Just in case you end up wandering in this corner of the interweb I have to warn you that this is the wrong mountain to climb.
I've decided to use the wordpress.com platform and you should go this blog instead: Climbing the Mountain.
If you by any chance are interested in Hadronic Physics and/or are interested in a kind of a log in my research on that area you should also take a look at this other blog of mine: ateixeira.
Take care and visit those two links often.
As an application of the previous theorem 33 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 the previous theorem 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)} $ we have $ {\displaystyle \lim_{x \rightarrow +\infty} \log x = \mathrm{sup} [\mathbb{R}] = +\infty } $ and $ {\displaystyle \lim_{x \rightarrow 0} \log x = \mathrm{inf} [\mathbb{R}] = -\infty} $.
If it is the case that in the previous definition that $ {g(x) \neq 0 } $, we have the following equivalences:
This notions work exactly as they worked for sequences and they give the same type of information about the behavior of the functions in question.
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)} $.
Pay special attention to the previous example cause I've lost count of the number of times I see people keep the leading term in a polynomial function near the origin and get the answers all wrong when they are solving exercises.
Another example that has a lot of interest to us is:
$ \displaystyle \sin x ~ x \,\, (x \rightarrow 0) $
We can see that it is so because of $ {\displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1} $
| Remark 2
And it is time for us to introduce the concept of limit using the much loved and talked about epsilon-delta ($ { \epsilon - \delta } $) condition. Once again we are walking into regions of greater and greater rigor (so that we are more certain of what we say) 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 as near, input signals, output signals, or the somewhat weak definition of limit we been using so far.
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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. It tell us that, if if we restrict ourselves to points near $ {c} $ than the images of those points are ll near $ {a} $, if the function indeed have limit $ {a} $ in point $ {c} $.
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 36
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 35 with $ {\delta=1} $ there exists $ {\varepsilon > 0} $ such as
$ \displaystyle 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[ $ 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)} $ $ QED $
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| Example 10
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.
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After this one we 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) $
| Theorem 37
Let $ {D \subset \mathbb{R}} $, $ {f:D \rightarrow \mathbb{R}} $, and $ {c \in D \prime } $. If $ {\displaystyle \lim_{x \rightarrow c^-}f(x)=\lim_{x \rightarrow c^+}f(x)=a} $, then $ {\displaystyle \lim_{x \rightarrow c}f(x)=a} $. Proof: Let $ {\delta > 0} $. By the $ {\varepsilon-\delta} $ condition it is:
$ \displaystyle \exists \varepsilon_1>0:x \in V(c,\varepsilon_1)\cap D_{c^+} \Rightarrow f(x) \in V(a,\delta) $
$ \displaystyle \exists \varepsilon_2>0:x \in V(c,\varepsilon_2)\cap D_{c^-} \Rightarrow f(x) \in V(a,\delta) $ Thus by taking $ {\varepsilon =\mathrm{min} \left\lbrace \varepsilon_1, \varepsilon_2 \right\rbrace } $ it follows $ {x \in V(c,\varepsilon) \cap (D \setminus \left\lbrace c \right\rbrace ) \Rightarrow x \in V(c,\varepsilon) \cap D_{c^+}} $ or $ {x \in V(c,\varepsilon) \cap D_{c^- }\Rightarrow f(x) \in V(a,\delta)} $ In conclusion: $ { \forall \delta > 0 \exists \varepsilon > 0: x \in V(c,\varepsilon)\cap (D\setminus \left\lbrace c \right\rbrace ) \Rightarrow f(x) \in V(a,\delta) } $ which is equivalent to saying that $ {\displaystyle \lim_{x \rightarrow c} f(x)=a} $ $ QED $
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| Example 11
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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} $. This fact causes people to sometimes confuse both concepts. And this confusion isn't all that alarming because both concepts are indeed related. But we can see both concepts as a kind of a measure of how well behaved a function is.
To finish this I'll just state a theorem that sheds some light on the connections of these two concepts:
| Theorem 39
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. $ QED $
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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
As promised in the last post I'll start by explaining a little bit more carefully what we are trying to convey with the formalization of the concept of limit.
The first thing I want to say is that the concept of limit is a local one. 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 very good for common day to day knowledge but it is not good enough for Mathematics. In Mathematics we want to be the most rigorous and formal so that very few doubts are left in the end (this is an oversimplification). So, with the concept of limit what we are doing is formalizing what do we mean with the expressions far away, 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 isn't very sophisticated but it serves 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 } $. So $ { \displaystyle \lim_{x \in \mathbb{Q} \rightarrow c} f(x) \neq \lim_{x \in \mathbb{R}\setminus \mathbb{Q}\rightarrow c} f(x)} $ we can conclude that this limit doesn't exist. For $ {c=0} $ it is possible to prove 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: 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} $.
Now getting back into our normal course we'll continue generalizing the theorems we proved for sequences into the real functions:
The previous theorem states a very straightforward fact, but nevertheless, 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. But, 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 30 we where particularly interested in the cases when the limit is $ { \pm \infty} $ but we already seen in Theorem 28 that limit weakens order relationships. In this case if we have $ {f(x)\leq g(x)} $, for some neighborhood 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 neighborhood 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} $
We have $ {\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} $
| Example 8
$ \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 } $
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After this simple example we'll introduce a theorem that will state a somewhat obvious result. But we have to put our intuitions on a firm ground and this is exactly what this theorem will do. 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 either always to the right of $ {c} $, or always to the left of $ {c} $, that values of the images of those points have to converge to the same value.
| Theorem 27
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 28. 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. $ QED $
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| Example 9
$ \displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x} $ It is easy to see that this limit doesn't exist using the previous theorem. 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.
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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. We can say, in a pretty relaxed way, that we used a reasoning by contradiction. Theorem 27 is an implication theorem. By that I mean a theorem that states a relationship between two propositions where the fact of one them being true implies that the other one is also true.
So, we have proposition $ {P} $ and proposition $ {Q} $. And an implication between those two propositions can be stated as : "The validity of $ {Q} $ follows from certainty from 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 in Planet Earth).
In Theorem 27 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 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 meaningless statement for any $ {a} $ and so $ {\lim_{x \rightarrow 0}\dfrac{1}{x}} $ doesn't exist.
We will now state a group of theorems that generalize what we already saw for sequences.
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 while using concepts that are more in the realm of analysis.
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 between the elements of two sets (it transforms an input signal from the first set into an output signal in the second set). 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, a bit more formally we introduce:
Sometimes we may be interested not in how the function maps the whole of $ { D } $ but just on a particular subset of $ { D } $. So it makes sense to introduce:
| Definition 21
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} $
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As we did for sequences we can too define what is a bounded from above function, a bounded from below function, a bounded function and etc. As an example we'll give:
| Definition 22 $ {f} $ is said to be bounded iff $ {\exists \, \alpha > 0 : |f(x)| \leq \alpha \forall x \in D } $
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We will now introduce some light topological notions in order to shed some light into the study of limits and continuity. It won't be nothing too serious for now, but on multivariable calculus things we'll be a little more serious.
Once again so that we don't let things get too abstract let us give a concrete 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} $.
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 } $
We will also say that $ { +\infty } $ is a limit point of $ {E} $ if $ {E} $ isn't bounded above in $ { \mathbb{R} } $. And we'll say that $ { -\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 remember that if $ {E} $ isn't bounded above than it means that $ { \exists x_n \in E: \quad \lim x_n = +\infty } $. And this is just the definition of limit point.
| Definition 25
$ {c} $ is said to be a limit point of $ {E} $ if
$ \displaystyle \forall \delta > 0 \quad V(c,\delta) \cap E \setminus \left\lbrace c \right\rbrace \neq \emptyset \ \ \ \ \ (2) $
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| Definition 26
Given $ {D \subset \mathbb{R} } $, $ {f : D \rightarrow \mathbb{R}} $, $ {c \in D \prime } $ and $ { a \in \mathbb{R} } $. We say that $ {f} $ has limit $ {a} $ in point $ {c} $ if for all sequences $ {x_n \in D \setminus \left\lbrace c \right\rbrace } $ such as $ {\lim x_n = c} $ we have $ {\lim f(x_n) = a} $.
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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.
This a rough draft of the notion of limit but for now it shows in a pretty intuitive way what we mean by the concept of limit. Later on this rough draft of ours will be polished into an $ { \epsilon - \delta } $ condition and things will be more rigorous.
| Example 6
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} $
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| Example 7
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}} $ and so $ {+\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 we can conclude that $ {\displaystyle \lim_{x \rightarrow +\infty} \sin x } $ doesn't exist.
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In order for us to proceed deeper in the study of limits and continuity we have now to introduce the notions of one-sided limit.
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