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 8

Recall

Assume $\underset{x\to a}{\text{lim}}f\left(x\right)$ and $\underset{x\to a}{\text{lim}}g\left(x\right)$ exist. Then

(a)	$\underset{x\to a}{\text{lim}}f\left(x\right)+g\left(x\right)=\underset{x\to a}{\text{lim}}f\left(x\right)+\underset{x\to a}{\text{lim}}g\left(x\right)\text{.}$
(b)	$\underset{x\to a}{\text{lim}}f\left(x\right)g\left(x\right)=\left(\underset{x\to a}{\text{lim}}f\left(x\right)\right)\left(\underset{x\to a}{\text{lim}}g\left(x\right)\right)\text{.}$
(c)	If $c\in ℝ$ then $\underset{x\to a}{\text{lim}}cf\left(x\right)=c\underset{x\to a}{\text{lim}}f\left(x\right)\text{.}$
(d)	If $f\left(x\right)\le g\left(x\right)$ then $\underset{x\to a}{\text{lim}}f\left(x\right)\le \underset{x\to a}{\text{lim}}g\left(x\right)\text{.}$
(e)	If $g\left(x\right)\ne 0$ then $\underset{x\to a}{\text{lim}}\frac{1}{g\left(x\right)}=\frac{1}{\underset{x\to a}{\text{lim}}g\left(x\right)}\text{.}$
(f)	If $\underset{x\to a}{\text{lim}}g\left(x\right)=\ell$ then $\underset{y\to \ell }{\text{lim}}f\left(y\right)=\underset{x\to a}{\text{lim}}f\left(g\left(x\right)\right)\text{.}$

For lecture 8, 17 March presentation

(a)	Let $x\in ℂ\text{.}$ Then $$\underset{n\to \infty }{\text{lim}}{x}^n=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}\left|x\right|<1,\\ \text{diverges,}& \text{if}\hspace{0.17em}\left|x\right|>1,\\ 1,& \text{if}\hspace{0.17em}x=1,\\ \text{diverges,}& \text{if}\hspace{0.17em}\left|x\right|=1\hspace{0.17em}\text{and}\hspace{0.17em}x\ne 1\text{.}\end{array}$$
(b)	Let $x\in ℂ\text{.}$ Then $$\underset{n\to \infty }{\text{lim}}1+x+\cdots +x^n=\underset{n\to \infty }{\text{lim}}\frac{1-x^{n+1}}{1-x}=\{\begin{array}{cc}\frac{1}{1-x},& \text{if}\hspace{0.17em}\left|x\right|<1,\\ \text{diverges,}& \text{if}\hspace{0.17em}\left|x\right|\ge 1\text{.}\end{array}$$

(a)	Let $n\in {\mathbb{Z}}_{>0}$ and $a\in ℂ\text{.}$ Then $$\underset{x\to a}{\text{lim}}{x}^n=a^n\text{.}$$
(b)	Let $a\in ℂ\text{.}$ Then $$\underset{x\to a}{\text{lim}}{e}^x=e^a\text{.}$$

Our definitions

$\underset{x\to a}{\text{lim}}f\left(x\right)=\ell$ if $f\left(x\right)$ satisfies: if $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $d(x,a)<\delta$ then $d(f\left(x\right),\ell )<\epsilon \text{.}$

In ${ℝ}^n,$ $d:{ℝ}^n\times {ℝ}^n\to {ℝ}_{\ge 0}$ is given by $$d(x,y)=\left|\sqrt{{(x_1-y_1)}^2+\cdots +{(x_n-y_n)}^2}\right|$$ where $$\left|a\right|=\text{sup}(a,-a),\text{for}a\in ℝ\text{.}$$ We defined $\sqrt{x}$ as the inverse expression to $x^2$ so that branches are possible and $\sqrt{9}=-3$ is possible.

We defined $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$ $$\text{sin}\hspace{0.17em}x=\frac{e^{ix}-e^{-ix}}{2i}\text{and}\text{cos}\hspace{0.17em}x=\frac{e^{ix}+e^{-ix}}{2}\text{.}$$

The limits $\underset{n\to \infty }{\text{lim}}{x}^n$ and $\underset{n\to \infty }{\text{lim}}1+x+x^2+\cdots +x^n$

(a)	Let $x\in ℂ\text{.}$ $$\underset{n\to \infty }{\text{lim}}{x}^n=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}\left|x\right|<1,\\ \text{diverges,}& \text{if}\hspace{0.17em}\left|x\right|>1,\\ 1,& \text{if}\hspace{0.17em}x=1,\\ \text{diverges,}& \text{if}\hspace{0.17em}\left|x\right|=1\hspace{0.17em}\text{and}\hspace{0.17em}x\ne 1\text{.}\end{array}$$
(b)	Let $x\in ℂ\text{.}$ $$\begin{array}{ccc}\underset{n\to \infty }{\text{lim}}1+x+x^2+\cdots +x^n& =& \underset{n\to \infty }{\text{lim}}\frac{1-x^{n+1}}{1-x}\\ & =& \{\begin{array}{cc}\frac{1}{1-x},& \text{if}\hspace{0.17em}\left|x\right|<1,\\ \text{diverges,}& \text{if}\hspace{0.17em}\left|x\right|\ge 1\text{.}\end{array}\end{array}$$

		Proof.
	If $x=1$ then the sequence $a_n=x^n$ is $a_n=1$ and $\underset{n\to \infty }{\text{lim}}{1}^n=1\text{.}$ If $x=1$ then the $\underset{n\to \infty }{\text{lim}}1+1^2+\cdots +1^n=\underset{n\to \infty }{\text{lim}}n,$ which diverges. The remaining statements in (b) follow from (a). $\square$

Let $x\in ℂ$ with $\left|x\right|<1\text{.}$ Prove that $\underset{n\to \infty }{\text{lim}}{x}^n=0\text{.}$

		Proof.
	Let $N\in {\mathbb{Z}}_{>0}$ such that $\left|x\right|<1-\frac{1}{N+1}\text{.}$ $$\begin{array}{ccc}{\displaystyle \underset{n\to \infty }{\text{lim}}|x^n-0|}& =& {\displaystyle \underset{n\to \infty }{\text{lim}}{\left|x\right|}^n}\\ & \le & {\displaystyle \underset{n\to \infty }{\text{lim}}{(1-\frac{1}{N+1})}^n}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{\left(\frac{N+1-1}{N+1}\right)}^n}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}{\left(\frac{N}{N+1}\right)}^n}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{1}{{(1+\frac{1}{N})}^n}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{1}{1+n\frac{1}{N}+\cdots +{\left(\frac{1}{N}\right)}^n}}\\ & \le & {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{1}{1+\frac{n}{N}}}\\ & =& {\displaystyle \underset{n\to \infty }{\text{lim}}\frac{N}{n+N}}\\ & =& {\displaystyle N\underset{n\to \infty }{\text{lim}}\frac{1}{n+N}}\\ & =& {\displaystyle N·0}\\ & =& {\displaystyle 0\text{.}}\end{array}$$ $\square$

Let $x\in ℂ$ with $\left|x\right|>1\text{.}$ Prove that $\underset{n\to \infty }{\text{lim}}{x}^n$ diverges.

		Proof.
	Let $N\in {\mathbb{Z}}_{>0}$ be such that $\left|x\right|>1+\frac{1}{N}\text{.}$ Then $$\begin{array}{ccc}{\displaystyle {\left|x\right|}^n}& >& {\displaystyle {(1+\frac{1}{N})}^n=1+n\left(\frac{1}{N}\right)+\cdots +{\left(\frac{1}{N}\right)}^n}\\ & =& {\displaystyle n\left(\frac{1}{N}\right)=\frac{n}{N}\text{.}}\end{array}$$ Since $\frac{n}{N}$ is unbounded as $n$ gets larger and larger, ${\left|x\right|}^n$ is unbounded as $n\to \infty \text{.}$ so $\underset{n\to \infty }{\text{lim}}{x}^n$ diverges. $\square$

Let $x\in ℝ$ with $\left|x\right|<1\text{.}$ Let $a_n=x^n\text{.}$ Find $\underset{n\to \infty }{\text{lim}}{a}_n\text{.}$ $$\underset{n\to \infty }{\text{lim}}{a}_n=\underset{n\to \infty }{\text{lim}}{x}^n$$

$\underset{n\to \infty }{\text{lim}}{a}_n=\underset{n\to \infty }{\text{lim}}{x}^n$ and the graphs of $y=x,$ $y=x^2,$ $y=x^3,$ $y=x^4,$ $\dots$ are $$\begin{array}{ccc} y=x y=x3 y=x5 - 1 1 x - 1 1 y & & y=x2 y=x4 y=x6 - 1 1 x - 1 1 y \end{array}$$ So $\underset{n\to \infty }{\text{lim}}{x}^n=0$ where $\left|x\right|<1\text{.}$

$\underset{n\to \infty }{\text{lim}}{x}^n$ diverges when $\left|x\right|>1\text{.}$

$\underset{n\to \infty }{\text{lim}}{1}^n=1$ and $\underset{n\to \infty }{\text{lim}}{(-1)}^n$ diverges.

Let $x\in ℝ\text{.}$ Find $\underset{n\to \infty }{\text{lim}}1+x+x^2+\cdots +x^n\text{.}$ $$\underset{n\to \infty }{\text{lim}}1+x+x^2+\cdots +x^n=\underset{n\to \infty }{\text{lim}}\frac{1-x^{n+1}}{1-x}=\frac{1}{1-x}\text{if}\left|x\right|<1\text{.}$$ For example, if $x=\frac{1}{2}$ $$\underset{n\to \infty }{\text{lim}}1+\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^3+\cdots +{\left(\frac{1}{2}\right)}^n=\underset{n\to \infty }{\text{lim}}\frac{1-{\left(\frac{1}{2}\right)}^{n+1}}{1-\frac{1}{2}}=\frac{1}{1-\frac{1}{2}}=2\text{.}$$

Let $n\in {\mathbb{Z}}_{>0}\text{.}$ Let $a\in ℝ\text{.}$ Prove that $\underset{x\to a}{\text{lim}}{x}^n=a^n\text{.}$

		Proof.
	To show: $\underset{y\to 0}{\text{lim}}|{(y+a)}^n-a^n|=0\text{.}$ $$\begin{array}{ccc}\underset{y\to 0}{\text{lim}}|{(y+a)}^n-a^n|& =& \underset{y\to 0}{\text{lim}}\mid y^n+n{y}^{n-1}a+\cdots +n{a}^{n-1}y+a^n-a^n\mid \\ & =& \underset{y\to 0}{\text{lim}}\mid y^n+n{y}^{n-1}a+\cdots +n{a}^{n-1}y\mid \\ & =& \underset{y\to 0}{\text{lim}}\mid y(y^{n-1}+a{y}^{n-2}n+\cdots +n{a}^{n-1})\mid \\ & =& \underset{y\to 0}{\text{lim}}\left|y\right|\mid y^{n-1}+a{y}^{n-2}n+\cdots +n{a}^{n-1}\mid \\ & \le & \underset{y\to 0}{\text{lim}}\left|y\right|({\left|y\right|}^{n-1}+\left|a\right|n{\left|y\right|}^{n-2}+\cdots +\left|n{a}^{n-1}\right|)\\ & \le & \underset{y\to 0}{\text{lim}}\left|y\right|n{\left|a\right|}^{n-1}=0·n·{\left|a\right|}^{n-1}=0\text{.}\end{array}$$ $\square$

So $f\left(x\right)=x^n$ is continuous at $x=a\text{.}$

An alternative proof is that

(a)	$f\left(x\right)=x$ (the identity function) is continuous,
(b)	the product is continuous (since $ℝ$ is a topological field)

and therefore $\underset{x\to a}{\text{lim}}{x}^n=a^n\text{.}$

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

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

Prove that $\underset{x\to a}{\text{lim}}{e}^x=e^a\text{.}$

		Proof.
	$$\begin{array}{ccc}\underset{x\to a}{\text{lim}}{e}^x& =& \underset{y\to 0}{\text{lim}}{e}^{y+a}=\underset{y\to 0}{\text{lim}}{e}^a{e}^y\\ & =& e^a\underset{y\to 0}{\text{lim}}{e}^y=e^a·e^0=e^{a+0}=e^a\text{.}\end{array}$$ $\square$

Hence $e^x$ is continuous at $x=a\text{.}$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100317suggLect8.pdf and was given on 17 March 2010.

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