.
Since we never made any assumption about $ {c}$ other than $ {c \in {\mathbb R}}$ we conclude that $ {f(x)=\alpha}$ is continuous in all points of its domain.
Let us now look at $ {f(x)=x}$. Again we'll look at continuity for point $ {c}$ ($ {f(c)=c}$):
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |x-c| &< \delta \end{aligned}}$
The last expression is just we want at this stage since want to have something of the form $ {x-c}$ (the first part of the $ {\epsilon}$ $ {\delta}$ criterion).
If we let $ {\epsilon=\delta}$ it is $ {|x-c| < \epsilon}$ and this completes our proof that $ {f(x)=x}$ is continuous at point $ {c}$.
And again since we never made any assumption about $ {c}$ other than $ {c \in {\mathbb R}}$ we conclude that $ {f(x)=\alpha}$ is continuous in all points of its domain.
Now we let $ {f(x)=\alpha x + \beta}$ and will see if $ {f(x)}$ is continuous at $ {c}$.
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |\alpha x + \beta-(\alpha c + \beta)| &< \delta \\ |\alpha x -\alpha c| &< \delta \\ |\alpha||x-c| &< \delta \\ |x-c| &< \dfrac{\delta}{|\alpha|} \end{aligned}}$
Hence if we let $ {\epsilon=|\delta|/ |\alpha|}$ it is $ {|x-c|< \epsilon}$ and $ {f(x)=\alpha x + \beta}$ is continuous at $ {c}$.
As a final example of Heine's criterion of continuity we'll look into $ {f(x)=\sin x}$.
$ {\begin{aligned} |f(x)-f(c)| &< \delta \\ |\sin x-\sin c| &< \delta \end{aligned}}$
Since we want something like $ {|x-c| < g(\delta)}$ the last expression isn't very useful to us.
In this case we'll take an alternative approach which nevertheless works and has exactly the same spirit of what we've using so far.
Please look at every step I make with a critical eye and see if you can really understand what's going on with this deduction.
$ {\begin{aligned} |\sin x-\sin c| &= 2\left| \cos\left( \dfrac{x+c}{2}\right)\right| \left| \sin\left( \dfrac{x-c}{2}\right)\right|\\ &< 2\left| \sin\left( \dfrac{x-c}{2}\right)\right| \end{aligned}}$
Since $ {x \rightarrow c}$ we know that at some point $ {\dfrac{x-c}{2}}$ will be in the first quadrant. Thus
$ {\begin{aligned} 2\left| \sin\left( \dfrac{x-c}{2}\right)\right| &< 2\left|\dfrac{x-c}{2}\right| \\ &= |x-c|\\ &< \epsilon \end{aligned}}$
Where the last inequality follows by hypothesis.
That is to say that if we let $ {\epsilon=\delta}$ it is $ {|x-c|<\epsilon \Rightarrow | \sin x - \sin x | < \delta}$ which is the epsilon delta definition of continuity.
— $ {\epsilon}$ $ {\delta}$ for Limits —
After looking into some simple $ {\epsilon}$ $ {\delta}$ proofs for continuity we'll take a look at $ {\epsilon}$ $ {\delta}$ for limits.
The procedure is the same, but we'll state it explicitly so that people can see it in action.
Let $ {f(x)=2}$. We want to show that it is $ {\displaystyle \lim_{x \rightarrow 1}f(x)=2}$.
$ {\begin{aligned} |f(x)-2| &< \delta \\ |2-2| &< \delta \\ 0 &< \delta \end{aligned}}$
Which is trivially true for any value of $ {\delta}$, hence $ {\epsilon}$ can be any positive real number.
Let $ {f(x)=2x+3}$. We want to show that it is $ {\displaystyle \lim_{x \rightarrow 1}f(x)=5}$.
$ {\begin{aligned} |f(x)-5| &< \delta \\ |2x+3-5| &< \delta \\ |2x-2| &< \delta \\ 2|x-1| &< \delta \\ |x-1| &< \dfrac{\delta}{2} \end{aligned}}$
With $ {\epsilon=\delta/2}$ we satisfy the $ {\epsilon}$ $ {\delta}$ for limit.
As a final example let us look at the modified Dirichlet function that was introduced at this post.
$ \displaystyle f(x) = \begin{cases} o \quad x \in \mathbb{Q}\\ x \quad x \in \mathbb{R}\setminus \mathbb{Q} \end{cases}$
At that post it was proved that for $ {a \neq 0}$ $ {\displaystyle\lim_{x \rightarrow a}f(x)}$ didn't exist and it was promised that in a later date I'd show that $ {\displaystyle\lim_{x \rightarrow 0}f(x)=0}$ using the epsilon delta condition.
Since we now know what the epsilon delta condition is and already have some experience with it will tackle this somewhat more abstruse problem.
$ {\begin{aligned} |f(x)-f(0)| &< \delta \\ |f(x)-0| &< \delta \end{aligned}}$
Since $ {f(x)=0}$ or $ {f(x)=x}$ we have two cases to look at.
In the first case it is $ {|0-0| < \delta}$ which is trivially valid, hence $ {\epsilon}$ can be any real positive number.
In the second case it is $ {|x-0| < \delta}$. Hence letting $ {\epsilon=\delta}$ gets the job done.
Since we proved that $ {\displaystyle\lim_{x \rightarrow 0}f(x)=0=f(0)}$ the conclusion is that the modified Dirichlet function that was presented is only continuous at $ {x=0}$.
As was said previously, they don't make local concepts more local than that.
1.
a) Calculate $ { \displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k)}$ and $ {\displaystyle\sum_{k=p}^{m}(u_k - u_{k+1}}$
$ {\displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k)=u_{p+1}-u_{p}+u_{p+2}-u_{p+1}+\ldots +u_{m+1}-u_{m}}$
As we can see the first term cancels out with the fourth, the third with the sixth, and so on and all we are left with is the second and second last terms:
$ {\displaystyle \sum_{k=p}^{m}(u_{k+1}-u_k) = u_{m+1}-u_p}$
$ {\begin{aligned} \displaystyle \sum_{k=p}^{m}(u_k - u_{k+1})&= - \sum_{k=p}^{m}(u_{k+1}-u_k)\\ &= - (u_{m+1}-u_p)\\ &= u_p-u_{m+1} \end{aligned}}$
b) Calculate $ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}}$ using the previous result.
$ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}= \lim \sum_{k=1}^n \left( \frac{1}{k}-\frac{1}{k+1} \right) }$
Defining $ {u_k=1/k}$ the previous sum can be written as
$ {\begin{aligned} \displaystyle \lim \sum_{k=1}^n \left( u_k-u_{k+1} \right)&=\lim (u_1 - u_{n+1})\\ &= \lim \left(1-\frac{1}{n+1}\right)\\ &=1 \end{aligned}}$
This last result apparently has a funny story. Mengoli was the first one to calculate $ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}=1}$.
At the time this happened people did research in mathematics (I'm using this term rather abusively) in a somewhat different vein. They didn't rush to print what they found like today.
Many times people held out their results for years while tormenting their rivals about what they found.
This is exactly what Mengoli did. In the times he was around the theory of series wasn't much developed, thus this result, that we can calculate without being particularly brilliant in Mathematics, was something to take note of.
So, he wrote some letters to people saying that $ {\displaystyle \lim \sum_{k=1}^n\dfrac{1}{k(k+1)}=1}$, but not how he concluded that.
The other mathematicians he sent the result too didn't know about his methods and all they could do was to add numbers up explicitly and the only thing they could see was that even though they could sum more and more terms the result was always less than $ {1}$ and was got nearer and nearer to $ {1}$.
Of course this didn't prove nothing since summing up a billion terms isn't the same as summing an infinite number of terms and everyone but Mengoli was dumbfounded with that surprising result.
c) Calculate $ {\displaystyle \sum_{k=0}^{n-1}(2k+1) }$
In this exercise what we are calculating is the sum of $ {n}$ consecutive odd numbers. This result was already known to the ancient Greeks and the result wasn't nothing short to astounding to them.
But enough with the talk already:
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1)&=\sum_{k=0}^{n-1}\left[ (k+1)^2-k^2\right]\\ &= \sum_{k=0}^{n-1}(u_{k+1}-u_k) \end{aligned}}$
With $ {u_k=k^2}$
Using the now familiar formula
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}(2k+1) &= (n-1+1)^2-0^2\\ &= n^2 \end{aligned}}$
An astounding result indeed!
Just look to $ {\displaystyle \sum_{k=0}^{n-1}(2k+1)=n^2}$, interpret the result and try not to be as surprised as the ancient Greeks were.
2.
a) Using 1.a) and $ {a^k=a^k\dfrac{a-1}{a-1}\quad (a \neq 1)}$ calculate $ {\displaystyle \sum_{k=0}^{n-1} a^k }$
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1} a^k &= \displaystyle\sum_{k=0}^{n-1} \left[ a^k\frac{a-1}{a-1}\right]\\ &= \displaystyle \frac{1}{a-1}\sum_{k=0}^{n-1}\left( a^{k+1}-a^k\right)\\ &= \displaystyle\frac{1}{a-1}(a^n-1)\\ &= \displaystyle\frac{a^n-1}{a-1} \end{aligned}}$
b) Using a) establish the Bernoulli inequality $ {a^n-1 \geq n(a-1)}$ if $ {a > 0}$ and $ {n \in \mathbb{Z}^+}$
If $ {a=1}$ it is $ {1-1=n(1-1) \Rightarrow 0=0}$ which is trivially true.
If $ {n=1}$ it is $ {a-1=a-1}$ which is trivially true.
For $ {n \geq 2 }$ and $ {a>1}$ it is:
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}a^k&= 1+a+a^2+\ldots+a^{n-1}\\ &> 1+1+\ldots+1\\ &= n \end{aligned}}$
Thus
$ {\begin{aligned} \dfrac{a^n-1}{a-1} &> n \\ a^n-1 &> n(a-1) \end{aligned}}$
Since $ {a > 1}$
Finally if $ {0 < a <1 }$ it is
$ {\begin{aligned} \displaystyle \sum_{k=0}^{n-1}a^k&= 1+a+a^2+\ldots+a^{n-1}\\ &< 1+1+\ldots+1\\ &= n \end{aligned}}$
Thus
$ {\begin{aligned} \dfrac{a^n-1}{a-1} & < n \\ a^n - 1 & > n(a-1) \end{aligned}}$
Since $ {a < 1}$
c) Use b) to calculate $ {\lim a^n}$ if $ {a > 1}$ and then conclude that $ {\lim a^n=0}$ if $ {|a| < 1}$.
By b) it is
$ {\begin{aligned} a^n &> n(a-1)+1 \\ \lim a^n &\geq \lim \left( n(a-1)+1 \right)= +\infty \end{aligned}}$
Hence $ {\lim a^n = +\infty \quad (a>1)}$
For the second part of the exercise we will calculate instead $ {\lim |a^n|}$ since that we know that $ { u_n \rightarrow 0 \Leftrightarrow |u_n| \rightarrow 0}$
Let us make a change of variable $ {t=1/a}$. Thus $ {|a|=|1/t|}$ and
$ {\begin{aligned} \lim |a^n| &= \lim |1/t|^n\\ &= \dfrac{1}{\lim |t|^n}\\ &= \dfrac{1}{+\infty}\\ &=0 \end{aligned}}$
3. Consider the sequences $ {u_n=\left( 1+\dfrac{1}{n} \right)^n }$ and $ {v_n=\left( 1+\dfrac{1}{n} \right)^{n+1}}$
a) Calculate $ {\dfrac{v_n}{v_{n+1}}}$ and $ {\dfrac{u_{n+1}}{u_n}}$. Then use Bernoulli's inequality to show that $ {v_n}$ is strictly decreasing and that $ {u_n}$ is strictly increasing.
$ {\begin{aligned} \dfrac{v_n}{v_{n+1}} &= \dfrac{\left( 1+1/n \right)^{n+1}}{\left(1+1/(n+1)\right)^{n+2}}\\ &=\dfrac{\left(\dfrac{n+1}{n}\right)^{n+1}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}}\\ &= \dfrac{n}{n+1}\dfrac{\left(\dfrac{n+1}{n}\right)^{n+2}}{\left( \dfrac{n+2}{n+1} \right)^{n+2}}\\ &=\dfrac{n}{n+1}\left( \dfrac{(n+1)^2}{n(n+2)} \right)^{n+2}\\ &= \dfrac{n}{n+1}\left( \dfrac{n^2+2n+1}{n(n+2)} \right)^{n+2}\\ &=\dfrac{n}{n+1}\left( \dfrac{n(n+2)+1}{n(n+2)} \right)^{n+2}\\ &= \dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} \end{aligned}}$
After having calculated $ {v_n/v_{n+1}}$ we can use Bernoulli's inequality, with $ {a=1+\dfrac{1}{n(n+2)}}$ , to conclude that $ {v_n}$ is strictly decreasing.
$ {\begin{aligned} \dfrac{n}{n+1}\left( 1+\dfrac{1}{n(n+2)} \right)^{n+2} &> \dfrac{n}{n+1}\left(1 + \dfrac{n+2}{n(n+2)} \right)\\ &= \dfrac{n}{n+1}(1+1/n)\\ &= \dfrac{n}{n+1}\dfrac{n+1}{n}\\ &= 1 \end{aligned}}$
Thus $ {v_n}$ is strictly decreasing.
With a similar technique we can prove that
$ { \displaystyle u_{n+1}/u_n=\dfrac{n+1}{n}\left( 1- \dfrac{1}{(n+1)^2}\right)^{n+1}}$
After that by using Bernoulli's inequality like in the previous example one can show that $ {u_{n+1}/u_n>1}$ and thus $ {u_n}$ is strictly increasing.
c) Using a) and b) and $ {\lim u_n = e}$ prove the following inequalities $ {(1+1/n)^n < e <(1+n)^{n+1}}$.
$ {\begin{aligned} \lim v_n&= \lim(1+1/n)^n(1+1/n)\\ &= e\times 1\\ &= e \end{aligned}}$
We already know that $ {v_n}$ is decreasing so it is $ {v_n<(1+1/n)^{n+1}}$
On the other hand $ {u_n}$ is increasing and $ {\lim u_n=e}$ so $ {(1+1/n)^n<e}$.
Hence $ {(1+1/n)^n<e<(1+1/n)^{n+1}}$
d) Use c) to prove that $ { \displaystyle \frac{1}{n+1}<\log (n+1)-\log n <\frac{1}{n}}$
$ { \begin{aligned} (1+1/n)^n &< e \\ n \log \left( \dfrac{n+1}{n} \right) &< 1 \\ \log(n+1) - \log n &< \dfrac{1}{n} \end{aligned} }$
And now for the second part of the inequality:
$ { \begin{aligned} e &< \left(1+\dfrac{1}{n}\right)^{n+1} \\ 1 &< (n+1)\log \left(\dfrac{n+1}{n}\right) \\ \dfrac{1}{n+1} &< \log (n+1) -\log n \end{aligned}}$
In conclusion it is $ { \dfrac{1}{n+1}<\log (n+1)- \log n < \dfrac{1}{n} }$
4.
a) Using 3d) show that
$ { \displaystyle 1+\log k < (k+1)\log (k+1)-k\log k < 1+ \log(k+1) }$
From
$ { \begin{aligned} \dfrac{1}{k+1} &< \log (k+1) - \log k \\ 1 &< (k+1)\log(k+1) - (k+1)\log k \\ 1+ \log k &< (k+1)\log(k+1)-k \log k \end{aligned}}$
With a similar reasoning we can also prove that $ {(k+1)\log(k+1)-l\log k < 1+ \log(k+1)}$.
Thus it is $ {1+\log k < (k+1)\log(k+1)-k\log k < 1+ \log(k+1)}$
b) Sum the previous inequalities between $ {1 \leq k \leq n-1}$.
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1}(1+ \log k) &< \sum_{k=1}^{n-1} ((k+1)\log(k+1)-k \log k)\\ &< \displaystyle \sum_{k=1}^{n-1}(1+\log(k+1)) \end{aligned}}$
Now
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1} (1+ \log k) &= \sum_{k=1}^{n-1}1+\sum_{k=1}^{n-1}\log k\\ &= n-1 +\sum_{k=1}^{n-1}\log k \end{aligned}}$
And
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1}\log k &= \log 1 + \log2 +\ldots+\log(n-1)\\ &=\log((n-1)!) \end{aligned}}$
It also is
$ {\begin{aligned} \displaystyle \sum_{k=1}^{n-1}((k+1)\log(k+1) - k\log k)&= m\log n -\log 1\\ &=n\log n \end{aligned}}$
And $ {\displaystyle \sum_{k=1}^{n-1}(1+\log(k+1))=n-1+\log n!}$
Thus it is $ {n-1+\log(n-1)! < n\log n < n-1 \log n!}$
c) Conclude the following inequalities $ { n \log n -n +1 < \log n! < n \log n -n+1+\log n}$ and establish Stirling's approximation $ { \displaystyle \log n! = n\log n -n +r_n}$ with $ {e < C_n < en}$
$ { \begin{aligned} n-1 + \log (n-1)! &< n\log n \\ \log (n-1)! &< n\log n -n+1 \\ \log n! &< n\log n -n +1+\log n \end{aligned}}$
On the other hand
$ {\begin{aligned} n\log n &< n-1 + \log n! \\ n\log n -n +1 &< \log n! \end{aligned} }$
Thus
$ {\begin{aligned} n\log n -n +1 &< \log n! \\ &< n\log n -n +1 +\log n \end{aligned}}$
And from this follows $ {1 < \log n! -n\log n+n < 1+\log n}$
Defining $ {r_n=\log n! -n\log n+n}$ it is $ {\log n! = n\log n-n+r_n}$ with $ {1 < r_n < 1+\log n}$
5.
Show that $ {\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$ and that $ {\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$
We know that
$ { \begin{aligned} \dfrac{1}{n+1} &< \log(n+1)-\log n < \dfrac{1}{n} \\ \dfrac{1}{n+1} &< \log\left( \dfrac{n+1}{n}\right) < \dfrac{1}{n} \\ \dfrac{1}{n+1} &< \log\left( 1+\dfrac{1}{n}\right) <\dfrac{1}{n} \\ \dfrac{1/(n+1)}{1/n} &< \dfrac{\log (1+1/n)}{1/n}<1 \\ \lim \dfrac{n}{n+1} &\leq \lim \dfrac{\log (1+1/n)}{1/n} \leq \lim 1 \\ 1 &\leq \lim \dfrac{\log (1+1/n)}{1/n} \leq 1 \end{aligned}}$
Thus $ {\lim \dfrac{\log (1+1/n)}{1/n}=1}$ and this equivalent to saying that $ {\log \left(1+\dfrac{1}{n}\right)\sim \dfrac{1}{n}}$
Let $ {u_n = \dfrac{\log (1+1/n)}{1/n}}$. In this case it is $ {\dfrac{\log (1+1/n^2)}{1/n^2}=u_{n^2}}$.
Since $ {u_{n^2}}$ is a subsequence of $ {u_n}$ we know that $ {\lim u_{n^2}= \lim u_n}$ and so it also is $ {\log \left(1+\dfrac{1}{n^2}\right)\sim \dfrac{1}{n^2}}$.
6. Show that $ {u_n \sim v_n}$ and $ {v_n \sim w_n \Rightarrow u_n \sim w_n }$
By hypothesis it is $ {u_n=h_n v_n}$, $ {v_n=t_n w_n}$ with $ {h_n,t_n \rightarrow 1}$.
Substituting the second equality in the first we obtain $ {u_n = h_n t_n w_n}$.
Let $ {s_n = h_n t_n}$ and we write $ {u_n =s_n w_n }$ with $ {\lim s_n = \lim h_n \lim t_n =1\times 1=1}$.
Thus $ {u_n \sim w_n}$
7. Let $ {u_n = O\left(1/n\right)}$ and $ {v_n = O (1/ \sqrt{n})}$. Show that $ {u_n v_n = o ( 1/n^{4/3})}$.
$ {u_n = h_n 1/n}$ and $ {v_n = t_n 1/ \sqrt{n}}$ with $ {h_n}$ and $ {t_n}$ bounded sequences. Now
$ {\begin{aligned} u_n v_n &= \dfrac{h_n}{n} \dfrac{t_n}{\sqrt{n}}\\ &= \dfrac{h_n t_n}{n^{3/2}}\\ &=\dfrac{h_n t_n}{n^{1/6}}\dfrac{1}{n^{4/3}} \end{aligned}}$
Let $ {s_n = \dfrac{h_n t_n}{n^{1/6}}}$ it is $ {\lim s_n = \lim \dfrac{h_n t_n}{n^{1/6}} = 0}$ since $ {h_n t_n}$ is bounded.
Thus $ {u_n v_n = o (1/n^{4/3})}$
8. Using Stirling's approximation show that $ {\log n! = n\log n -n + O(\log n)}$
We know that it is $ {\log n! = n\log n -n + +r_n}$ with $ { 1< r_n < 1+\log n}$. Thus
$ {\begin{aligned} 0 &<\dfrac{1}{\log n}\\ &< \dfrac{r_n}{\log n}\\ &< \dfrac{1}{\log n} +1\\ &\leq \dfrac{1}{\log 2}+1 \end{aligned}}$
Where we used the fact that $ { \dfrac{1}{\log n}+1}$ is decreasing function.
Thus $ {\dfrac{r_n}{\log n}}$ is bounded and so $ {r_n=O(\log n)}$ as desired.
As an application of theorem 35 let us look into the functions $ {f(x)=e^x}$ and $ {g(x)=\log x}$.
Now $ {f:\mathbb{R} \rightarrow \mathbb{R^+}}$ and is a strictly increasing function, and $ {g:\mathbb{R^+} \rightarrow \mathbb{R}}$ also is a strictly increasing function.
By theorem 35 it is $ {\displaystyle \lim_{x \rightarrow +\infty}\exp x = \mathrm{sup} [\mathbb{R^+}] = +\infty}$ and $ {\displaystyle \lim_{x \rightarrow -\infty} \exp x= \mathrm{inf} [\mathbb{R^+}] = 0}$.
As for $ {g(x)}$ it is $ {\displaystyle \lim_{x \rightarrow +\infty} \log x=\sup [\mathbb{R}]=+\infty}$ and $ {\displaystyle \lim_{x \rightarrow 0} \log x = \inf [\mathbb{R}]=-\infty}$.
If in the previous definition $ {g(x)}$ doesn't equal zero:
- $ { f(x) \sim g(x) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 1}$.
- $ { f(x) = o (g(x)) \,\, (x \rightarrow c) \Leftrightarrow \displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = 0}$.
- $ { f(x) = O(g(x)) \,\, (x \rightarrow c) \Leftrightarrow \dfrac{f(x)}{g(x)} }$ is bounded in some neighborhood of $ {c}$.
As an example of the previous definitions we can say, with full generality, that for any polynomial function we can keep track of the term with the leading degree if we are interested in how it behaves for larger and larger values.
But on the other hand if we are interested on how the polynomial function behaves near the origin we have to keep track of the term with the smaller degree. To see that this is indeed so let us introduce the following example:
$ \displaystyle f(x) = x^2+x $
Now $ {x^2+x=(x+1)x}$. If we take $ {h(x)=x+1}$ it is $ {\displaystyle \lim_{x \rightarrow 0} h(x)=1}$ and so it is $ {x^2+x=O(x) \,\, (x \rightarrow 0)}$.
Another example that has a lot of interest to us is:
$ \displaystyle \sin x \sim x \,\, (x \rightarrow 0) $
We can see that it is so because of $ {\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1}$
— 6.6. Epsilon-delta condition —
And it is time for us to introduce the concept of limit using the $ { \epsilon - \delta }$ condition.
Once again we are walking into regions of greater and greater rigor at the expense of having to use more abstract concepts while we are doing it. Things are going to get a little harder for people that aren't used to this types of reasoning but please bear with me and you'll find it rewarding when you get used to it.
The point of the $ { \epsilon - \delta }$ condition is to avoid using fuzzy concepts near, input signals, output signals, or the somewhat weak definition of limit we been using so far.
In case you are wondering what that means the straightforward answer is that it means exactly what you're idea of a function having a limit in a given point is (I'm assuming you have the right idea). It tell us that if a function indeed has limit $ {a}$ in point $ {c}$ then, if we restrict ourselves to points near $ {c}$, the images of those points are all near $ {a}$.
Once again I tell the reader to look at this as if it were a game played between two (slightly odd) people.
One of them is choosing the $ {\delta}$ and the the other is choosing the $ {\varepsilon}$. But this game isn't just about choosing. The first player gets to choose any $ {\delta}$ he wants, but the second has to choose the right $ {\varepsilon}$that makes the condition hold.
If he can prove that he has an $ {\varepsilon}$ for every $ {\delta}$ that the other player chooses than he succeeds in the game and the function does have limit $ {a}$ at point $ {c}$.
| Theorem 38 Let $ {D \subset \mathbb{R}}$, $ {f: D \rightarrow \mathbb{R}}$, and $ {c \in D^\prime}$. If $ {\displaystyle \lim_{x \rightarrow c} f(x)}$ exists and is finite, than there exists a neighborhood of $ {c }$ where $ {f(x)}$ is bounded. Proof: Let $ {\displaystyle \lim_{x \rightarrow c} f(x) = a \in \mathbb{R}}$. By theorem 37 with $ {\delta=1}$ there exists $ {\varepsilon > 0}$ such as $ {\begin{aligned} x \in V(c,\varepsilon)\cap(D\setminus\left\lbrace c \right\rbrace ) &\Rightarrow f(x) \in V(a,1) \\ &\Rightarrow f(x) \in \left] a-1, a+1\right[ \end{aligned}}$ Thus $ {x\in V(c,\varepsilon)\cap(D\setminus\left\lbrace c \right\rbrace)\Rightarrow a-1 < f(x) < a+1}$. So $ {x \in V(c,\varepsilon) \cap D \Rightarrow f (x) \begin{cases} \leq \mathrm{max} \left\lbrace a+1,f(c)\right\rbrace \\ \geq \mathrm{max}\left\lbrace a+1,f(c)\right\rbrace \end{cases} }$ and $ {f(x)}$ is bounded in $ {V(c,\varepsilon)}$ $ \Box$ |
If $ {\displaystyle \lim_{x \rightarrow c} f(x)/g(x)}$ exists, then $ {f(x)= O(g(x))\,\, (x \rightarrow c)}$ since in this case it is $ {h(x)=f(x)/g(x)}$ and there exists some neighborhood of $ {c}$ where $ {h(x)}$ is bounded.
After this one may be interested in knowing how we can translate $ {\displaystyle \lim_{x \rightarrow c^+} f(x) = a}$ to a $ {\varepsilon - \delta}$ condition.
In this case we are considering $ {f(x)}$ only in the set $ {D_{c^+}}$ and so what we get is:
$ \displaystyle \forall \delta > 0 \exists \varepsilon > 0: \, x \in V(c,\varepsilon)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta)$
A few examples to clarify definition 34
-
$ \displaystyle f(x)=|x| \quad \forall x \in \mathbb{R}$
Let $ {c \in \mathbb{R}}$ and $ {x_n}$ a sequence such as $ {x \rightarrow c}$. Then $ {f(x_n)=|x_n|}$ and $ {\lim f(x_n) = \lim |x_n| = |c|}$. In conclusion $ {f(x_n) \rightarrow f(c)}$ which is equivalent to saying that $ {f}$ is continuous in $ {c}$. Since $ {c}$ can be any given point $ {f(x)=|x|}$ is continuous in $ {\mathbb{R}}$.
- Let $ {f(x)= \sin x}$ and $ {x_n}$ a sequence such as $ {x_n \rightarrow \theta}$. It is $ {\lim \sin x= \sin \theta}$ and by the same reasoning $ {\sin x}$ is also continuous.
- In general if $ {x_n \rightarrow c}$ it is $ {\lim f(x_n)=f(c)=f(\lim x_n)}$. So for $ {\exp (x)}$ it is $ {\lim \exp (x_n)=\exp (\lim x_n)}$.
If $ {x_n \rightarrow +\infty }$ it follows that $ {\lim \exp(x_n)=+\infty }$ and for $ {x_n \rightarrow -\infty}$ it follows that $ {\lim \exp(x_n)=0}$.
Thus if we define $ {\exp (+\infty)=+\infty}$ and $ {\exp (-\infty)=0}$ it follows that it always is $ {\lim \exp (x_n)=\exp (\lim x_n)}$.
- Analogously we can define $ {\log +\infty= +\infty}$ and $ {\log 0 = -+\infty}$ and it always is $ {\lim \log x_n = \log (\lim x_n)}$.
As can be seen the $ {\varepsilon - \delta}$ condition for continuity in point $ {c}$ is very similar to the one for limit $ {a}$ in point $ {c}$.
To finish this post I'll just state a theorem that sheds some light on the connections of these two concepts:
| Theorem 41 Let $ {D \subset \mathbb{R}}$, $ {f:D \rightarrow \mathbb{R}}$ and $ {c \in D \cap D^\prime}$. Then $ {f}$ it's continuous in point $ {c}$ if and only if $ {\displaystyle \lim_{x \rightarrow c} f(x) = c}$. Proof: Omitted. $ \Box$ |
So as this theorem shows the connection between continuity and limit is indeed a deep one, but we can look at the concept of limit as being an auxiliary tool to determine if a function is continuous or not and we should not confuse them.
In the next post I intend to write a little bit more about continuity but in the mean time a very good text about it can be found here
The first thing I want to say is that the concept of limit is a local concept.
In mathematical lingo what this means is that for a function to have a limit in a given point, $ {\displaystyle \lim_{x \rightarrow c} f(x) = a}$, it doesn't matter how the function behaves when we are far away from the point in question, what matters is just how the function behaves in the vicinity of the point.
This all very good for common day to day knowledge but it is not good enough for Mathematics.
So, with the concept of limit what we are doing is formalizing what we mean with the expressions far away and vicinity.
For an example let me introduce the function
$ \displaystyle f(x) = \begin{cases} o \quad x \in \mathbb{Q}\\ x \quad x \in \mathbb{R}\setminus \mathbb{Q} \end{cases}$
This function's not a very sophisticated but it's good enough for what I'm trying to convey.
First of all let us plot this function to see what it looks like.
Where we have drawn the $ {x \in \mathbb{Q}}$ case in blue and the $ {x \in \mathbb{R}\setminus \mathbb{Q}}$ case in red.
It is easy to see that for all $ {c}$ different than $ {0}$ the function has no limit.
For $ {c \neq 0}$ $ { \displaystyle \lim_{x \in \mathbb{Q} \rightarrow c} f(x) = 0 }$ and $ { \displaystyle\lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x) = c }$. Thus $ { \displaystyle \lim_{x \in \mathbb{Q} \rightarrow c} f(x) \neq \lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x)}$, and we can conclude that this limit doesn't exist.
For $ {c=0}$ it is possible to prove (I'll do that when the concept of limit is formalized using an $ { \epsilon-\delta }$ condition) that $ { \displaystyle \lim_{x \rightarrow 0} f(x) = 0}$.
They don't make concepts more local than this! This function only has a limit at point $ {0}$.
In an intuitive way we can understand this result like this: we can think that the concept of limit is a measure of how good behaved a function is.
Since this function is always jumping from point to point as we move from rational numbers to irrational numbers we can say that it isn't a well behaved one.
The former statement is true almost everywhere in the domain of the function. The only point where it breaks down is at point $ {0}$.
This is so because even though the function is a badly behaved one it misbehaves less and less while $ {x \rightarrow 0}$.
The previous theorem states a very straightforward fact, but, as always, what matters is that this result can be proven. In more prosaic terms this theorem expresses the conditions that need to fulfilled for us to know the limits of some functions just by knowing the limit of another one.
It may well be the case that one limit may be very easy to calculate while the other is not.
If we can establish an order relationship and calculate one of the limits it is possible for us to conclude something about the limit of the other function. In Theorem 32 we where particularly interested in the cases when the limit is $ { \pm \infty}$ but we already seen in Theorem 30 that limit weakens order relationships.
In this case if it is $ {f(x)\leq g(x)}$, for some neighbourhood around a point $ {c}$, then we know that $ {\displaystyle \lim_{x \rightarrow c} f(x)\leq \lim_{x \rightarrow c} g(x)}$ also. Now, if $ {\displaystyle\lim_{x \rightarrow c}f(x)=+\infty}$ $ {g(x)}$ has no choice but to go to positive infinity as we move closer to $ {c}$ since it has to be larger than $ {f(x)}$.
In the case $ {\displaystyle \lim_{x \rightarrow c} g(x) = -\infty}$ a similar reasoning applies.
$ {f(x)}$ is smaller than $ {g(x)}$ and if $ {g(x)}$ gets to smaller and smaller values as we approach $ {c}$ than $ {f(x)}$ also has to get smaller and smaller values.
This theorem continues the trend of computing limits of functions without computing them!
In here if we can box the function in a neighbourhood of a point by two functions, and if we compute the limits of the boxing functions and come to the conclusion that they are equal we are able to know that the boxed function has the same limit.
As an example let us see the limit:
$ \displaystyle \lim_{x \rightarrow +\infty} \frac{\sin x}{x}$
It is $ {-1 \leq \sin x \leq 1 \quad \forall x \in \mathbb{R}}$. Thus $ {\displaystyle -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \quad \forall x > 0}$.
Since $ {\displaystyle \lim_{x \rightarrow +\infty}-\frac{1}{x}=\lim_{x \rightarrow +\infty}\frac{1}{x}= 0}$ it also is $ {\displaystyle \lim_{x \rightarrow +\infty}-\frac{\sin x}{x}=0}$.
As a second example let us now look into:
$ \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}$
Since $ {\displaystyle \cos x < \frac{\sin x}{x} < 1\quad \forall x \in \left] -\frac{\pi}{2},0 \right[ \cup \left] 0,\frac{\pi}{2}\right[ }$
It is $ {\displaystyle \lim_{x \rightarrow 0}1=1}$ and $ {\displaystyle \lim_{x \rightarrow 0} \cos x = \cos 0 = 1}$. Thus it also is $ {\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}=1}$
— 6.5. Algebraic Properties of the Limit of Functions —
Just like we did for sequences we'll derive the algebraic rules that allows to manipulate the limits of some more complex expressions.
$ \displaystyle \lim_{x \rightarrow 0^+} \frac{1}{x}$
In this case it is $ {D_{0^+} = \left] 0, +\infty\right[ }$ and $ { 0^+ \in D_{0^+} }$.
If $ {x_n}$ is a sequence of points in $ {D_{0^+}}$ such as $ {x_n \rightarrow 0^+}$ it follows $ { \lim f(x_n) = \lim \dfrac{1}{x_n} = \dfrac{1}{0^+} = + \infty }$.
After this simple example we'll introduce a theorem that will state a somewhat obvious result.
In layman terms what it expresses is that if a function has a limit in a given point $ {c}$ than the one-sided limits have to be equal and equal to the limit of the function.
In a more kinematic way it tell us if we approach $ {c}$ by points of the domain from the right of $ {c}$, or from the left of $ {c}$, the images of those points have to converge to the same value.
| Theorem 29 Let $ {D \subset \mathbb{R}}$, $ {f: D \rightarrow \mathbb{R} }$, $ {c \in D^\prime}$ and let us suppose that $ {\displaystyle \lim_{x \rightarrow c} f(x) = a}$. Then, if $ {c \in D^\prime_{c^+}}$ it also is $ {\displaystyle \lim_{x \rightarrow c^+} f(x) = a}$; and if $ {c \in D^\prime_{c^-}}$ it also is $ {\displaystyle \lim_{x \rightarrow c^-} f(x) = a}$. Proof: Let $ {x_n}$ be a sequence of points in $ {D_{c^+} }$ such as $ {x_n \rightarrow c}$. Since $ {x_n}$ is a sequence of points in $ {D \setminus \left\lbrace c \right\rbrace}$ (by our choice of $ {x_n}$) and $ {\displaystyle \lim_{x \rightarrow c} f(x)=a }$ (by the hypothesis of the theorem) it follows from the definition of limit that $ { \lim f(x_n) = a }$. But this is just $ { \displaystyle \lim_{x \rightarrow c^+} f(x) = a}$ by definition 32. The case $ { \displaystyle \lim_{x \rightarrow c^-} f(x) }$ is proved in the same way with the due modifications and is left as an exercise for the reader. $ \Box$ |
$ \displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x} $
Using the previous theorem it is easy to see that this limit doesn't exist. We already know that $ {\displaystyle\lim_{x \rightarrow 0^+}\dfrac{1}{x}=+\infty}$ and that $ {\displaystyle \lim_{x \rightarrow 0^-} \dfrac{1}{x} = - \infty }$.
Since the limit from the right of $ {0}$ is different from the limit of the left of $ {0}$ we can conclude that this limit doesn't exist.
Just in case the previous example has caused some doubts on the reader I'll now try to explain in a more clear way the reasoning behind it.
Theorem 29 is an implication theorem. By that I mean a theorem that states a relationship between two propositions where the fact of the first proposition is true implies that the second proposition is also true.
So, we have proposition $ {P}$ and proposition $ {Q}$. And an implication between those two propositions can be for example: "The validity of $ {Q}$ depends on the validity of $ {P}$".
What this means is that if $ {P}$ is a true statement than the validity of $ {Q}$ will follow.
And that if $ {Q}$ is a false statement than $ {P}$ will also be a false statement.
In mathematical notation: (where $ {\neg P}$ means not $ {P}$) $ {P \Rightarrow Q \Leftrightarrow \neg Q \Rightarrow \neg P}$.
Note that if we have $ {\neg P}$ we can't conclude anything about the logical value of $ {Q}$ and that if we have $ {Q}$ we can't conclude anything about the value of $ {P}$.
An everyday situation may helps us here:
Imagine that you are waiting for and old friend from an uncle of yours called Pierre.
You have never known Pierre and the only thing that you know about him is that he only speaks French.
So a fellow comes to you and starts asking for directions in English. At that moment you can conclude that the fellow in question isn't Pierre ($ { \neg Q \Rightarrow \neg P }$).
If by chance some fellow comes near you speaking French than you can't conclude anything (remember that Pierre isn't the only French speaking guy on Planet Earth).
In Theorem 29 we had $ { \displaystyle \lim_{x \rightarrow c} f(x)=a \Rightarrow\lim_{x \rightarrow c^+} f(x)=a \land \lim_{x \rightarrow c^-} f(x)=a }$.
In this case $ {P}$ is $ { \displaystyle \lim_{x \rightarrow c} f(x) = a }$ and Q is $ { \lim_{x \rightarrow c^+} f(x) = a \land \lim_{x \rightarrow c^-} f(x) = a }$.
So by showing that $ {\displaystyle \lim_{x \rightarrow 0^+} \dfrac{1}{x} \neq \lim_{x \rightarrow 0^-} \dfrac{1}{x}}$ we arrived at the conclusion that we have $ {\neg Q}$ and so $ {\neg P}$ has to follow.
In this case $ {\neg P}$ is just the statement that $ {\lim_{x \rightarrow 0}\dfrac{1}{x} = a}$ is a meaningless statement for any $ {a\in{\mathbb R}}$ and so $ {\lim_{x \rightarrow 0}\dfrac{1}{x}}$ doesn't exist.
— 6.4. Limits of Functions and Inequalities —
We will now state a group of theorems that generalize what we already saw for sequences.| Theorem 30 (Limit of Inequalities) Let $ {D \subset \mathbb{R}}$, $ {f,g : D \rightarrow \mathbb{R}}$, $ {c \in D^\prime}$ and let us suppose that there exists $ {r > 0}$ such as $ {f(x) < g(x)\quad \forall x \in V(c,r) \cap (D\setminus \left\lbrace c \right\rbrace ) }$. If $ {\displaystyle \lim_{x \rightarrow c} f(x)}$ and $ {\displaystyle \lim_{x \rightarrow c} g(x)}$ exist it is $ {\displaystyle \lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} g(x)}$ Proof: Let $ {x_n}$ be a sequence of points in $ {D \setminus \left\lbrace c \right\rbrace }$ such as $ {x_n \rightarrow c}$. By definition 18 $ {\exists k \in \mathbb{N}:\quad n \geq k \Rightarrow x_n \in V(c,r) \Rightarrow x_n \in V(c,r) \cap D\setminus \left\lbrace c \right\rbrace }$. Since $ {x \in V(c,r) \cap D \setminus \left\lbrace c \right\rbrace \Rightarrow f(x) \leq g(x)}$. Thus $ {n \geq k}$ implies that $ {f(x_n) \leq g(x_n)}$. By Theorem 14 we know that it is $ {\displaystyle \lim f(x_n) \leq \lim g(x_n)}$. Since $ {\displaystyle \lim_{x \rightarrow c} f(x) = f(x_n)}$ and $ {\displaystyle \lim_{x \rightarrow c} g(x) = g(x_n)}$ it follows $ {\displaystyle \lim_{x \rightarrow c} f(x) \leq \lim_{x \rightarrow c} g(x)}$ $ \Box$ |
— 6. Limits and Continuity —
After introducing sequences and gaining some knowledge of some of their properties (I, II, III, and IV) we are ready to embark on the study of real analysis.
— 6.1. Preliminary Definitions —
Physics is expressed best and most powerfully in the language of mathematics and a very useful mathematical concept for physics is the concept of a function.
Generally speaking a function is an association (it transforms an input signal from the first set into an output signal of a second set) between the elements of two sets.
The sequences we studied are a special case of functions: they take natural numbers (or a subset of them) as their input signals and map them to real numbers.
Now, more formally we introduce:
Sometimes we may not be interested in how the function maps the whole of $ {D}$ but just on a particular subset of $ {D}$. So it makes sense to introduce:
| Definition 24 Given $ {E \subset D}$ it is $ {f\left[ E \right] = \left\lbrace f(x):x \in E \right\rbrace }$ is the image of $ {f}$ by $ {E}$. |
As we did for sequences we can too define what is a bounded above function, a bounded below function, a bounded function and etc.
As an example we'll give:
| Definition 25 $ {f}$ is said to be bounded iff $ {\exists \, \alpha > 0 : |f(x)| \leq \alpha \forall x \in D }$ |
— 6.2. Introduction to Topology —
Once again so that we don't let things get too abstract let us give an example:
$ \displaystyle E = \left] 0,1\right[ \cup \left\lbrace 2 \right\rbrace $
It is easy to see (and we won't give a rigorous proof of that) that $ {E^\prime= \left[ 0,1 \right] }$ and that $ {2}$ is the only isolated point of $ {E}$.
| Definition 28 The symbol $ {D_{c^+}}$ will be used to denote $ {D \cap \left] c, \infty \right[ }$ and the symbol $ {D_{c^-}}$ will denote $ {D \cap \left] - \infty , c \right[ }$ |
As an example let us calculate
$ \displaystyle \lim _{x \rightarrow 0^+} \frac{1}{x} $
In this case it is $ {D_{0^+} = \left] 0, \infty \right[ }$ and $ {0^+ \in D^\prime_{c^+}}$ so that the limit we intend to calculate indeed makes sense.
If $ {x_n}$ is a sequence of points in $ {D^\prime_{c^+}}$ such as $ {x_n \rightarrow 0^+}$ then it follows that $ {\lim f(x_n)=\lim \dfrac{1}{x_n}=\dfrac{1}{0^+}=+\infty }$
As an application of theorem 28 let us calculate the following limit
$ \displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x} $
It is easy to see that this limit doesn't exist. Let $ {f(x)=\dfrac{1}{x}}$ it is $ {\displaystyle \lim_{x \rightarrow 0^+} f(x) = +\infty}$ and $ {\displaystyle \lim_{x \rightarrow 0^-} f(x) = -\infty}$.
Since the limit from the left is different from the limit from the right we can conclude that $ {\displaystyle\lim_{x \rightarrow 0}\dfrac{1}{x}}$ doesn't exist.
| Definition 29 $ { +\infty }$ is a limit point of $ {E}$ if $ {E}$ isn't bounded above in $ { \mathbb{R} }$. $ { -\infty }$ is a limit point of $ {E}$ if $ {E}$ isn't bounded below in $ {\mathbb{R}}$. |
If you're having trouble understanding these definitions just think that if $ {E}$ isn't bounded above than it means that $ { \exists x_n \in E: \quad \lim x_n = +\infty }$.
Which is just the definition of limit point.
| Definition 30 $ {c}$ is said to be a limit point of $ {E}$ if $ \displaystyle \forall \delta > 0 \; V(c,\delta) \cap E \setminus \left\lbrace c \right\rbrace \neq \emptyset$ |
We'll only define the limit of a function in limit points of the domain. Notice that by this way we can too define the limit of points that don't belong in the domain of the function.
As always a few examples will be provided in order for us to test our knowledge.
- Calculate the limit of $ {\displaystyle \lim_{x \rightarrow + \infty} \dfrac{1}{x} }$.
$ { D = \mathbb{R} \setminus \left\lbrace 0 \right\rbrace }$ and $ { + \infty \in D^\prime }$ since $ {D}$ isn't bounded above in $ { \mathbb{R} }$. Thus the limit we set ourselves to calculate makes sense in our theory of limits.
Let $ {x_n}$ be a sequence of points in $ {D}$ such as $ { x_n \rightarrow + \infty }$ and $ {f(x)=\dfrac{1}{x}}$, then $ {f(x_n)=\dfrac{1}{x_n}}$ and it always is $ {\lim f(x_n)=0}$.
- Calculate the limit of $ {\displaystyle \lim_{x \rightarrow + \infty} \sin x }$
Choosing $ {f(x)= \sin x}$ we see that the domain is $ {D = \mathbb{R}}$. Thus $ {+\infty \in D^\prime}$
Let us choose $ {x_n = n \pi}$. Thus $ {x_n \rightarrow +\infty }$ and $ {f(x_n)=\sin x_n = 0}$.
In this case it trivially is $ {\lim f(x_n)=0}$.
Now if we choose $ {y_n=\pi/2 + 2n\pi}$ it also is $ {y_n \rightarrow + \infty}$ but $ {f(y_n)= \sin (\pi/2+2n\pi)=1}$ and so $ {\lim f(y_n)=1}$.
Thus we were able to find $ {x_n}$, $ {y_n}$ such as $ {\lim x_n = \lim y_n = + \infty}$ but $ {\lim f(x_n) \neq \lim f(y_n)}$. Thus $ {\displaystyle \lim_{x \rightarrow +\infty} \sin x }$ doesn't exist.
Formalizing the previous notions:
1.
a) Given the sequence $ { \dfrac{n^2+1}{2n^2-1}}$ prove that there exists an order $ { k}$ where $ { \left | u_n - \dfrac{1}{2} \right |<10^{-3}}$ is valid.
$ { \begin{aligned} \left | \dfrac{n^2+1}{2n^2-1} - \dfrac{1}{2} \right | &< 10^{-3} \\ \left | \dfrac{2n^2+2-2n^2+1}{2(2n^2-1)} \right | &< 10^{-3} \\ \left | \dfrac{3}{2(2n^2-1)} \right | &< 10^{-3} \\ \dfrac{|3|}{|2(2n^2-1)|} &< 10^{-3} \end{aligned}}$
Since $ { 2(2n^2-1)>0}$ what follows is
$ { \begin{aligned} \dfrac{3}{2(2n^2-1)} &< 10^{-3} \\ 3/2 \times 10^3 &< 2n^2-1 \\ 3/4\times 10^3+1/2 &< n^2 \\ \sqrt{3/4\times 10^3 + 1/2} &< n \end{aligned}}$
So by taking $ { k > \left \lfloor \sqrt{3/4\times 10^3 + 1/2}\right \rfloor +1}$ we have the intended result.
b) Prove by definition that $ { u_n \rightarrow 1/2}$
By the definition of limit, and using a), we have $ { n > \sqrt{\dfrac{3}{4 \delta}+1/2}}$. If we take $ { k= \left \lfloor \sqrt{\dfrac{3}{4 \delta}+1/2} \right\rfloor+1}$ the difference between $ { u_n}$ and $ { 1/2}$ is always less than $ { \delta}$.
2. Prove that $ { \lim u_n = 0 \Leftrightarrow \lim |u_n| = 0}$
The gist of this result is that sometimes is easier to prove that the modulus of a sequence tends to zero than to prove the sequence tends to zero. Since one result implies the other we may avoid doing some boring calculations.
We say that $ { u_n \rightarrow a}$ iff $ { \forall \delta > 0 \, \exists k \in \mathbb{N}: \quad n>k \Rightarrow |u_n - a| < \delta}$
Thus $ { \lim |u_n - a| = 0}$ iff
$ {\forall \delta > 0\,\exists k\in\mathbb{N}:\; n > k\Rightarrow||u_n-a|-0| < \delta}$
$ {\Leftrightarrow \forall \delta > 0 \, \exists k \in \mathbb{N}:\; n > k \Rightarrow |u_n - a| < \delta}$
Hence with $ { a=0}$ the conditions $ { \lim u_n = 0}$ and $ { \lim |u_n| = 0}$ are indeed equivalent.
3. Calculate $ { \lim \sqrt{n+1}-\sqrt{n}}$
This limit we are interested in calculating can be seen as $ { \lim u_n - v_n}$ where $ { u_n = \sqrt{n+1}}$ and $ { v_n = \sqrt{n}}$. We know that $ { \lim u_n = \lim \sqrt{n+1} = +\infty}$ and $ { \lim v_n = \lim \sqrt{n} = +\infty}$.
So what we are trying to measure is how fast these sequences diverge. If this limit is $ { a \in \mathbb{R}^+}$ then $ { u_n}$ grows slightly faster, if it is $ { a \in \mathbb{R}^-}$ then it is $ { v_n}$ that grows slightly faster.
In the case of $ { \pm \infty}$ we have that $ { u_n}$, for the $ { +}$ sign ($ { v_n}$ for the $ { -}$ sign) tends infinitely faster to the given limit.
On with the calculations now:
$ {\begin{aligned} \lim \sqrt{n+1}-\sqrt{n} &= \lim \dfrac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}} \\ &= \lim \dfrac{n+1-n}{\sqrt{n+1} + \sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n+1}+\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n(1+1/n)}+\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n}\sqrt{1+1/n}\sqrt{n}} \\ &= \lim \dfrac{1}{\sqrt{n}\left( \sqrt{1+1/n}+1 \right)} \\ &= \lim\dfrac{1}{\left( \sqrt{1+1/n}+1 \right) } \lim \dfrac{1}{\sqrt{n}} \\ &= \lim\dfrac{1}{2 \sqrt{n}} \\ &= 0 \end{aligned}}$
4. Calculate $ { \lim \left( \sqrt{n^2+n} - \sqrt{n^2+1} \right)}$
$ {\begin{aligned} \lim \left( \sqrt{n^2+n} - \sqrt{n^2+1} \right)&=\lim \dfrac{n^2+n-n^2-1}{\sqrt{n^2+n} + \sqrt{n^2+1}} \\ &=\lim \dfrac{n-1}{\sqrt{n^2\left(1+\frac{1}{n}\right)} + \sqrt{n^2\left(1+\frac{1}{n^2}\right)}} \\ &=\lim \dfrac{n-1}{n \sqrt{1+\frac{1}{n}} + n\sqrt{1+\frac{1}{n^2}}} \\ &=\lim \dfrac{n-1}{n\left( \sqrt{1+\frac{1}{n}} + \sqrt{1+\frac{1}{n^2}} \right)} \\ &=\lim \dfrac{n-1}{2n} \\ &=\dfrac{1}{2} \end{aligned}}$
5. Calculate $ { \lim \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2}}$.
Let us write out a few terms of this sum so that we can gain some intuition of what's going on
$ { \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} = \dfrac{1}{(n+1)^2}+\dfrac{1}{(n+2)^2}+\cdots + \dfrac{1}{(2n)^2}}$
So by making $ { n \rightarrow \infty}$ what we get is a bigger number of increasingly small numbers to add.
The result of this limit will tell us what of these two contradictory effects wins.
Since we are adding up $ { n}$ terms that get increasingly small we have
$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \leq \dfrac{n}{(n+1)^2}$
And also
$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \geq \dfrac{n}{4n^2}$
Hence $ { \dfrac{n}{4n^2} \leq \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} \leq \dfrac{n}{(n+1)^2}}$ with $ { \lim \dfrac{n}{4n^2} = \lim \dfrac{n}{(n+1)^2} = 0}$
Thus $ { \lim \displaystyle \sum _{k=1}^n \dfrac{1}{(n+k)^2} = 0}$ also.
Hence the fact that the fractions tend to $ { \,0}$ is more relevant to the value of the limit than the the fact that we get an infinite number of fractions to add.
6. Calculate $ { \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}}}$
Now we have a similar situation to the previous exercise but this time the numbers that are being added are bigger than the previous ones. So it begs the question: What will be the limit this time?
Like previously we are summing $ { n}$ increasingly small terms and so
$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \geq \dfrac{n}{\sqrt{2n}}$
And
$ \displaystyle \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \leq \dfrac{n}{\sqrt{n+1}}$
Thus $ { \dfrac{n}{\sqrt{2n}} \leq \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} \leq \dfrac{n}{\sqrt{n+1}}}$.
Since $ { \lim \dfrac{n}{\sqrt{2n}} = \lim \dfrac{1}{\sqrt{2}}\dfrac{n}{\sqrt{n}}= \lim \dfrac{1}{\sqrt{2}}\sqrt{n} = + \infty}$ and $ { \lim \dfrac{n}{\sqrt{n+1}} = \lim \dfrac{n}{\sqrt{n}}\dfrac{1}{\sqrt{1+1/n}} = \lim \sqrt{n}\dfrac{1}{\sqrt{1+1/n}} = +\infty}$ it follows that $ { \displaystyle \sum _{k=1}^n \dfrac{1}{\sqrt{n+k}} = + \infty}$
This time the fact that the number of fractions to add grows without a bound is more relevant than the fact that those numbers tend to $ { \,0}$.
7. Calculate $ { \lim \dfrac{n^n}{n!}}$
Let us do this visually:
$ { n^{n-1} = n \times n \times n \ldots \times n}$ with $ { n-1}$ terms.
$ { n! = 1 \times 2 \times 3 \times \ldots \times n = 2 \times 3 \times \ldots \times n}$ with $ { n-1}$ terms.
So $ { \lim \dfrac{n^n}{n!} \geq \lim \dfrac{n^n}{n^{n-1}} = + \infty}$. Then also $ { \lim \dfrac{n^n}{n!} = +\infty}$
From this we can see that $ { n^n}$ grows to infinity faster than $ { n!}$
8. Give examples of sequences that
a) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n=0}$
$ { u_n = n}$ and $ { v_n = -n}$
b) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n=10}$
$ { u_n = n+10}$ and $ { v_n = -n}$
c) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n=+\infty}$
$ { u_n = 2n}$ and $ { v_n = -n}$
d) $ { u_n \rightarrow +\infty}$ and $ { v_n \rightarrow - \infty}$: $ { u_n+v_n}$ doesn't exist.
$ { u_n = n+(-1)^n}$ and $ { v_n = -n}$
e) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n = a \in \mathbb{R}}$
$ { u_n = \dfrac{a}{n}}$ and $ { v_n = n}$
f) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n = 0}$
$ { u_n = \dfrac{1}{n^2}}$ and $ { v_n = n}$
g) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n = +\infty}$
$ { u_n = \dfrac{1}{n}}$ and $ { v_n = n^2}$
h) $ { u_n \rightarrow 0}$ and $ { v_n \rightarrow \infty}$: $ { u_n v_n}$ doesn't exist
$ { u_n = \dfrac{\sin n}{n}}$ and $ { v_n = n}$
After having stated and/or proved some important theorems about sequences in the previous post we will know introduce some auxiliary notions that will help us continuing our study of sequences.
— 5.3. Relationships Between Sequences —
As an example let us consider the sequence $ { u_n=3n^2-5n+1}$. It is easy to see that in this case we have $ { u_n \sim 3n^2}$.
We can write $ { 3n^2-5n+1=3n^2\left(1-\dfrac{5}{3n}+\dfrac{1}{3n^2}\right)}$.
In this case it is $ { h_n=1-\dfrac{5}{3n}+\dfrac{1}{3n^2}}$ and we have $ { \lim h_n = \lim \left( 1-\dfrac{5}{3n}+\dfrac{1}{3n^2} \right) = 1}$.
Let us now try to give a more intuitive meaning to these three notions introduced so far:
First of the notion $ { u_n \sim v_n}$ expresses the fact the difference between $ { u_n}$ and $ { v_n}$ tends to $ { 0\,}$ as $ { n \rightarrow \infty}$. That is to say that the two sequences get closer and closer together.
The notion of $ { u_n = O(v_n)}$ expresses the fact the both sequences differ only by a scale factor. That is to say that they have the same kind of behavior at $ { \infty}$.
The meaning of the sentence the same kind of behavior will be made clearer as real analysis gets unfolded in this blog.
The notion of $ { u_n = o(v_n)}$ tell us at the $ { u_n}$ gets smaller and smaller when compared to $ { v_n}$ when we get to $ { \infty}$. In a more formal way: if $ { v_n \neq 0 \quad \lim \dfrac{u_n}{v_n}=0}$.
Let us now give some examples in order to make things a little bit easier to grasp:
$ \displaystyle \dfrac{1}{n^3}=o \left(\dfrac{1}{n}\right)$
This is easy to see if we write $ { \dfrac{1}{n^3}=\dfrac{1}{n^2}\dfrac{1}{n}}$. Taking $ { h_n = \dfrac{1}{n^2}}$ we see that it is effectively $ { \lim h_n=0}$.
$ \displaystyle \dfrac{\sin n}{n}=O\left(\dfrac{1}{n}\right)$
In this case we write $ { \dfrac{\sin n}{n}=\sin n \dfrac{1}{n}}$ and take $ { h_n=\sin n}$. Since $ { \sin n}$ is a bounded function we get the intended result.
— 5.4. Final Comments on Sequences —
| Definition 21 We'll say that $ { u_{\alpha_n}}$ is a subsequence of $ { u_n}$ whenever $ { \alpha_n}$ is a sequence that tends to $ { \infty}$. |
Roughly speaking a subsequence, $ { u_{\alpha_n}}$, of a given sequence, $ { u_n}$, is sequence that doesn't consider some of the indexes of the initial sequence.
A few examples of subsequences would be $ { u_{2n}}$ (where we don't take into account the odd numbered indexes of the initial sequence), $ { u_{n^2}}$ (only taking into account the the perfect square indexes of the initial sequence).
We already saw that $ { u_n = \left (1+\dfrac{1}{n} \right )^n}$ was a converging sequence, then even though $ { v_n = \left (1+\dfrac{1}{n^2} \right )^{n^2}}$ appears to be a harder sequence we can say, without any effort, that $ { \lim \left (1+\dfrac{1}{n} \right )^n = \lim \left (1+\dfrac{1}{n^2} \right )^{n^2}}$ if we note that it is actually $ { v_n=u_{n^2}}$ and so $ { v_n}$ is a subsequence of a converging sequence.
As an application from the previous corollary we have $ { u_n = (-1)^n}$.
$ { u_{2n}= (-1)^{2n}=1}$ and it is $ { \lim u_{2n}=1}$.
$ { u_{2n+1}=(-1)^{2n+1}=-1}$ and it is $ { \lim u_{2n+1}=-1}$.
In conclusion $ { u_n=(-1)^n}$ is a divergent sequence.
| Theorem 26 (Bolzano-Weierstrass) Each bounded sequence has a converging subsequence in $ { \mathbb{R}}$. Proof: This is only the sketch of a proof. One way to do this is first to prove that all sequences have a monotone subsequence. Applying this result to a bounded sequence we'd have that the bounded sequence have a subsequence that is monotone and bounded (since the sequence is bounded). But by the Corollary 21 we know that a bounded and monotone sequence is convergent. $ \Box$ |
| Definition 22 Let $ { X \subset \mathbb{R}}$. We'll say that $ { X}$ is a compact interval if it is bounded and closed. |
| Corollary 27 Let $ { X}$ be a compact interval and $ { u_n : \mathbb{N} \rightarrow X}$. Then $ { \exists \, u_{\alpha_n}: \quad \lim u_{\alpha_n}=x \in X}$ where $ { u_{\alpha_n}}$ is a subsequence of $ { u_n}$. Proof: Let $ { X= \lbrack a, b \rbrack}$ be the interval and $ { u_n}$ be a sequence of points in $ { X}$. Since $ { a \leq u_n \leq b}$, $ { u_n}$ is bounded. From the theorem 26 $ { u_n}$ has a converging subsequence $ { u_{\alpha_n}}$. For $ { u_{\alpha_n}}$ it also is $ { a \leq u_{\alpha_n} \leq b}$. This implies $ { \lim a \leq \lim u_{\alpha_n} \leq \lim b \Rightarrow a \leq \lim u_{\alpha_n} \leq b\Rightarrow \lim u_{\alpha_n} \in X}$ $ \Box$ |
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