## Real Analysis – Limits and Continuity VI

— More properties of continuous functions —

 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 ${\displaystyle\lim_{x \rightarrow 0} \sin x/x=1}$ 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 ${\displaystyle\lim_{x\rightarrow 0^+}f(x)=+\infty}$ and ${\displaystyle\lim_{x\rightarrow 0^-}f(x)=-\infty}$ we can’t define ${\tilde{f}}$ for ${1/x}$.

Finally if we let ${f(x)=1/x^2}$ we have ${\displaystyle\lim_{x\rightarrow 0^+}f(x)=\displaystyle\lim_{x\rightarrow 0^-}f(x)=+\infty}$. 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 ${\displaystyle\lim_{x \rightarrow c}f(x)}$ 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 ${g(c)}$, 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, ${\displaystyle \lim_{x \rightarrow +\infty} \sin (1/x)=\displaystyle\lim_{t \rightarrow 0}\sin t =0}$.

In general if ${\displaystyle \lim_{x \rightarrow c} g(x)= a \in \mathbb{R}}$ it is ${\displaystyle \lim_{x \rightarrow c} \sin (g(x))=\displaystyle\lim_{t \rightarrow a} \sin t = \sin a}$. 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 ${\displaystyle \lim_{x \rightarrow c^+}f(g(x))=\displaystyle\lim_{t \rightarrow 0}\tilde{f}(t)=1}$.

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 ${\displaystyle \lim_{x \rightarrow c}e^{g(x)}=e^{\displaystyle\lim_{x \rightarrow c}g(x)}}$.

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 ${\displaystyle \lim_{x \rightarrow +\infty}a^x =\displaystyle\lim_{x \rightarrow +\infty}e^{x\log a}=e^{\displaystyle\lim_{x \rightarrow +\infty} x\log a}=+\infty }$ (since ${\log a>0}$).

On the other hand for ${\alpha > 0}$ it also is ${\displaystyle \lim_{x \rightarrow +\infty}x^\alpha =\displaystyle\lim_{x \rightarrow +\infty}e^{\alpha \log x}= e^{\displaystyle \lim_{x \rightarrow +\infty}\alpha \log x}=+\infty}$.

The question we want to answer is ${\displaystyle \lim_{x \rightarrow +\infty}\dfrac{a^x}{x^\alpha} }$ 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 ${\displaystyle\lim_{x \rightarrow \infty}\frac{a^x}{x^\alpha}=\displaystyle\lim_{t \rightarrow +\infty}t^{2\alpha}=+\infty}$ $\Box$

 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 an 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 ${\displaystyle \lim_{x \rightarrow c} g(x)=0}$