Definition 35 Let $ {D \subset \mathbb{R}}$; $ {f: D\rightarrow \mathbb{R}}$ and $ {c \in D'\setminus D}$. If $ {\displaystyle \lim_{x\rightarrow c}f(x)=a\in \mathbb{R}}$, we can define $ {\tilde{f}}$ as: $ \displaystyle \tilde{f}(x)=\begin{cases} f(x) \quad x \in D \\ a \quad x=c \end{cases} \ \ \ \ \ (16)$

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



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.

Theorem 43 Let $ {D, E \subset \mathbb{R}}$, $ {g: D\rightarrow E}$, $ {f: E \rightarrow \mathbb{R}}$ and $ {c \in D}$. If $ {g}$ is continuous in $ {c}$ and $ {f}$ is continuous in point $ {g(c)}$, then the composite function $ {f \circ g (x)=f(g(x)) }$ is continuous in point $ {c}$. Proof: Let $ {x_n}$ be a sequence of points in $ {D}$ with $ {x_n \rightarrow c}$. Hence $ {\lim g(x_n)=g(c)}$. If $ {f}$ is continuous in $ {g(c)}$ it also is $ {\lim f(g(x_n))=f(g(c))}$. This is $ {\lim (f \circ g)(x_n)= (f \circ g)(c)}$. Thus $ {f \circ g}$ is continuous in $ {c}$. $ \Box$

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}^+}$.

Theorem 44 Let $ {D, E \subset \mathbb{R}}$, $ {g: D\rightarrow E}$, $ {f: E \rightarrow \mathbb{R}}$ and $ {c \in D'}$. Suppose that $ {\displaystyle \lim_{x \rightarrow c}g(x)=a}$ and that $ {\displaystyle \lim_{t \rightarrow a}f(t)}$ exists. If $ {f}$ is continuous it follows $ {\displaystyle \lim_{x \rightarrow c}f(g(x))=\lim_{t \rightarrow a}f(t)}$. Proof: Omitted. $ \Box$

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



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



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 $ \displaystyle \lim_{\infty}\frac{a^x}{x^\alpha}=+\infty \ \ \ \ \ (17)$ 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 $ \displaystyle \frac{a^x}{x^\alpha}=\left( \frac{b^x}{\sqrt{x}} \right)^{2\alpha} \ \ \ \ \ (18)$ 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$

Corollary 46 Let $ {\alpha > 0}$, then $ \displaystyle \lim_{x \rightarrow +\infty}\frac{x^\alpha}{\log x}=+\infty $ Proof: Left as an exercise for the reader (remember to make the convenient change of variable). $ \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 $ \displaystyle \lim_{x \rightarrow +\infty}\left( 1+\frac{1}{x}\right)^x=e \ \ \ \ \ (19)$ Proof: Omitted. $ \Box$

Theorem 49 $ \displaystyle \lim_{x \rightarrow 0}\frac{\log (1+x)}{x}=1 \ \ \ \ \ (20)$ Proof: Will be proven as an exercise. $ \Box$

Corollary 50 $ \displaystyle \lim_{x \rightarrow 0}\frac{e^x-1}{x} \ \ \ \ \ (21)$ 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}$