Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 8 July 2014

Lecture 9

Limit examples

Prove that $\underset{x\to 0}{\text{lim}}\frac{e^x-1}{x}=1\text{.}$

		Proof.
	$$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}|\frac{e^x-1}{x}-1|}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\left|\frac{e^x-1-x}{x}\right|}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\mid \frac{1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots -1-x}{x}\mid }\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\mid \frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\cdots \mid }\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\left|x\right|\mid \frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}+\cdots \mid }\\ & \le & {\displaystyle \underset{x\to 0}{\text{lim}}\left|x\right|(\frac{1}{2!}+\frac{x}{3!}+\frac{x^2}{4!}+\cdots )}\\ & \le & {\displaystyle \underset{x\to 0}{\text{lim}}\left|x\right|(1+\left|x\right|+{\left|x\right|}^2+\cdots )}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\left|x\right|\frac{1}{1-\left|x\right|}=0·1=0\text{.}}\end{array}$$ So $$\underset{x\to 0}{\text{lim}}\frac{e^x-1}{x}=1\text{.}$$ $\square$

Evaluate $\underset{x\to 0}{\text{lim}}\frac{\text{cos}\hspace{0.17em}x-1}{x^2}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{cos}\hspace{0.17em}x-1}{x^2}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{(\frac{e^{ix}+e^{-ix}}{2}-1)}{x^2}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{1}{2}\left(\frac{e^{ix}+e^{-ix}-2}{x^2}\right)}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{1}{2}{e}^{-ix}\frac{(e^{2ix}-2e^{ix}+1)}{x^2}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}-\frac{1}{2}{e}^{-ix}\frac{{(e^{ix}-1)}^2}{{\left(ix\right)}^2}}\\ & =& {\displaystyle -\frac{1}{2}·e^0·1^2}\\ & =& {\displaystyle -\frac{1}{2}\text{.}}\end{array}$$

Find $\underset{x\to 2}{\text{lim}}\frac{3x^2+8}{x^2-x}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 2}{\text{lim}}\frac{3x^2+8}{x^2-x}}& =& {\displaystyle \left(\underset{x\to 2}{\text{lim}}3x^2+8\right)\left(\underset{x\to 2}{\text{lim}}\frac{1}{x^2-x}\right)}\\ & =& {\displaystyle (3·2^2+8)\frac{1}{2^2-2}}\\ & =& {\displaystyle (3·4+8)\frac{1}{2}}\\ & =& {\displaystyle \frac{20}{2}}\\ & =& {\displaystyle 10,}\end{array}$$ since $\underset{x\to 2}{\text{lim}}3x^2+8$ and $\underset{x\to 2}{\text{lim}}\frac{1}{x^2-x}$ exist (because $f\left(x\right)=3x^2+8$ and $f\left(x\right)=\frac{1}{x^2-x}$ are continuous at $x=2$ – the graph doesn't jump at $x=2\text{).}$

Find $\underset{x\to 0}{\text{lim}}\frac{5x}{x}\text{.}$ $$\underset{x\to 0}{\text{lim}}\frac{5x}{x}=\left(\underset{x\to 0}{\text{lim}}5x\right)\left(\underset{x\to 0}{\text{lim}}\frac{1}{x}\right)$$ NO. We can only do this if $\underset{x\to 0}{\text{lim}}5x$ exists and $\underset{x\to 0}{\text{lim}}\frac{1}{x}$ exists. A more sensible approach is to use some algebra, $\underset{x\to 0}{\text{lim}}\frac{5x}{x}=\underset{x\to 0}{\text{lim}}\frac{5}{1}=5\text{.}$

Evaluate $\underset{x\to 0}{\text{lim}}\frac{\sqrt{1+x}-1}{x}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{(\sqrt{1+x}-1)}{x}\frac{(\sqrt{1+x}+1)}{(\sqrt{1+x}+1)}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{1+x-1}{(\sqrt{1+x}+1)}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{1}{\sqrt{1+x}+1}}\\ & =& {\displaystyle \frac{1}{\sqrt{1+0}+1}}\\ & =& {\displaystyle \frac{1}{2}\text{.}}\end{array}$$

Evaluate $\underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}x}{x}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{sin}\hspace{0.17em}x}{x}}& =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{e^{ix}-e^{-ix}}{2ix}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{\frac{e^{ix}-1}{ix}-\frac{e^{-ix}-1}{ix}}{2}}\\ & =& {\displaystyle \underset{x\to 0}{\text{lim}}\frac{1}{2}(\frac{e^{ix}-1}{ix}+\frac{e^{-ix}-1}{-ix})}\\ & =& {\displaystyle \frac{1}{2}(1+1)=1\text{.}}\end{array}$$

$\underset{x\to 0}{\text{lim}}\frac{\text{log}(1+x)}{x}$

Let $x=e^y-1\text{.}$ Then $y=\text{log}(x+1)$ and $y\to 0$ as $x\to 0\text{.}$ So $$\begin{array}{ccc}{\displaystyle \underset{x\to 0}{\text{lim}}\frac{\text{log}(1+x)}{x}}& =& {\displaystyle \underset{y\to 0}{\text{lim}}\frac{\text{log}(1+e^y-1)}{e^y-1}}\\ & =& {\displaystyle \underset{y\to 0}{\text{lim}}\frac{\text{log}\left(e^y\right)}{e^y-1}}\\ & =& {\displaystyle \underset{y\to 0}{\text{lim}}\frac{y}{e^y-1}}\\ & =& {\displaystyle \underset{y\to 0}{\text{lim}}\frac{1}{\frac{e^y-1}{y}}}\\ & =& {\displaystyle \frac{1}{1}=1\text{.}}\end{array}$$

Evaluate $\underset{n\to \infty }{\text{lim}}{(1+\frac{34}{n})}^n\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{n\to \infty }{\text{lim}}{(1+\frac{34}{n})}^n}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{\text{log}{(1+\frac{34}{n})}^n}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{n\text{log}(1+\frac{34}{n})}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{e}^{34·\frac{\text{log}(1+\frac{34}{n})}{34/n}}}\\ & =& {\displaystyle e^{34·1}=e^{34}\text{.}}\end{array}$$

Let $a_n$ and $b_n$ be sequences in $ℝ\text{.}$ Assume that $\underset{n\to \infty }{\text{lim}}{a}_n$ exists and $\underset{n\to \infty }{\text{lim}}{b}_n$ exists. If $a_n\le b_n$ then $\underset{n\to \infty }{\text{lim}}{a}_n\le \underset{n\to \infty }{\text{lim}}{b}_n\text{.}$

		Proof.
	Proof by contradiction. Let ${\ell }_1=\underset{n\to \infty }{\text{lim}}{a}_n$ and ${\ell }_2=\underset{n\to \infty }{\text{lim}}{b}_n\text{.}$ Assume ${\ell }_1>{\ell }_2\text{.}$ Let $\epsilon ={\ell }_1-{\ell }_2\text{.}$ Let $N_1\in {\mathbb{Z}}_{>0}$ be such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N_1$ then $|a_n-{\ell }_1|<\epsilon /2\text{.}$ Let $N_2\in {\mathbb{Z}}_{>0}$ be such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N_2$ then $|b_n-{\ell }_1|<\epsilon /2\text{.}$ Let $N\in {\mathbb{Z}}_{>0}$ be such that $N>N_1$ and $N>N_2\text{.}$ Then $$a_N>{\ell }_1-\epsilon /2={\ell }_2+\epsilon -\epsilon /2={\ell }_2+\epsilon /2>b_N\text{.}$$ This is a contradiction to $a_N\le b_N\text{.}$ So $\underset{n\to \infty }{\text{lim}}{a}_n\le \underset{n\to \infty }{\text{lim}}{b}_n\text{.}$ $\square$

Assume that $\underset{x\to a}{\text{lim}}g\left(x\right)=\ell$ and $\underset{y\to \ell }{\text{lim}}f\left(y\right)$ exists. Then $$\underset{y\to \ell }{\text{lim}}f\left(y\right)=\underset{x\to a}{\text{lim}}f\left(g\left(x\right)\right)\text{.}$$

		Proof.
	Let $L=\underset{y\to \ell }{\text{lim}}f\left(y\right)\text{.}$ To show: $\underset{x\to a}{\text{lim}}f\left(g\left(x\right)\right)=L\text{.}$ To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $|x-a|<\delta$ then $|f\left(g\left(x\right)\right)-L|<\epsilon \text{.}$ Assume $\epsilon \in {ℝ}_{>0}\text{.}$ To show: There exists $\delta \in {ℝ}_{>0}$ such that if $|x-a|<\delta$ then $\left|f\left(g\left(x\right)\right)0L\right|<\epsilon \text{.}$ Let ${\delta }_1\in {ℝ}_{>0}$ be such that if $|y-\ell |<{\delta }_1$ then $|f\left(y\right)-L|<\epsilon \text{.}$ Let $\delta \in {ℝ}_{>0}$ be such that if $|x-a|<\delta$ then $|g\left(x\right)-\ell |<{\delta }_1\text{.}$ So if $|x-a|<\delta$ then $$|g\left(x\right)-\ell |<{\delta }_1\text{and}|f\left(g\left(x\right)\right)-L|<\epsilon \text{.}$$ $\square$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100319suggLect9.pdf and was given on 19 March 2010.

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