To show: If $\epsilon \in {ℝ}_{\ge 0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $x\in B_{\delta }\left(a\right)$ then $\left(f\left(x\right)+g\left(x\right)-\left(l_1+l_2\right)\right)<\epsilon$.

Assume $\epsilon \in {ℝ}_{>0}$.

We know: there exists ${\delta }_1\in {ℝ}_{>0}$ such that if $x\in B_{{\delta }_1}\left(c\right)$ then $\left(f\left(x\right)-l_1\right)<\frac{\epsilon }{2}.$

We know: there exists ${\delta }_2\in {ℝ}_{>0}$ such that if $x\in B_{{\delta }_2}\left(c\right)$ then $\left(f\left(x\right)-l_2\right)<\frac{\epsilon }{2}$.

Let $\delta =\min \left({\delta }_1,{\delta }_2\right)$.

To show: $\left(f\left(x\right)+g\left(x\right)-\left(l_1+l_2\right)\right)<\epsilon .$

$\begin{array}{rcl}\left(f\left(x\right)+g\left(x\right)-\left(l_1+l_2\right)\right)& =& \left(\left(f\left(x\right)-l_1\right)+\left(g\left(x\right)-l_2\right)\right)\\ & \le & \left(f\left(x\right)-l_1\right)+\left(g\left(x\right)-l_2\right)\\ & \le & \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .\end{array}$

So $\underset{x\to c}{\lim }f\left(x\right)+g\left(x\right)=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right).$$\square$

Let $l=\underset{x\to a}{\lim }f\left(x\right).$

To show: $\underset{x\to a}{\lim }cf\left(x\right)=cl.$

To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $d\left(x,a\right)<\delta$ then $d\left(cf\left(x\right),cl\right)<\epsilon .$

Assume $\epsilon \in {ℝ}_{>0}.$

To show: There exists $\delta \in {ℝ}_{>0}$ such that if $d\left(x,a\right)<\delta$ then $d\left(f\left(x\right),l\right)<\frac{\epsilon }{\left|c\right|}.$

To show: If $d\left(x,a\right)<\delta$ then $d\left(cf\left(x\right),cl\right)<\epsilon .$

Assume $d\left(x,a\right)<\delta .$

To show: $d\left(cf\left(x\right),cl\right)<\epsilon .$$$\begin{array}{rcl}d\left(cf\left(x\right),cl\right)& =& \left|cf\left(x\right)-cl\right|\\ & =& \left|c\right|\left|f\left(x\right)-l\right|\\ & <& \left|c\right|\frac{\epsilon }{\left|c\right|}=\epsilon .\square \end{array}$$

1. Let $l_1=\underset{x\to a}{\lim }f\left(x\right)$ and $l_2=\underset{x\to a}{\lim }g\left(x\right).$
2. To show: $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=l_1{l}_2$.
3. To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $\delta \in {ℝ}_{>0}$ such that if $\left|x-a\right|<\delta$ then $\left|f\left(x\right)g\left(x\right)-l_1{l}_2\right|<\epsilon$.
4. Assume $\epsilon \in {ℝ}_{>0}$.
5. Let ${\epsilon }_1=\min \left(\frac{\epsilon }{\left|l_1\right|+\left|l_2\right|+1},1\right)$.
6. Since $\underset{x\to a}{\lim }f\left(x\right)=l_1,$ there exists ${\delta }_1\in {ℝ}_{>0}$ such that if $\left|x-a\right|<{\delta }_1$ then $\left|f\left(x\right)-l_1\right|<{\epsilon }_1$.
7. Since $\underset{x\to a}{\lim }g\left(x\right)=l_2,$ there exists ${\delta }_2\in {ℝ}_{>0}$ such that if $\left|x-a\right|<{\delta }_2$ then $\left|f\left(x\right)-l_2\right|<{\epsilon }_1.$
8. Let $\delta =\min \left({\delta }_1,{\delta }_2\right)$. Then $$\begin{array}{rcl}\left|f\left(x\right)g\left(x\right)-l_1{l}_2\right|& =& \left|\left(f\left(x\right)-g\left(x\right)\right)g\left(x\right)+l_1\left(g\left(x\right)-l_2\right)\right|\\ & \le & \left|\left(f\left(x\right)-g\left(x\right)\right)g\left(x\right)\right|+\left|l_1\left(g\left(x\right)-l_2\right)\right|,\text{by the triangle inequality,}\\ & =& \left|\left(f\left(x\right)-l_1\right)\left(g\left(x\right)-l_2\right)+\left(f\left(x\right)-l_1\right)l_2\right|+\left|l_1\right|\left|g\left(x\right)-l_2\right|\\ & \le & \left|\left(f\left(x\right)-l_1\right)\left(g\left(x\right)-l_2\right)\right|+\left|\left(f\left(x\right)-l_1\right)l_2\right|+\left|l_1\right|\left|g\left(x\right)-l_2\right|\\ & \le & \left|\left(f\left(x\right)-l_1\right)\right|\left|\left(g\left(x\right)-l_2\right)\right|+\left|\left(f\left(x\right)-l_1\right)l_2\right|+\left|l_1\right|\left|g\left(x\right)-l_2\right|\\ & \le & {\epsilon }_1^2+{\epsilon }_1\left|l_2\right|+\left|l_1\right|{\epsilon }_1\\ & =& {\epsilon }_1\left(\left|l_1\right|+\left|l_2\right|+{\epsilon }_1\right)\\ & \le & {\epsilon }_1\left(\left|l_1\right|+\left|l_2\right|+1\right)\\ & \le & \epsilon .\end{array}$$

$\square$

$\square$

$\square$

$\square$

1. To show: If $\epsilon \in {ℝ}_{>0}$ then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{{>}_0}$ and $n>N$ then ${\displaystyle \left|\frac{1}{n^p}\right|<\epsilon }$.
2. Assume $\epsilon \in {ℝ}_{{>}_0}$.
3. Let ${\displaystyle N={\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}}$.
4. To show: If $n\in {\mathbb{Z}}_{{>}_0}$ and $n>N$ then ${\displaystyle \left|\frac{1}{n^p}\right|<\epsilon }$.
5. Assume $n\in {\mathbb{Z}}_{{>}_0}$ and $n>N$.
6. To show: ${\displaystyle \left|\frac{1}{n^p}\right|<\epsilon }$.
7. ${\displaystyle \left|\frac{1}{n^p}\right|<\left|\frac{1}{N^p}\right|=\frac{1}{{\left({\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}\right)}^p}=\frac{1}{\left(\frac{1}{\epsilon }\right)}=\epsilon }$.

$\square$

$\square$

1. To show: $\underset{n\to \infty }{\lim }{n}^{\frac{1}{n}}=1$.
2. To show: $\underset{n\to \infty }{\lim }\left(n^{\frac{1}{n}}-1\right)=0$.
3. We know: $$\begin{array}{rcl}n& =& {\left(n^{\frac{1}{n}}\right)}^n\\ & =& {\left(\left(n^{\frac{1}{n}}-1\right)+1\right)}^{\mathrm{n}}\\ & =& 1+n\left(n^{\frac{1}{n}}-1\right)+\frac{n\left(n-1\right)}{2}{\left(n^{\frac{1}{n}}-1\right)}^2+\cdots \\ & \ge & \frac{n\left(n-1\right)}{2}{\left(n^{\frac{1}{n}}-1\right)}^2\text{.}\end{array}$$
4. So ${\displaystyle {\left(n^{\frac{1}{n}}-1\right)}^2\le \frac{2}{n-1}}$.
5. So ${\displaystyle 0\le \underset{n\to \infty }{\lim }{\left(n^{\frac{1}{n}}-1\right)}^2\le \underset{n\to \infty }{\lim }\frac{2}{n-1}=0}$.
6. So $\underset{n\to \infty }{\lim }\left(n^{\frac{1}{n}}-1\right)=0$.

$\square$

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

Let ${\delta }_1\in {ℝ}_{>0}$ be such that $$\text{if }\left|y-l\right|<{\delta }_1\text{then }\left|f\left(y\right)-L\right|<\epsilon .$$

Let $\delta \in {ℝ}_{>0}$ be such that $$\text{if }\left|x-a\right|<\delta \text{then }\left|g\left(x\right)-l\right|<{\delta }_1.$$

$\square$

Proof by contradiction.

Let $l_1=\underset{n\to \infty }{\lim }{a}_n$ and $l_2=\underset{n\to \infty }{\lim }{b}_n.$

Assume $l_1>l_2.$ Let $\epsilon =l_1-l_2.$ Let $N_1\in {\mathbb{Z}}_{>0}$ be such that $$\text{if }n\in {\mathbb{Z}}_{>0}\text{ and }n>N_1\text{ then }\left|a_n-l_1\right|<\frac{\epsilon }{2}.$$ Let $N_2\in {\mathbb{Z}}_{>0}$ be such that $$\text{if }n\in {\mathbb{Z}}_{>0}\text{ and }n>N_2\text{ then }\left|b_n-l_1\right|<\frac{\epsilon }{2}.$$ Let $N\in {\mathbb{Z}}_{>0}$ be such that $N>N_1$ and $N>N_2.$

Then $$a_N>l_1-\frac{\epsilon }{2}=l_2+\epsilon -{\epsilon }_2=l_2+\frac{\epsilon }{2}>b_N.$$

This is a contradiction to $a_N\le b_N.$

Thus $\underset{n\to \infty }{\lim }{a}_n\le \underset{n\to \infty }{\lim }{b}_n.$

$\square$

If $x=1$ then the sequence $a_n=x^n$ is $a_n=1$ and $\underset{n\to \infty }{\lim }{1}^n=1.$

If $x=1$ then the $\underset{n\to \infty }{\lim }1+1^2+\dots +1^n=\underset{n\to \infty }{\lim }n,$ which diverges.

The remaining statements in (b) then follow from (a).

$\square$

Let $x\in ℂ$ with $\left|x\right|<1.$ Let $N\in {\mathbb{Z}}_{>0}$ such that $\left|x\right|<1-\frac{1}{N+1}.$ $$\begin{array}{ccc}\underset{n\to \infty }{\lim }\left|x^n-0\right|& =& \underset{n\to \infty }{\lim }{\left|x\right|}^n\le \underset{n\to \infty }{\lim }{\left(1-\frac{1}{N+1}\right)}^n\\ & =& \underset{n\to \infty }{\lim }{\left(\frac{N+1-1}{N+1}\right)}^n=\underset{n\to \infty }{\lim }{\left(\frac{N}{N+1}\right)}^n=\underset{n\to \infty }{\lim }\frac{1}{{\left(1+\frac{1}{N}\right)}^n}\\ & =& \underset{n\to \infty }{\lim }\frac{1}{1+n\frac{1}{N}+\dots +{\left(\frac{1}{N}\right)}^n}\le \underset{n\to \infty }{\lim }\frac{1}{1+\frac{n}{N}}\\ & =& \underset{n\to \infty }{\lim }\frac{N}{n+N}=N\underset{n\to \infty }{\lim }\frac{1}{n+N}=N\times 0=0.\end{array}$$ Thus $\underset{n\to \infty }{\lim }{x}^n=0.$

$\square$

Let $x\in ℂ$ with $\left|x\right|>1.$ Let $N\in {\mathbb{Z}}_{>0}$ be such that $\left|x\right|>1+\frac{1}{N}.$ Then $$\begin{array}{ccc}{\left|x\right|}^n& >& {\left(1+\frac{1}{N}\right)}^n=1+n\left(\frac{1}{N}\right)+\dots +{\left(\frac{1}{N}\right)}^n\\ & >& \frac{n}{N}\end{array}$$ Since $\frac{n}{N}$ is unbounded as $n$ gets larger and larger, ${\left|x\right|}^n$ is unbounded as $n\to \infty .$

Thus $\underset{n\to \infty }{\lim }{x}^n$ diverges.

$\square$

$\begin{array}{ccc}\underset{x\to 0}{\lim }\left|\left(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \right)-1\right|& =& \underset{x\to 0}{\lim }\left|x\left(1+\frac{x}{2}+\frac{x^2}{3!}+\dots \right)\right|\\ & \ge & \underset{x\to 0}{\lim }\left|x\right|\left(1+\frac{\left|x\right|}{2}+\frac{{\left|x\right|}^2}{3!}+\dots \right)\\ & \ge & \underset{x\to 0}{\lim }\left|x\right|\left(1+\left|x\right|+{\left|x\right|}^2+\dots \right)\\ & =& \underset{x\to 0}{\lim }\left|x\right|\times \frac{1}{1-\left|x\right|}=0\times 1=0.\end{array}$

$\square$

$\square$

To show: $\underset{y\to 0}{\lim }\left|{\left(y+a\right)}^n-a^n\right|=0.$ $$\begin{array}{ccc}\underset{y\to 0}{\lim }\left|{\left(y+a\right)}^n-a^n\right|& =& \underset{y\to 0}{\lim }\left|y^n+n{y}^{n-1}a+\dots +n{a}^{n-1}y+a^n-a^n\right|\\ & =& \underset{y\to 0}{\lim }\left|y^n+n{y}^{n-1}a+\dots +n{a}^{n-1}y\right|\\ & =& \underset{y\to 0}{\lim }\left|y\left(y^{n-1}+a{y}^{n-2}n+\dots +n{a}^{n-1}\right)\right|\\ & =& \underset{y\to 0}{\lim }\left|y\right|\left|y^{n-1}+a{y}^{n-2}n+\dots +n{a}^{n-1}\right|\\ & \ge & \underset{y\to 0}{\lim }\left|y\right|\left({\left|y\right|}^{n-1}+\left|a\right|n{\left|y\right|}^{n-2}+\dots +\left|n{a}^{n-1}\right|\right)\\ & \ge & \underset{y\to 0}{\lim }\left|y\right|n{\left|a\right|}^{n-1}=0\times n\times {\left|a\right|}^{n-1}=0.\end{array}$$ So $x^n=f\left(x\right)$ is continuous at $x=a.$

$\square$