True or False The limit of a function as $x$ approaches $c$ depends on the value of the function at $c$.

True or False In the $\epsilon$-$\delta$ definition of a limit, we require $0 \lt \vert x-c\vert$ to ensure that $x\neq c.$

True or False In an $\epsilon$-$\delta$ proof of a limit, the size of $\delta$ usually depends on the size of $\epsilon.$

True or False When proving $\lim\limits_{x\rightarrow c}f( x) =L$ using the $\epsilon$-$\delta$ definition of a limit, we try to find a connection between $\vert f( x) -c\vert$ and $\vert x-c\vert$.

True or False Given any $\epsilon \gt0$, suppose there is a $\delta \gt0$ so that whenever $0\lt\vert x-c\vert \lt\delta$, then $\vert f(x)-L\vert \lt\epsilon$. Then $\lim\limits_{x\rightarrow c}f(x)=L$.

True or False A function $f$ has a limit $L$ at infinity, if for any given $\epsilon \gt0$, there is a positive number $M$ so that whenever $x\gt M$, then $\vert f(x)-L\vert \gt\epsilon$.

$\lim\limits_{x\rightarrow 1}(2x)=2, \quad \epsilon =0.01$

$\lim\limits_{x\rightarrow 2}(-3x)=-6, \quad \epsilon =0.01$

$\lim\limits_{x\rightarrow 2}(6x-1)=11$ $\epsilon =\dfrac{1}{2}$

$\lim\limits_{x\rightarrow -3}(2-3x)=11$ $\epsilon =\dfrac{1}{3}$

$\lim\limits_{x\rightarrow 2}\left( -\dfrac{1}{2}x+5\right) =4$ $\epsilon =0.01$

$\lim\limits_{x\rightarrow \frac{5}{6}}\left( 3x+\dfrac{1}{2}\right) =3$ $\epsilon =0.3$

For the function $f(x)=4x-1$, we have $\lim\limits_{x \rightarrow 3}f(x)=11$. For each $\epsilon \gt0$, find a $\delta \gt0$ so that $$\hbox{whenever }\quad 0 \lt\vert x-3\vert \lt\delta\qquad \hbox{then } \vert(4x-1)-11\vert \lt\epsilon$$

For the function $f(x)=2-5x$, we have $\lim\limits_{x \rightarrow -2}f(x)=12$. For each $\epsilon \gt0$, find a $\delta >0$ so that $$\hbox{whenever } 0 \lt\vert x+2\vert \lt\delta \qquad \hbox{then } \vert (2-5x)-12\vert \lt\epsilon$$

For the function $f(x)=\dfrac{x^{2}-9}{x+3}$, we have $\lim\limits_{x\rightarrow -3} f(x)=-6$. For each $\epsilon \gt0$, find a $\delta >0$ so that $$\hbox{whenever }\quad 0 \lt\vert x+3\vert \lt\delta \qquad \hbox{then } \left\vert \dfrac{x^{2}-9}{x+3}- ( -6) \right\vert \lt\epsilon$$

For the function $f(x)=\dfrac{x^{2}-4}{x-2}$, we have $\lim\limits_{x\rightarrow 2}f(x)=4$. For each $\epsilon \gt0$, find a $\delta \gt0$ so that $$\hbox{whenever } 0 \lt\vert x-2\vert \lt\delta \qquad \hbox{then } \left\vert \dfrac{x^{2}-4}{x-2}-4\right\vert \lt\epsilon$$

$\lim\limits_{x\rightarrow 2}(3x)=6$

$\lim\limits_{x\rightarrow 3}(4x)=12$

$\lim\limits_{x\rightarrow 0}(2x+5)=5$

$\lim\limits_{x\rightarrow -1}(2-3x)=5$

$\lim\limits_{x\rightarrow -3}(-5x+2)=17$

$\lim\limits_{x\rightarrow 2}(2x-3)=1$

$\lim\limits_{x\rightarrow 2}(x^{2}-2x)=0$

$\lim\limits_{x\rightarrow 0}(x^{2}+3x)=0$

$\lim\limits_{x\rightarrow 1}\dfrac{1+2x}{3-x}=\dfrac{3}{2}$

$\lim\limits_{x \rightarrow 2}\dfrac{2x}{4+x}=\dfrac{2}{3}$

$\lim\limits_{x\rightarrow 0}\sqrt[3]{{x}}=0$

$\lim\limits_{x\rightarrow 1}\sqrt{2-x}=1$

$\lim\limits_{x\rightarrow -1}x^{2}=1$

$\lim\limits_{x\rightarrow 2}x^{3}=8$

$\lim\limits_{x\rightarrow 3}\dfrac{1}{x}=\dfrac{1}{3}$

$\lim\limits_{x\rightarrow 2}\dfrac{1}{x^{2}}=\dfrac{1}{4}$

Use the $\epsilon$-$\delta$ definition of a limit to show that the statement $\lim\limits_{x\rightarrow 3}(3x-1)=12$ is false.

Use the $\epsilon$-$\delta$ definition of a limit to show that the statement $\lim\limits_{x\rightarrow -2}(4x)=-7$ is false.

Show that $\left\vert \dfrac{{1}}{{x^{2}\,+\,9}}-\dfrac{1}{18} \right\vert \lt\dfrac{{7}}{{234}}\,\vert x-3\vert$ if $2 \lt x\lt4$. Use this to show that ${\lim\limits_{x\rightarrow 3}}{\dfrac{{1}}{{ x^{2}+9}}}=\dfrac{1}{18}.$

Show that $\vert (2+x)^{2}-4\vert \leq 5\,\vert x\vert$ if -1 $\lt x\lt 1$. Use this to show that $\lim\limits_{x\rightarrow 0}(2+x)^{2}=4$.

Show that ${\left\vert {{\dfrac{{1}}{{x^{2}+9}}}-{\dfrac{{1}}{{ 13}}}}\right\vert }\leq {\dfrac{{1}}{{26}}\,}\left\vert \,{x-2}\right\vert$ if $1 \lt x \lt3$. Use this to show that ${\lim\limits_{x\rightarrow 2}\,}{\dfrac{{1}}{{x^{2}+9}}}={\dfrac{{1}}{{13}}}.$

Use the $\epsilon-\delta$ definition of a limit to show that ${\lim\limits_{x\rightarrow 1}}x^{2}\neq 1.31.$ (Hint: Use $\epsilon = 0.1$.)

If $m$ and $b$ are any constants, prove that $$\lim\limits_{x\rightarrow c}( {mx+b}) =mc+b$$

Verify that if $x$ is restricted so that $\vert x-2\vert \lt\dfrac{1}{3}$ in the proof of $\lim\limits_{x\rightarrow 2}x^{2}=4$, then the choice for $\delta$ would be less than or equal to the smaller of $\dfrac{1}{3}$ and $\dfrac{3\epsilon }{13}$; that is, $\delta \leq \min \left\{ \dfrac{1}{3},\dfrac{3\epsilon }{13}\right\}$.

For $x\neq 3$, how close to $3$ must $x$ be to guarantee that $2x-1$ differs from 5 by less than 0.1?

For $x\neq 0$, how close to $0$ must $x$ be to guarantee that $3^{x}$ differs from $1$ by less than 0.1?

Prove that if $\lim\limits_{x\rightarrow c}f(x)\lt0$, then there is an open interval around $c$ for which $f(x)\lt0$ everywhere in the interval, except possibly at $c$.

Use the $\epsilon-\delta$ definition of a limit at infinity to prove that $$\lim\limits_{x\rightarrow -\infty }\dfrac{1}{x}=0.$$

Use the $\epsilon$-$\delta$ definition of a limit at infinity to prove that $$\lim\limits_{x\rightarrow \infty }\left( -\dfrac{1}{\sqrt{x}}\right) =0.$$

For $\lim\limits_{x\rightarrow -\infty }\dfrac{1}{x^{2}} =0$, find a value of $N$ that satisfies the $\epsilon$ - $\delta$ definition of limits at infinity for $\epsilon = 0.1$.

Use the $\epsilon$ - $\delta$ definition of limit to prove that no number $L$ exists so that $\lim\limits_{x\rightarrow 0}$ $\dfrac{1 }{x}=L$.

Explain why in the $\epsilon$ - $\delta$ definition of a limit, the inequality $0 \lt \vert x-c\vert \lt\delta$ has two strict inequality symbols.

The $\epsilon$ - $\delta$ definition of a limit states, in part, that $f$ is defined everywhere in an open interval containing $c$, except possibly at $c$. Discuss the purpose of including the phrase, except possibly at c, and why it is necessary.

In the $\epsilon$ - $\delta$ definition of a limit, what does $\epsilon$ measure? What does $\delta$ measure? Give an example to support your explanation.

Discuss $\lim\limits_{x\rightarrow 0}f( x)$ and $\lim\limits_{x\rightarrow 1}f( x)$ if $$\begin{equation*} f(x) =\left\{ \begin{array}{l@{\quad}l} x^{2} & \hbox{if \(x\) is rational} \\ 0 & \hbox{if \(x\) is irrational} \end{array}. \right. \end{equation*}$$

Discuss $\lim\limits_{x\rightarrow 0}f( x)$ if $$\begin{equation*} f( x) =\left\{ \begin{array}{c@{\quad}l} x^{2} & \hbox{if \(x\) is rational} \\ \tan x & \hbox{if \(x\) is irrational} \end{array}. \right. \end{equation*}$$

Use the $\epsilon$- $\delta$ definition of limit to prove that $\lim\limits_{x\rightarrow 1}(4x^{3}+3x^{2}-24x+22)=5$.

If $\lim\limits_{x\rightarrow c}f(x)=L$ and $\lim\limits_{x\rightarrow c}g(x)=M$, prove that $\lim\limits_{x\rightarrow c}[f(x)+g(x)]=L+M$. Use the $\epsilon$ - $\delta$ definition of limit.

For $\lim\limits_{x\rightarrow \infty }\dfrac{2-x}{\sqrt{ 5+4x^{2}}}=-\dfrac{1}{2}$, find a value of $M$ that satisfies the $\epsilon$ - $\delta$ definition of limits at infinity for $\epsilon =0.01$.

Use the $\epsilon$ - $\delta$ definition of a limit to prove that the linear function $f(x)=ax+b$ is continuous everywhere.

Show that the function $f( x) =\left\{ \begin{array}{l@{\quad}ll} 0 & \hbox{if} & x\hbox{ is rational} \\ x & \hbox{if} & x\hbox{ is irrational} \end{array} \right.$ is continuous only at $x=0$.

Suppose that $f$ is defined on an interval $(a,b)$ and there is a number $K$ so that $\left\vert f(x)-f(c)\right\vert \leq K\vert x-c\vert$ for all $c$ in $(a,b)$ and $x$ in $(a,b)$. Such a constant $K$ is called a Lipschitz constant. Find a Lipschitz constant for $f(x)=x^{3}$ on $(0,2)$.