Limits

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

Last updates: 27 May 2012

Limits

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

$l=\underset{n\to \infty }{\lim }{a}_n$    means:    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 $d(a_n,l)<\epsilon$.

Let $X$ be a metric space. Let $f:X\to ℝ$ and $g:X\to ℝ$ be functions and let $a\in X$. Assume that $\underset{x\to a}{\lim }f\left(x\right)$ and $\underset{x\to a}{\lim }g\left(x\right)$ exist. Then

1. (a) $\underset{x\to a}{\lim }\left(f\right(x)+g(x\left)\right)=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right)$,
2. (b) If $c\in ℝ$ then $\underset{x\to a}{\lim }cf\left(x\right)=c\underset{x\to a}{\lim }f\left(x\right)$,   and
3. (c) $\underset{x\to a}{\lim }f\left(x\right)g\left(x\right)=\left(\underset{x\to a}{\lim }f\right(x\left)\right)\left(\underset{x\to a}{\lim }g\right(x\left)\right)$.

		Proof of part (a).
	Let $l_1=\underset{x\to a}{\lim }f\left(x\right)$ and $l_2=\underset{x\to a}{\lim }g\left(x\right)$. To show: $\underset{x\to a}{\lim }f\left(x\right)+g\left(x\right)=l_1+l_2$. 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$\left(f\right(x)+g(x)-(l_1+l_2\left)\right)<\epsilon$. Assume $\epsilon \in {ℝ}_{>0}$. We know: there exists ${\delta }_1\in {ℝ}_{>0}$ such that if $x\in B_{{\delta }_1}\left(a\right)$ then $\left(f\right(x)-l_1)<\frac{\epsilon }{2}$. We know: there exists ${\delta }_2\in {ℝ}_{>0}$ such that if $x\in B_{{\delta }_2}\left(a\right)$ then $\left(f\right(x)-l_2)<\frac{\epsilon }{2}$. Let $\delta =\min ({\delta }_1,{\delta }_2)$. To show: $\left(f\right(x)+g(x)-(l_1+l_2\left)\right)<\epsilon$. $$\begin{array}{rcl}\left(f\right(x)+g(x)-(l_1+l_2\left)\right)& =& \left(\right(f\left(x\right)-l_1)+g(x)-l_2))\\ & \le & \left(f\right(x)-l_1)+\left(g\right(x)-l_2)\\ & \le & \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .\end{array}$$ So $\underset{x\to a}{\lim }\left(f\right(x)+g(x\left)\right)=\underset{x\to a}{\lim }f\left(x\right)+\underset{x\to a}{\lim }g\left(x\right).$ $\square$

		Proof of part (b):
	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 $d(x,a)<\delta$ then $d\left(cf\right(x),cl)<\epsilon$. Assume $\epsilon \in {ℝ}_{>0}$. We know: There exists $\delta \in {ℝ}_{>0}$ such that if $d(x,a)<\delta$ then $d\left(f\right(x),l)<\frac{\epsilon }{\left|c\right|}$. To show: If $d(x,a)<\delta$ then $d\left(cf\right(x),cl)<\epsilon$. Assume $d(x,a)<\delta$. To show: $d\left(cf\right(x),cl)<\epsilon$. $$\begin{array}{rcl}d(cf,cl)& =& \left|cf\right(x)-cl|\\ & =& \left|c\right|\cdot \left|f\right(x)-l|\\ & <& \left|c\right|\cdot \frac{\epsilon }{\left|c\right|}=\epsilon .\square \end{array}$$

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

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

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

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

		Proof.
	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 $$\text{if}n\in {\mathbb{Z}}_{>0}\text{and}n>N_1\text{then}|a_n-l_1|<\frac{\epsilon }{2}.$$ Let $N_2\in {\mathbb{Z}}_{>0}$ be such that $$\text{if}n\in {\mathbb{Z}}_{>0}\text{and}n>N_2\text{then}|b_n-l_1|<\frac{\epsilon }{2}.$$ 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 $\underset{n\to \infty }{\lim }{a}_n\le \underset{n\to \infty }{\lim }{b}_n$. $\square$

The functions $f\left(x\right)=x^n$ and $f\left(x\right)=e^x$

1. (a) Let $n\in {\mathbb{Z}}_{>0}$ and $a\in ℝ$. Then $x^n$ is continuous at $x=a$ (i.e. $\underset{x\to a}{\lim }{x}^n=a^n$).
2. (b) $e^x$ is continuous at $x=a$.
3. (c) Prove that $\underset{x\to 0}{\lim }{e}^x=e^0.$
4. (d) Prove that $\underset{x\to a}{\lim }{e}^x=e^a$.

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

		First part of proof of part (b).
	$$\begin{array}{ccc}\underset{x\to 0}{\lim }|(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots )-1|& =& \underset{x\to 0}{\lim }\left|x\right|(1+\frac{x}{2}+\frac{x^2}{3!}+\dots )\\ & \ge & \underset{x\to 0}{\lim }\left|x\right|{\displaystyle (1+\frac{\left|x\right|}{2}+\frac{{\left|x\right|}^2}{3!}+\dots )}\\ & \ge & \underset{x\to 0}{\lim }\left|x\right|(1+\left|x\right|+{\left|x\right|}^2+\dots )\\ & =& \underset{x\to 0}{\lim }\left|x\right|·{\displaystyle \frac{1}{1-\left|x\right|}}=0·1=0.\end{array}$$ $\square$

		Proof of part (d).
	$$\begin{array}{ccc}\underset{x\to a}{\lim }{e}^x& =& \underset{y\to 0}{\lim }{e}^{y+a}=\underset{y\to 0}{\lim }{e}^a{e}^y\\ & =& e^a\underset{y\to 0}{\lim }{e}^y=e^a{e}^0=e^{a+0}=e^a.\end{array}$$ $\square$

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

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

		Proof of part (a) Case $\left|x\right|>1$.
	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 $$\begin{array}{ccc}{\left|x\right|}^n& >& {(1+\frac{1}{N})}^n=1+n\left(\frac{1}{N}\right)+\dots +{\left(\frac{1}{N}\right)}^n\\ & >& \frac{n}{N}.\end{array}$$ 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. $\square$

		Proof of part (b).
	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). $\square$

Useful limits

1. (a) If $n\in {\mathbb{Z}}_{>0}$ then $\underset{x\to \infty }{\lim }{x}^n{e}^{-x}=0$.
2. (b) If $\alpha \in {ℝ}_{>0}$ then $\underset{x\to \infty }{\lim }{x}^{-\alpha }\log x=0$.
3. (c) Let $p\in {ℝ}_{>0}$. Then ${\displaystyle \underset{n\to \infty }{\lim }\frac{1}{n^p}=0}$.
4. (d) Let $p\in {ℝ}_{>0}$. Then $\underset{n\to \infty }{\lim }{p}^{1/n}=1$.
5. (e) $\underset{n\to \infty }{\lim }{n}^{1/n}=1$.
6. (f) Let $\alpha \in ℝ$ and $p\in {ℝ}_{>0}$. Then ${\displaystyle \underset{n\to \infty }{\lim }\frac{n^{\alpha }}{{(1+p)}^n}=0}$.
7. (g) If $\left|x\right|<1$ then $\underset{n\to \infty }{\lim }{x}^n=0$.
8. (h) If $0<\alpha <1$ then $\underset{x\to 0}{\lim }{x}^{\alpha }=0$.
9. (i) ${\displaystyle \underset{x\to 0}{\lim }\frac{e^x-1}{x}=1}$.
10. (j) ${\displaystyle \underset{x\to 0}{\lim }\frac{\sin x}{x}=1}$.
11. (k) ${\displaystyle \underset{x\to 0}{\lim }\frac{\cos x-1}{x^2}=-\frac{1}{2}}$.
12. (l) ${\displaystyle \underset{x\to 0}{\lim }\frac{\log (1+x)}{x}=1}$.

		Proof of part (l):
	$$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\log (1+x)}{x}& =& \underset{y\to 0}{\lim }\frac{\log (1+e^y-1)}{e^y-1}\\ & =& \underset{y\to 0}{\lim }\frac{y}{e^y-1}\\ & =& \underset{y\to 0}{\lim }\frac{1}{\left(\frac{e^y-1}{y}\right)}\\ & =& \frac{1}{1}\\ & =& 1\text{.}\end{array}$$ $\square$

		Proof of part (j):
	$$\begin{array}{rcl}\underset{x\to 0}{\lim }\frac{\sin x}{x}& =& \underset{x\to 0}{\lim }\frac{e^{ix}-e^{-ix}}{2ix}\\ & =& \underset{x\to 0}{\lim }\frac{e^{ix}-1}{2ix}-\frac{e^{-ix}-1}{2ix}\\ & =& \underset{x\to 0}{\lim }\frac{1}{2}\left(\frac{e^{ix}-1}{ix}\right)+\frac{1}{2}\left(\frac{e^{-ix}-1}{-ix}\right)\\ & =& \frac{1}{2}·1+\frac{1}{2}·1\\ & =& 1\text{.}\end{array}$$ $\square$

		Proof of part (a):
	Assume $n\in {\mathbb{Z}}_{>0}$. $$\begin{array}{rcl}0\le \underset{x\to \mathrm{\infty }}{\lim }{x}^n{e}^{-x}& =& \underset{x\to \mathrm{\infty }}{\lim }\frac{x^n}{e^x}\\ & \le & \underset{x\to \mathrm{\infty }}{\lim }\frac{x^n}{\frac{1}{(n+1)!}{x}^{n+1}}\\ & =& \underset{x\to \mathrm{\infty }}{\lim }\frac{(n+1)!}{x}\\ & =& (n+1)!\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{x}\\ & =& 0\text{.}\end{array}$$ $\square$

		Proof of part (b):
	Let $0<\epsilon <\alpha$. $$\begin{array}{rcl}0\le \underset{x\to \mathrm{\infty }}{\lim }{x}^{-\alpha }\log x& =& \underset{x\to \mathrm{\infty }}{\lim }\left(x^{-\alpha }{\int }_1^x\frac{1}{t}dt\right)\\ & \le & \underset{x\to \mathrm{\infty }}{\lim }\left(x^{-\alpha }{\int }_1^x{t}^{\epsilon -1}dt\right)\\ & =& \underset{x\to \mathrm{\infty }}{\lim }{x}^{-\alpha }\left(\frac{x^{\epsilon }-1^{\epsilon }}{\epsilon }\right)\\ & =& \underset{x\to \mathrm{\infty }}{\lim }\left(\frac{x^{\epsilon -\alpha }-x^{-\alpha }}{\epsilon }\right)\\ & \le & \underset{x\to \mathrm{\infty }}{\lim }\frac{x^{\epsilon -\alpha }}{\epsilon }\\ & =& 0\text{.}\end{array}$$ $\square$

		Proof of part (c).
	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 ${\displaystyle \left|\frac{1}{n^p}\right|<\epsilon }$. Assume $\epsilon \in {ℝ}_{{>}_0}$. Let ${\displaystyle N={\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}}$. To show: If $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then ${\displaystyle \left|\frac{1}{n^p}\right|<\epsilon }$. Assume $n\in {\mathbb{Z}}_{>0}$ and $n>N$. To show: ${\displaystyle \left|\frac{1}{n^p}\right|<\epsilon }$. $$\left|\frac{1}{n^p}\right|<\left|\frac{1}{N^p}\right|=\frac{1}{{\left({\left(\frac{1}{\epsilon }\right)}^{\frac{1}{p}}\right)}^p}=\frac{1}{\left(\frac{1}{\epsilon }\right)}=\epsilon .$$ $\square$

		Proof of part (d):
	$$\begin{array}{rcl}\underset{n\to \infty }{\lim }{p}^{\frac{1}{n}}& =& \underset{n\to \infty }{\lim }{\left(e^{\log p}\right)}^{\frac{1}{\mathrm{n}}}\\ & =& \underset{n\to \infty }{\lim }{e}^{\frac{1}{n}\log p}\\ & =& e^{\log p\underset{n\to \infty }{\lim }\frac{1}{n}}\\ & =& e^{\log p·0}\\ & =& e^0\\ & =& 1\end{array}$$ $\square$

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

Notes and References

This section proves the fundamental limit theorems used most often for computations. THe presentation is focused on Theorem 1.1, establishing, without much effort, the primary tools that are used for evaluation of limits in examples.

It feels a bit funny not to include the f(x)/g(x) case in Theorem 1.1. SHOULD IT BE PUT IN? WHY? WHY NOT?

It is interesting to note that in practice, once Theorems 1.1 and 1.2 are proved, it is hardly ever necessary to use an $\epsilon \text{-}\delta$ proof to evaluate a limit.

A traditional reference for this material is [Ru1, Ch. 3 and 4] with Theorem 1.1 being [Ru1, Theorem 4.4] and much of Theorem ??? coinciding with [Ru1, Theorem 3.20].

References

[Bou] N. Bourbaki, Algèbre, Chapitre ?: ??????????? MR?????.

[Ru1] W. Rudin, Principles of mathematical analysis, McGraw-Hill, 1976. MR0385023

[Ru2] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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