Limits and Continuity

Maggie Schildt

22 Limits and Continuity

We covered much about limits and continuity in calculus but we need to apply it to analysis. Lets first give the formal definition for functional limits under analysis.

Definition. Functional Limits

Let $f: A\rightarrow \mathbb{R}$ , and let $c$ be a limit point of the domain $A$ . The limit, $\lim_{x\rightarrow c}{f(x)}=L$ if, for all $\epsilon > 0$ , there exists a $\delta 0$ such that whenever $0<|x-c|<\delta$ for $x\in A$ it follows that $|f(x)-L|<\epsilon$ .

We have already been familiarized with $\epsilon$ and we can say that $\delta$ carries the same meaning for neighborhood but we utilize both because the open neighborhood for $x$ will not always be the same as the open neighborhood for $f(x)$ . Lets look at an example of how we can use the value for $\epsilon$ to determine the largest value for $\delta$ which will thus return the largest open neighborhood that can possibly satisfy the limit.

Example:

$\lim_{x\rightarrow 3}{(5x-6)=9}$ , where $\epsilon = 1$ . Find the largest $\delta$ which satisfies the limit [3].

Proof: Let $f(x)=(5x-6)$ . Since $\epsilon = 1$ then $|f(x)-9|<1$ . We first need to simplify this inequality. Firstly we can rewrite $|f(x)-9|$ as follows:

$\begin{equation*} \begin{split} |f(x)-9|&=|5x-6-9|\\ &=|5x-15|\\ &=5|x-3| \end{split} \end{equation*}$

Thus, $5|x-3|<1$ and if we divide both sides by $5$ we get $|x-3|<\frac{1}{5}$ . Now, let $\delta = \frac{1}{5}$ then $0<|x-3|<\frac{1}{5}$ such that $|f(x)-9|<1$ is the same as $5|x-3|<\frac{1}{5}\times 5$ which simplifies to $5|x-3|<1$ and since $\epsilon=1$ then $5|x-3|<\epsilon$ . Thus for $\lim_{x\rightarrow 3}{(5x-6)=9}$ with $\epsilon=1$ , we find the largest $\delta = \frac{1}{5}$ .

There is a really valuable visual explanation, given below, of the epsilon and delta neighborhoods and what we are focusing on within a function when we analyze these limits.

We can see form this illustration, how $\epsilon$ and $\delta$ neighborhoods associate with functions and point $c$ . Using neighborhoods to prove limits is like narrowing down the range and domain for which a limit point may be. Given a better understanding of the relationship between neighborhoods let us look at one more example where we have to determine an appropriate $\epsilon$ and $\delta$ .

Example:

Given $\lim_{x\rightarrow -3}{(x^{2}+2x-5)}=-2$ , prove the limit.

Proof: Let $f(x)=(x^{2}+2x-5)$ , Let $\epsilon >0$ then to find $\delta >0$ such that $0<|x-(-3)|<\delta$ we need $|f(x)-(-2)|<\epsilon$ . We first simplify $|f(x)-(-2)|$

$\begin{equation*} \begin{split} |f(x)-(-2)|&=|f(x)+2|\\ &=|x^{2}+2x-5+2|\\ &=|x^{2}+2x-3|\\ &=|x+3||x-1| \end{split} \end{equation*}$

Here we can see that $|x+3|$ can be easily bounded by $\delta$ as $x\rightarrow -3$ but $|x-1|$ will not be. If $|x+3|<1$ then $-1<x-1<1$ and if we remove $-1$ from $x-1$ by way of adding $1$ we get $0<x<2$ so $|x-1|\leq |x|-1<3$ if $|x+3|<1$ . Let us considered that $|x-1|<1$ then, since $|f(x)-(-2)|=|x+3||x-1|$ we have $|x+3||x-1|<3|x-1|<\epsilon$ . Thus $|3x-1|<\epsilon$ and if we divide both sides by $3$ we find $|x-1|<\frac{\epsilon}{3}$ . Now let $\delta =\min \left\lbrace 0,\frac{\epsilon}{3}\right\rbrace$ , if $0<|x+3|<\delta$ then $|f(x)-(-2)|<\frac{\epsilon}{3}\times 3$ which we simplify to $|f(x)-(-2)|<\epsilon$ . Thus we conclude that $\lim_{x\rightarrow -3}{(x^{2}+2x-5)}=-2$ .

Now we can expand on what we have already proven by introducing the definition of continuity as it pertains to analysis.

Definition. Continuity

A function $f:A\rightarrow \mathbb{R}$ is continuous at a point $c\in A$ if, for all $\epsilon >0$ , there exists a $\delta 0$ such that whenever $|x-c|<\delta$ for $x\in A$ it follows that $|f(x)-f(c)|<\epsilon$ [3].

This definition is very similar to our definition for functional limits with one major difference. Instead of finding neighborhoods that satisfy the limit of the function we are now looking at finding the neighborhood that satisfies the point $c$ for $f(x)$ .

Example:

Prove that $f(x)=x^{2}+3x+4$ is continuous on $\mathbb{R}$ [3].

Proof: Let $\epsilon >0$ then there is $|f(x)-f(c)|$ for all $x\in \left\lbrace c-\delta, c+\delta \right\rbrace$ .

Part 1) If $c=0$ then $f(x)-f(c)|<\epsilon$ . We rewrite $|f(x)-f(c)|$ by the following:

$\begin{equation*} \begin{split} |f(x)-f(c)|&=|x^{2}+3x+4-0^{2}-3\cdot 0-4|\\ &=|x^{2}+3x|\\ &=|x||x+3| \end{split} \end{equation*}$

Thus $|x||x+3|<\epsilon$ . $|x|$ can easily be bounded by $\delta$ as $x\rightarrow 0$ because $|x-0|=|x|$ where $|x|<\delta$ . If $|x|<1$ then $-1<x<1$ so $|x+3|\leq |x|+3<4$ if $|x|<1$ . Let $|x+3|<1$ then since $|f(x)-f(c)|=|x||x+3|$ and $|x||x+3|<4|x|<\epsilon$ we can divide both sides by $4$ to find $|x|<\frac{\epsilon}{4}$ . Now let $\delta = \min \left\lbrace 1, \frac{\epsilon}{4} \right\rbrace$ , if $0<|x|<\delta$ then $|f(x)-f(c)|<\frac{\epsilon}{4}\times 4$ which simplifies to $|f(x)-f(c)<\epsilon$ .

Part 2) If $c\neq 0$ then $|f(x)-f(c)|<\epsilon$ can be simplified by

$\begin{equation*} \begin{split} |f(x)-f(c)|&=|x^{2}+3x+4-c^{2}-3c-4|\\ &=|x^{2}-c^{2}+3x-3c|\\ &=|x-c||x+3+c| \end{split} \end{equation*}$

Where $|x-c||x+3+c|<\epsilon$ . $|x-c|$ can easily be bounded by $\delta$ . If $|x-c|<1$ then $-1<x-c<1$ can easily become $2<x-c+3<4$ by adding $3$ and then, by removing $c$ from the center and adding it to either side we find $2+c<x+3<4+c$ so $|x+c+3|\leq |x|+|c|+3<|c|+|c|+4$ if $|x-c|<1$ . Now $|f(x)-f(c)|=|x-c||x+c+3|$ and $|x-c||x+c+3|<(2|c|+4)|x-c|<\epsilon$ then we can simplify this inequality by dividing both sides by $(2|c|+4)$ to get $|x-c|<\frac{\epsilon}{2|c|+4}$ . Let $\delta =\min \left\lbrace 1, \frac{\epsilon}{2|c|+4} \right\rbrace$ , if $0<|x-c|<\delta$ then $|f(x)-f(c)|<\frac{\epsilon}{2|c|+4} \times 2|c|+4$ which simplifies to $|f(x)-f(c)|<\epsilon$ .

Thus we can conclude that $f(x)=x^{2}+3x+4$ is continuous on $\mathbb{R}$

Building on our understanding of continuity we introduce the following theorem [3].

Theorem: The Intermediate Value Theorem

Let $f:[a,b]\rightarrow \mathbb{R}$ be continuous. If $L$ is a real number satisfying $f(a)<L<f(b)$ of $f(a)>L>f(b)$ , then there exists a point $c\in (a,b)$ where $f(c)=L$ .

Example:

Let $f: [0,1]\rightarrow \mathbb{R}$ be continuous with $f(0)=f(1)$ . Show that there must exist $x,y\in[0,1]$ satisfying $|x-y|=\frac{1}{2}$ and $f(x)=f(y)$ [3].

Proof: Let $f(\frac{1}{2})>f(0)$ and let $g$ be a continuous function $g:[0,\frac{1}{2}]\rightarrow \mathbb{R}$ such that $g(x)=f(x+\frac{1}{2})-f(x)$ which leads us to conclude $g(0)=[f(\frac{1}{2})-f(0)]$ such that $g(0)<0$ and since $f(0)=f(1)$ then $g(\frac{1}{2})=f(1)-(\frac{1}{2})$ such that $g\left( \frac{1}{2}\right)<0$ . We can observe from these inequalities that at some point on the interval $g(x)=0$ because the value of $g(x)$ changes from greater than 0 to less than 0. Consider a point $c\in [0,\frac{1}{2}$ such that $g(x)=0$ then $f(x)=f(x+\frac{1}{2})$ . Thus there exists an $x,y\in [0,1]$ such that $|x-y|=\frac{1}{2}$ and $f(x)=f(y)$ .

Expanding even further on continuity we look, now, to uniform continuity which is a stricter class of continuity.

19.1 Uniform Continuity

Previously mentioned uniform continuity is a stricter class of continuity. What this means is that instead of proving continuity by points, uniform continuity simply utilizes the $\epsilon$ -neighborhood to prove continuity.

Definition. Uniform Continuity

The function $f: A\rightarrow \mathbb{R}$ is uniformly continuous on interval $I$ if for every $\epsilon >0$ , there exists a $\delta > 0$ such that for $|x-y|<\delta$ for $x\in A$ it follows that $|f(x)-f(y)|<\epsilon$ [3].

It goes without need for explanation that if $f$ is uniformly continuous then $f$ is continuous on the given interval.

Example:

Prove that $g(x)=\sqrt{x^{2}+1}$ is uniformly continuous on $(0,1)$ .

Proof:Let $x,y\in (0,1)$ then by the definition of uniform continuity

$\begin{equation*} \begin{split} |g(x)-g(y)|&=|\sqrt{x^{2}+1} -\sqrt{y^{2}+1}|\\ &=|(\sqrt{x^{2}+1} -\sqrt{y^{2}+1})\cdot \frac{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|\\ &=|\frac{x^{2}+1-y^{2}-1}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|\\ &=|\frac{x^{2}+y^{2}}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|\\ &=|\frac{(x-y)(x+y)}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|\\ &=|x-y||x+y|\frac{1}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}| \end{split} \end{equation*}$

Now we let $\delta>0$ such that $|x-y|<\delta$ then we can conclude that

$\begin{equation*} |x-y||x+y|\leq|x-y|\cdot (|x|+|y|)<2\delta \end{equation*}$

if $|x|<1$ and $|y|<1$ . Let $\delta >0$ such that

$\begin{equation*} |x-y||x+y|\frac{1}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|<|x-y|\cdot (|x|+|y|)<2\delta \end{equation*}$

Let $2\delta = \epsilon$ then, by dividing both sides by 2, we get $\delta = \frac{\epsilon}{2}$ such that

$\begin{equation*} |x-y||x+y|\frac{1}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|<|x-y|\cdot (|x|+|y|)<2\delta \end{equation*}$

Can be rewritten as

$\begin{equation*} |x-y||x+y|\frac{1}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|<|x-y|\cdot (|x|+|y|)<2\frac{\epsilon}{2} \end{equation*}$

Which is equivalent to

$\begin{equation*} |x-y||x+y|\frac{1}{\sqrt{x^{2}+1} +\sqrt{y^{2}+1}}|<|x-y|\cdot (|x|+|y|)<\epsilon \end{equation*}$

Thus if $|x-y|<\delta$ the $|f(x)-f(y)|<\epsilon$ . Thus we con conclude that $g(x)=\sqrt{x^{2}+1}$ is uniformly continuous on $(0,1)$ .

22 Limits and Continuity

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