Johann Heinrich Ludolf Steinike (1825-1909) was a German landscape painter.
We have sections
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Sequences.
Let \(S\) be a nonempty set of real numbers. We say that the real number \(u\) is an upper bound of \(S\) when \(u\geq x\) for all \(x\in S\). In this case, we also say that \(S\) is bounded above.
Definition. (Least Upper Bound). Let \(S\) be a nonempty set of real numbers which is bounded above. A number \(u^{*}\) is the least upper bound of \(S\) when the following conditions are satisfied.
- \(u^{*}\) is an upper bound for \(S\).
- we have \(u^{*}\leq u\) for any upper bound \(u\) of \(S\).
In this case, we denote by \(u^{*}=\sup S\). \(\sharp\)
For example, we have
\begin{align}
&\sup\left\{\frac{1}{2},\frac{2}{3},\frac{3}{4},\cdots ,\frac{n}{n+1},\cdots\right\}=1,\\
&\sup\left\{-\frac{1}{2},-\frac{1}{8},-\frac{1}{27},\cdots ,-\frac{1}{n^{3}},\cdots\right\}=0,\\
&\sup\{x:x^{2}<3\}=\sup\{x:-\sqrt{3}<x<\sqrt{3}\}=\sqrt{3}.\label{eq2}\tag{1}
\end{align}
We have the following properties:
(a) Every nonempty set of real numbers that has an upper bound has a least upper bound.
(b) Suppose that \(u^{*}\) is the least upper bound of the set \(S\), and that \(\epsilon\) is a positive number. Then, there is at least one number \(s\) in \(S\) satisfying
\[u^{*}-\epsilon <s\leq u^{*}.\]
Let \(S\) be a nonempty set of real numbers. We say that the real number \(l\) is an upper bound of \(S\) when \(l\leq x\) for all \(x\in S\). In this case, we also say that \(S\) is bounded below.
Definition. (Greatest Lower Bound). Let \(S\) be a nonempty set of real numbers which is bounded below. A number \(l^{*}\) is the greatest lower bound of \(S\) when the following conditions are satisfied.
- \(m\) is a lower bound for \(S\),
- \(m\geq k\), where \(k\) is any lower bound for \(S\), and denoted by \(\inf S=m\). $\sharp$
Then we have the following properties:
(a) Every nonempty set of real numbers that has a lower bound has a greatest lower bound.
(b) If \(m\) is the greatest lower bound of the set \(S\) and \(\epsilon\) is a positive number, then there is at least one number \(s\) in \(S\) satisfying $\latex m\leq s<m+\epsilon$.
Definition. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a sequence in \(\mathbb{R}\).
- The sequence \(\{a_{n}\}_{n=1}^{\infty}\) is said to be increasing when \(a_{n}<a_{n+1}\) for each positive integer \(n\),
- The sequence \(\{a_{n}\}_{n=1}^{\infty}\) is said to be nondecreasing when \(a_{n}\leq a_{n+1}\) for each positive integer \(n\),
- The sequence \(\{a_{n}\}_{n=1}^{\infty}\) is said to be decreasing when \(a_{n}>a_{n+1}\) for each positive integer \(n\),
- The sequence \(\{a_{n}\}_{n=1}^{\infty}\) is said to be increasing when \(a_{n}\geq a_{n+1}\) for each positive integer \(n\).
If any of these four properties holds, the sequence is said to be monotonic. $\sharp$
Example. We have the following interesting examples.
(i) The sequence \(\{a_{n}\}_{n=1}^{\infty}\) defined by \(a_{n}=n/(n+1)\) is increasing since
\[\frac{a_{n+1}}{a_{n}}=\frac{n+1}{n+2}\cdot\frac{n+1}{n}=\frac{n^{2}+2n+1}{n^{2}+2n}>1.\]
We also have \(\sup\{a_{n}\}_{n=1}^{\infty}=1\).
(ii) The sequence \(\{a_{n}\}_{n=1}^{\infty}\) defined by \(a_{n}=2^{n}/n!\) is nonincreasing. Moreover, we have \(a_{n}>a_{n+1}\) for \(n\geq 2\) since
\[\frac{a_{n+1}}{a_{n}}=\frac{2^{n+1}}{(n+1)!}\cdot\frac{n!}{2^{n}}=\frac{2}{n+1}<1\mbox{ if }n\geq 2.\]
(iii) Consider the sequence \(\{a_{n}\}_{n=1}^{\infty}\) defined by \(a_{n}=n/e^{n}\). Let \(f(x)=x/e^{x}\). Then \(f'(x)=(1-x)/e^{x}\). Since \(f'(x)<0\) for \(x>1\), \(f\) is a decreasing function on \([1,\infty )\). Therefore, we have \(f(1)>f(2)>\cdots\), i.e., $a_{1}>a_{2}>\cdots$, which says that \(\{a_{n}\}_{n=1}^{\infty}\) is decreasing.
(iv) Consider the sequence \(\{a_{n}\}_{n=1}^{\infty}\) defined by \(a_{n}=n^{1/n}\). Let $f(x)=x^{1/x}$. Since \(f(x)=e^{(1/x)\ln x}\), we have
\[f'(x)=e^{(1/x)\ln x}\frac{d}{dx}\left (\frac{\ln x}{x}\right )=x^{1/x}\left (\frac{1-\ln x}{x^{2}}\right ).\]
For \(x>e\), \(f'(x)<0\). This shows that \(f\) is a decreasing function on $[e,\infty )$. Since \(3>e\), \(f\) decreases on \([3,\infty )\). If follows that $\{a_{n}\}_{n=1}^{\infty}$ decreases for \(n\geq 3\). $\sharp$
Limits of Sequences.
Definition. Let \(\{a_{n}\}_{n=1}^{\infty}\) be a sequence of real numbers. We say that \(L\) is a limit of sequence \(\{a_{n}\}_{n=1}^{\infty}\), denoted by
\[\lim_{n\rightarrow\infty} a_{n}=L,\]
when, for each \(\epsilon >0\), there exists a positive integer $K$ such that \(n\geq K\) implies \(|a_{n}-L|<\epsilon\). A sequence that has a limit is said to be convergent, A sequence that has no limit is said to be divergent. $\sharp$
Example. We have the limits
\[\lim_{n\rightarrow\infty}\frac{1}{n}=0\mbox{ and }\lim_{n\rightarrow\infty}\frac{2n-1}{n}=2.\]
However, the limit
\[\lim_{n\rightarrow\infty}\frac{n}{\sqrt{n+1}}\]
does not exist, since
\[\frac{n}{\sqrt{n+1}}=\frac{1}{\frac{1}{n}\sqrt{n+1}}=\frac{1}{\sqrt{\frac{1}{n}+\frac{1}{n^{2}}}}.\]
Theorem. We have the following properties.
(i) (Uniqueness of Limit). Suppose that
\[\lim_{n\rightarrow\infty} a_{n}=L\mbox{ and }\lim_{n\rightarrow\infty} a_{n}=M.\]
Then, we have \(L=M\).
(ii)} Every convergent sequence is bounded, and every unbounded sequence is divergent.
(iii) A bounded nondecreasing sequence converges to its least upper bound, and a bounded nonincreasing sequence converges to its greatest lower bound. $\sharp$
Theorem. Let \(\alpha\) be a real number. Suppose that
\[\lim_{n\rightarrow\infty} a_{n}=L\mbox{ and }\lim_{n\rightarrow\infty} b_{n}=M\]
Then, we have
\begin{align*} & \lim_{n\rightarrow\infty} (a_{n}+b_{n})=L+M\\
& \lim_{n\rightarrow\infty} \alpha a_{n}=\alpha L\\
& \lim_{n\rightarrow\infty} a_{n}b_{n}=LM.\end{align*}
If, in addition, \(M\neq 0\) and \(b_{n}\neq 0\) for all \(n\), then
\[\lim_{n\rightarrow\infty}\frac{1}{b_{n}}\mbox{ and }\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\frac{L}{M}.\]
We also have
\[\lim_{n\rightarrow\infty} a_{n}=L\mbox{ if and only if }
\lim_{n\rightarrow\infty}(a_{n}-L)=0\mbox{ if and only if }
\lim_{n\rightarrow\infty}|a_{n}-L|=0.\]
We are now in a position to handle any rational sequence \(\{a_{n}\}_{n=1}^{\infty}\), where
\begin{equation}{\label{e10}}\tag{2}
a_{n}=\frac{\alpha_{k}n^{k}+\alpha_{k-1}n^{k-1}+\cdots +\alpha_{0}}{\beta_{j}n^{j}+\beta_{j-1}n^{j-1}+\cdots +\beta_{0}},\alpha_{k}\neq 0,\beta_{j}\neq 0.
\end{equation}
To determine the behavior of such a sequence we need only divide both numerator and denominator by the highest power of \(n\) that occurs.
Example. We have
\begin{align*}
& \frac{3n^{4}-2n^{2}+1}{n^{5}-3n^{3}}=\frac{3/n-2/n^{3}+1/n^{5}}{1-3/n^{2}}\rightarrow\frac{1}{0}=0\\
& \frac{1-4n^{7}}{n^{7}+12n}=\frac{1/n^{7}-4}{1+12/n^{6}}\rightarrow\frac{-4}{1}=-4\\
& \frac{n^{4}-3n^{2}+n+2}{n^{3}+7n}=\frac{1-3/n^{2}+1/n^{3}+2/n^{4}}{1/n+7/n^{3}}\rightarrow\infty .
\end{align*}
In general, if \(a_{n}\) is given by (\ref{e10}), then
\[\lim_{n\rightarrow\infty} a_{n}=\left\{\begin{array}{ll} 0 & \mbox{if \(k<j\)}\\
\alpha_{k}/\beta_{k} & \mbox{if \(k=j\)}\\ \pm\infty & \mbox{if \(k>j\)}.
\end{array}\right .\]
Theorem. (The Pinching Theorem for Sequences). Suppose that there is a positive integer \(K\) satisfying \(a_{n}\leq b_{n}\leq c_{n}\) for all \(n\geq K\). Then
\[\lim_{n\rightarrow\infty} a_{n}=L=\lim_{n\rightarrow\infty} c_{n}\mbox{ implies }\lim_{n\rightarrow\infty} b_{n}=L.\]
\begin{equation}{\label{c2}}\tag{3}\mbox{}\end{equation}
Corollary \ref{c2}. Suppose that there is a positive integer \(K\) satisfying $|b_{n}|\leq c_{n}$ for all \(n\geq K\). Then, we have \[\lim_{n\rightarrow\infty} c_{n}=0\mbox{ implies }\lim_{n\rightarrow\infty} |b_{n}|=0.\]
Example. We have
\[\lim_{n\rightarrow\infty} \cos n/n=0\]
since
\[\left |\frac{\cos n}{n}\right |\leq 1/n\mbox{ and }\lim_{n\rightarrow\infty} 1/n=0.\]
Example. Find the limit
\[\lim_{n\rightarrow\infty} \left (1+\frac{1}{n}\right )^{n}.\]
We have
\begin{equation}{\label{e100}}\tag{4}
\left (1+\frac{1}{n}\right )^{n}\leq e\leq\left (1+\frac{1}{n}\right )^{n+1}.
\end{equation}
for all positive integers \(n\). The proof is given as follows. Since
\[\ln\left (1+\frac{1}{n}\right )=\int_{1}^{1+\frac{1}{n}}\frac{dt}{t}\leq\int_{1}^{1+\frac{1}{n}}dt=\frac{1}{n},\]
it follows that
\[1+\frac{1}{n}\leq e^{\frac{1}{n}}\mbox{ implies }\left (1+\frac{1}{n}\right )^{n}\leq e.\]
Similarly, we have
\begin{align*} \ln\left (1+\frac{1}{n}\right ) & =\int_{1}^{1+\frac{1}{n}}\frac{dt}{t}\geq\int_{1}^{1+\frac{1}{n}}\frac{dt}{1+\frac{1}{n}}\\ & =\frac{1}{1+\frac{1}{n}} \cdot\frac{1}{n}=\frac{1}{n+1}\end{align*}
Therefore, we obtain that
\[e^{\frac{1}{n+1}}\leq 1+\frac{1}{n}\mbox{ implies }e\leq\left (1+\frac{1}{n}\right )^{n+1}.\]
Dividing the right-hand inequality in (\ref{e100}) by \(1+1/n\), we have
\[\frac{e}{1+1/n}\leq\left (1+\frac{1}{n}\right )^{n}.\]
Combining this with the left-hand inequality, we also have
\[\frac{e}{1+1/n}\leq\left (1+\frac{1}{n}\right )^{n}\leq e.\]
Using the pinching theorem , we obtain
\[\lim_{n\rightarrow\infty} \left (1+\frac{1}{n}\right )^{n}=e.\]
Theorem. Suppose that \(\lim_{n\rightarrow\infty} c_{n}=c\) and that, for each \(n\), $c_{n}$ is in the domain of \(f\). If \(f\) is continuous at \(c\), then
\[\lim_{n\rightarrow\infty} f(c_{n})=f(c).\]
Example. We have some interesting examples.
(i) Since \(\lim_{n\rightarrow\infty} \pi /n=0\) and the consine function is continuous at \(0\), we have
\[\lim_{n\rightarrow\infty} \cos (\pi /n)=\cos 0=1.\]
(ii) Since \(\lim_{n\rightarrow\infty} 1/n=0\) and the exponential function is continuous at \(0\), we have
\[\lim_{n\rightarrow\infty} e^{1/n}=e^{0}=1.\]
(iii) Since
\[\lim_{n\rightarrow\infty} \frac{n^{2}\pi^{2}-8}{16n^{2}}=
\lim_{n\rightarrow\infty} \frac{\pi^{2}-8/n^{2}}{16}=\frac{\pi^{2}}{16}\]
and the function \(f(x)=\tan\sqrt{x}\) is continuous at \(\pi^{2}/16\), we have
\[\tan\left (\sqrt{\frac{\pi^{2}n^{2}-8}{16n^{2}}}\right )=\tan\left (\frac{\pi^{2}}{16}\right )=\tan\frac{\pi}{4}=1.\]
(iv) Since the absolute-value function is everywhere continuous, we have that
\[\lim_{n\rightarrow\infty} a_{n}=L\mbox{ implies }\lim_{n\rightarrow\infty} |a_{n}|=|L|.\]
From Corollary \ref{c2}, we can also obtain $\lim_{n\rightarrow\infty} b_{n}=0$. $\sharp$
Example. For each real \(x\), find the limit
\[\lim_{n\rightarrow\infty} \frac{x^{n}}{n!}.\]
Fix any real number \(x\) and choose an intger \(k\) such that \(k>|x|\). For $n>k+1$,
\[\frac{k^{n}}{n!}=\left (\frac{k^{k}}{k!}\right )\left [\frac{k}{k+1}
\frac{k}{k+2}\cdots\frac{k}{n-1}\right ]\left (\frac{k}{n}\right )<
\left (\frac{k^{k+1}}{k!}\right )\left (\frac{1}{n}\right ),\]
where the middle term is less than \(1\). Since \(k>|x|\), we have
\[0<\frac{|x|^{n}}{n!}<\frac{k^{n}}{n!}<\left (\frac{k^{k+1}}{k!}\right )\left (\frac{1}{n}\right ).\]
Since \(k\) is fixed and \(\lim_{n\rightarrow\infty} 1/n=0\), using the pinching theorem, we obtain
\[\lim_{n\rightarrow\infty}\frac{|x|^{n}}{n!}=0,\mbox{ i.e., }\lim_{n\rightarrow\infty}\frac{x^{n}}{n!}=0.\]
Example. Find the limit
\[\lim_{n\rightarrow\infty} n^{1/n}.\]
We first have \(n^{1/n}=e^{(1/n)\ln n}\) and
\begin{align*} 0\leq\frac{\ln n}{n} & =\frac{1}{n}\int_{1}^{n}\frac{dt}{t}\leq\frac{1}{n}
\int_{1}^{n}\frac{dt}{\sqrt{t}}\\ & =\frac{2}{n}(\sqrt{n}-1)=2\left (\frac{1}{\sqrt{n}}-\frac{1}{n}\right ).\end{align*}
Since the exponential function is continuous at \(0\), using \(\lim_{n\rightarrow\infty} \ln n/n=0\) and the pinching theorem, we obtain
\[\lim_{n\rightarrow\infty} n^{1/n}=e^{0}=1.\]
Example. For each \(x\in\mathbb{R}\), find the limit
\[\lim_{n\rightarrow\infty} \left (1+\frac{x}{n}\right )^{n}.\]
For \(x=0\), the result is obvious. For \(x\neq 0\), we have
\[\ln\left (1+\frac{x}{n}\right )^{n}=n\ln\left (1+\frac{x}{n}\right )=x\left [\frac{\ln (1+x/n)-\ln 1}{x/n}\right ].\]
For each \(t>0\), we have
\[\lim_{h\rightarrow 0}\frac{\ln (t+h)-\ln t}{h}=\frac{d}{dt}(\ln t)=\frac{1}{t}.\]
Let \(h=x/n\) and write
\[\lim_{n\rightarrow\infty} \left [\frac{\ln (1+x/n)-\ln 1}{x/n}\right ]=\lim_{h\rightarrow 0} \left [\frac{\ln (1+h)-\ln 1}{h}\right ]=1.\]
It follows
\[\lim_{n\rightarrow\infty} \ln\left (1+\frac{x}{n}\right )^{n} =x,\]
which implies
\[\lim_{n\rightarrow\infty} \left (1+\frac{x}{n}\right )^{n}=\lim_{n\rightarrow\infty} e^{\ln (1+x/n)^{n}}=e^{x}.\]
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
The Indeterminate Form.
Theorem. (L’Hospital’s Rule $(0/0)$). Suppose that \(f\) and \(g\) are differentiable functions satisfying \(f(x)\rightarrow 0\) and \(g(x)\rightarrow 0\) as $x\rightarrow c^{+}$, \(x\rightarrow c^{-}\), \(x\rightarrow c\), $x\rightarrow\infty$, or \(x\rightarrow -\infty\). Then, we have that
\[\frac{f'(x)}{g'(x)}\rightarrow L\mbox{ implies }\frac{(x)}{g(x)}\rightarrow L.\]
We also have that
\[\frac{f'(x)}{g'(x)}\rightarrow\infty\mbox{ or } -\infty\mbox{ implies }\mbox{f(x)}{g(x)}\rightarrow\infty\mbox{ or } -\infty,\]
respectively. $\sharp$
For convenience, we use \(*\) to indicate the differentiation of numerator and denominator.
Example. We some interesting examples.
(i) We have
\[\lim_{x\rightarrow\pi /2}\frac{\cos x}{\pi -2x}\stackrel{*}{=}\lim_{x\rightarrow\pi /2}\frac{-\sin x}{-2}=\frac{1}{2}.\]
(ii) We have
\[\lim_{x\rightarrow 0^{+}}\frac{x}{\sin\sqrt{x}}\stackrel{*}{=}\lim_{x\rightarrow 0^{+}}\frac{2\sqrt{x}}{\cos\sqrt{x}}=0.\]
(iii) We have
\[\lim_{x\rightarrow 0}\frac{e^{x}-x-1}{x^{2}}\stackrel{*}{=}
\lim_{x\rightarrow 0}\frac{e^{x}-1}{2x}\stackrel{*}{=}\lim_{x\rightarrow 0}\frac{e^{x}}{2}=\frac{1}{2}.\]
(iv) Given the sequence
\[\left\{\frac{e^{2/n}-1}{1/n}\right\}_{n=1}^{\infty}.\]
Find the limit
\[\lim_{n\rightarrow\infty} \frac{e^{2/n}-1}{1/n}.\]
Applying the L’Hostital’s rule, we have
\[\lim_{x\rightarrow\infty}\frac{e^{2/x}-1}{1/x}\stackrel{*}{=}
\lim_{x\rightarrow\infty}\frac{e^{2/x}(-2/x^{2})}{-1/x^{2}}=2\lim_{x\rightarrow\infty} e^{2/x}=2.\]
This shows
\[\lim_{x\rightarrow\infty}\frac{e^{2/x}-1}{1/x}=2, \mbox{ i.e., }\lim_{n\rightarrow\infty}\frac{e^{2/n}-1}{1/n}=2.\]
Theorem. (L’Hospital’s Rule \((\infty /\infty )\)). Suppose that \(f\) and \(g\) are differentiable functions satisfying \(f(x)\rightarrow\pm\infty\) and \(g(x)\rightarrow\pm\infty\) as \(x\rightarrow c^{+}\), \(x\rightarrow c^{-}\), \(x\rightarrow c\), $x\rightarrow\infty$, or \(x\rightarrow -\infty\). Then, we have that
\[\frac{f'(x)}{g'(x)}\rightarrow L\mbox{ implies }\frac{f(x)}{g(x)}\rightarrow L.\]
We also have that
\[\frac{f'(x)}{g'(x)}\rightarrow\infty\mbox{ or }-\infty\mbox{ implies }\mbox{f(x)}{g(x)}\rightarrow\infty\mbox{ or }-\infty,\]
respectively. $\sharp$
Example. We have some interesting examples.
(i) Let \(\alpha\) be any positive number. We have
\[\lim_{x\rightarrow\infty}\frac{\ln x}{x^{\alpha}}\stackrel{*}{=}
\lim_{x\rightarrow\infty}\frac{1/x}{\alpha x^{\alpha -1}}=
\lim_{x\rightarrow\infty} \frac{1}{\alpha x^{\alpha}}=0.\]
(ii) Let \(k\) be any positive integer. We have
\begin{align*}\lim_{x\rightarrow\infty}\frac{x^{k}}{e^{x}} & \stackrel{*}{=}
\lim_{x\rightarrow\infty}\frac{kx^{k-1}}{e^{x}}\stackrel{*}{=}
\lim_{x\rightarrow\infty}\frac{k(k-1)x^{k-2}}{e^{x}}\\ & \stackrel{*}{=}
\cdots\stackrel{*}{=}\lim_{x\rightarrow\infty}\frac{k!}{e^{x}}=0.\end{align*}
(iii) Let \(k\) be any positive integer. Find the limit of the sequence \(\{a_{n}\}_{n=1}^{\infty}\) given by \(a_{n}=\frac{2^{n}}{n^{k}}\). Applying the L’Hospital’s rule, we have
\begin{align*}\lim_{x\rightarrow\infty} \frac{2^{x}}{x^{k}} & \stackrel{*}{=}
\lim_{x\rightarrow\infty} \frac{2^{x}\ln 2}{kx^{k-1}}\stackrel{*}{=}
\lim_{x\rightarrow\infty}\frac{2^{x}(\ln 2)^{2}}{k(k-1)x^{k-2}}\\ & \stackrel{*}{=}
\cdots\stackrel{*}{=}\lim_{x\rightarrow\infty} \frac{2^{x}(\ln 2)^{k}}{k!}=\infty,\end{align*}
which implies
\[\lim_{n\rightarrow\infty} \frac{2^{n}}{x^{n}}=\infty\]
The other indeterminate forms are classified as \(0\cdot\infty\), \(\infty -\infty\), \(0^{0}\), \(1^{\infty}\) and \(\infty^{0}\).
Example. We have some interesting examples.
(i) (\(0\cdot\infty\) form) Find the limit
\[\lim_{x\rightarrow 0^{+}}\sqrt{x}\ln x.\]
Rewritting the product as a quotient, we have
\begin{align*} \lim_{x\rightarrow 0^{+}}\sqrt{x}\ln x & =\lim_{x\rightarrow 0^{+}}
\frac{\ln x}{1/\sqrt{x}}\stackrel{*}{=}\lim_{x\rightarrow 0^{+}}\frac
{1/x}{(-1/2)x^{-3/2}}\\ & =\lim_{x\rightarrow 0^{+}}-2\sqrt{x}=0.\end{align*}
(ii) (\(\infty -\infty\) form) Find the limit
\[\lim_{x\rightarrow (\pi /2)^{-}} (\tan x-\sec x).\]
We rewrite the difference as a quotient
\[\tan x-\sec x=\frac{\sin x}{\cos x}-\frac{1}{\cos x}=\frac{\sin x-1}{\cos x}.\]
Applying the L’Hospital’s rule, we have
\[\lim_{x\rightarrow (\pi /2)^{-}}\frac{\sin x-1}{\cos x}\stackrel{*}{=}\lim_{x\rightarrow (\pi /2)^{-}}\frac{\cos x}{-\sin x}=0,\]
which implies
\[\lim_{x\rightarrow (\pi /2)^{-}} (\tan x-\sec x)=0.\]
(iii) (\(0^{0}\) form) Find the limit
\[\lim_{x\rightarrow 0^{+}} x^{x}.\]
We have \(x^{x}=e^{x\ln x}\). Applying the L’Hosital’s rule, we obtain
\begin{align*} \lim_{x\rightarrow 0^{+}} (x\ln x) & =\lim_{x\rightarrow 0^{+}}\frac{\ln x}
{1/x}\\ & \stackrel{*}{=}\lim_{x\rightarrow 0^{+}}\frac{1/x}{-1/x^{2}}=\lim_{x\rightarrow 0^{+}} (-x)=0,\end{align*}
which implies
\[\lim_{x\rightarrow 0^{+}} x^{x}=1.\]
(iv) (\(1^{\infty}\) form) Find the limit
\[\lim_{x\rightarrow 0^{+}} (1+x)^{1/x}.\]
We have \((1+x)^{1/x}=e^{\ln (x+1)/x}\). Applying the L’Hospital’s rule, we obtain
\[\lim_{x\rightarrow 0^{+}}\frac{\ln (1+x)}{x}\stackrel{*}{=}\lim_{x\rightarrow 0^{+}}\frac{1}{1+x}=1,\]
which implies
\[\lim_{x\rightarrow 0^{+}} (1+x)^{1/x}=1.\]
(v) (\(\infty^{0}\) form) Find the limit
\[\lim_{x\rightarrow\infty} (x^{2}+1)^{1/\ln x}.\]
We have \((x^{2}+1)^{1/\ln x}=e^{\ln (x^{2}+1)/\ln x}\). Applying the L’Hospital’s rule, we obtain
\[\lim_{x\rightarrow\infty}\frac{\ln (x^{2}+1)}{\ln x}\stackrel{*}{=}
\lim_{x\rightarrow\infty}\frac{2x/(x^{2}+1)}{1/x}=\lim_{x\rightarrow\infty}\frac{2x^{2}}{x^{2}+1}=2,\]
which implies
\[\lim_{x\rightarrow\infty} (x^{2}+1)^{1/\ln x}=e^{2}.\]
Now, we know
\[\lim_{x\rightarrow 0^{+}}\frac{1+x}{\sin x}=\infty\]
A misapplication of L’Hospital’s rule would lead to the limit
\[\lim_{x\rightarrow 0^{+}}\frac{1}{\cos x}=1,\]
and the incorrect conclusion given by
\[\lim_{x\rightarrow 0^{+}}\frac{1+x}{\sin x}=1.\]
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Improper Integrals.
We begin with a function \(f\) continuous on an unbounded interval $[a,\infty )$. For each number \(b>a\), we can form the definite integral $\int_{a}^{b} f(x)dx$. As \(b\) tends to \(\infty\), this integral tends to a finite limit \(L\) given by
\[\lim_{b\rightarrow\infty}\int_{a}^{b} f(x)dx=L.\]
In this case, we write
\[\int_{a}^{\infty} f(x)dx=L.\]
W also e say that
\[\mbox{the improper integral }\int_{a}^{\infty} f(x)dx\mbox{ converges to \(L\).}\]
Otherwise, we say that
\[\mbox{the improper integral }\int_{a}^{\infty} f(x)dx\mbox{ diverges.}\]
In a similar manner, when \(f\) is continuous on the unbounded interval $(-\infty ,b]$, for each number \(a<b\), we can form the definite integral \(\int_{a}^{b} f(x)dx\) and calculate
\[\lim_{a\rightarrow -\infty}\int_{a}^{b} f(x)dx.\]
When this limit exists and equals \(L\), we say that
\[\mbox{the improper integral }\int_{-\infty}^{b} f(x)dx\mbox{ converges to \(L\).}\]
Otherwise
\[\mbox{the improper integral }\int_{-\infty}^{b} f(x)dx\mbox{ diverges.}\]
Example. We have some interesting examples.
(i) We have
\begin{align*}\int_{0}^{\infty} e^{-2x}dx & =\lim_{b\rightarrow\infty}\int_{0}^{b}e^{-2x}dx=\lim_{b\rightarrow\infty}\left [-\frac{e^{-2x}}{2}
\right ]_{0}^{b}\\ & =\lim_{b\rightarrow\infty}\left (\frac{1}{2}-\frac{1}{e^{2b}}\right )=\frac{1}{2}.\end{align*}
(ii) We have
\[\int_{1}^{\infty}\frac{dx}{x}=\lim_{b\rightarrow\infty}\int_{1}^{b}\frac{dx}{x}=\lim_{b\rightarrow\infty}\ln b=\infty.\]
(iii) We have
\begin{align*} \int_{-\infty}^{1}\cos \pi xdx & =\lim_{a\rightarrow -\infty}\int_{a}^{1}
\cos \pi xdx=\lim_{a\rightarrow -\infty}\left [\frac{1}{\pi}\sin \pi x
\right ]_{a}^{1}\\ & =\lim_{a\rightarrow -\infty}\left (-\frac{1}{\pi}\sin \pi a\right ).\end{align*}
As \(a\) tends to \(-\infty\), we see that \(\sin \pi a\) oscillates between \(-1\) and \(1\). Therefore, the integral oscillates between \(1/\pi\) and \(-1/\pi\) and does not converge. $\sharp$
(Comparison Test). Suppose that \(f\) and \(g\) are continuous and \(0\leq f(x) \leq g(x)\) for all \(x\in [a,\infty )\).
\[\mbox{If }\int_{a}^{\infty} g(x)dx\mbox{ converges, then }\int_{a}^{\infty}f(x)dx\mbox{ converges.}\]
\[\mbox{If }\int_{a}^{\infty} f(x)dx\mbox{ diverges, then }\int_{a}^{\infty}g(x)dx\mbox{ diverges.}\]
Example. We have some interesting examples.
(i) The improper integral
\[\int_{1}^{\infty} \frac{dx}{\sqrt{1+x^{3}}}\]
is convergent since
\[\frac{1}{\sqrt{1+x^{3}}}<\frac{1}{x^{3/2}}\mbox{ for }x\in [1,\infty )
\mbox{ and }\int_{1}^{\infty}\frac{dx}{x^{3/2}}\mbox{ is convergent.}\]
In contrast, if we try to evaluate
\[\lim_{b\rightarrow\infty} \int_{1}^{b}\frac{dx}{\sqrt{1+x^{3}}}\]
directly, we would have to calculate the indefinite integral $\int\frac{dx}{\sqrt{1+x^{3}}}$ and this cannot be done by any of the methods we have developed so far.
(ii) The improper integral
\[\int_{1}^{\infty}\frac{dx}{\sqrt{1+x^{2}}}\]
is divergent since
\[\frac{1}{1+x}\leq\frac{1}{\sqrt{1+x^{2}}}\mbox{ for }x\in [1,\infty )
\mbox{ and }\int_{1}^{\infty}\frac{dx}{1+x}\mbox{ is divergent.}\]
Suppose now that \(f\) is continuous on \((-\infty ,\infty )\). The improper integral
\[\int_{-\infty}^{\infty} f(x)dx\]
is said to be convergent when
\[\int_{-\infty}^{0} f(x)dx\mbox{ and }\int_{0}^{\infty} f(x)dx\]
are both convergent. In this case, we set
\[\int_{-\infty}^{\infty} f(x)dx=L+M,\]
where
\[\int_{-\infty}^{0} f(x)dx=L\mbox{ and }\int_{0}^{\infty} f(x)dx=M.\]
Example. Find the improper integral
\[\int_{-\infty}^{\infty}\frac{e^{x}}{1+e^{2x}}dx.\]
We first consider the indefinite integral
\[\int\frac{e^{x}}{1+e^{2x}}dx.\]
Let \(u=e^{x}\) and \(du=e^{x}dx\). Then, we have
\[\int\frac{e^{x}}{1+e^{2x}}dx=\int\frac{du}{1+u^{2}}=\tan^{-1}u+C=\tan^{-1}(e^{x})+C.\]
Now, we obtain
\begin{align*} \int_{-\infty}^{0}\frac{e^{x}}{1+e^{2x}}dx & =\lim_{a\rightarrow -\infty}
\int_{a}^{0}\frac{e^{x}}{1+e^{2x}}dx=\lim_{a\rightarrow -\infty}\left [
\tan^{-1}(e^{x})\right ]_{a}^{0}\\ & =\tan^{-1}1-\lim_{a\rightarrow -\infty}\tan^{-1}(e^{a})=\frac{\pi}{4}\end{align*}
and
\begin{align*}
\int^{\infty}_{0}\frac{e^{x}}{1+e^{2x}}dx & =\lim_{b\rightarrow \infty}\int_{0}^{b}\frac{e^{x}}{1+e^{2x}}dx=\lim_{b\rightarrow\infty}\left [\tan^{-1}(e^{x})\right ]_{0}^{b}\\
& =\lim_{b\rightarrow\infty}\tan^{-1}(e^{b})-\tan^{-1}1=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4},
\end{align*}
which implies
\[\int_{-\infty}^{\infty}\frac{e^{x}}{1+e^{2x}}dx=\frac{\pi}{2}.\]
Improper integrals can also arise on bounded intervals. Suppose that \(f\) is continuous on the half-open interval \([a,b)\) and unbounded at \(b\). For each number \(c<b\), we can form the definite integral \(\int_{a}^{c}f(x)dx\). When
\[\lim_{c\rightarrow b^{-}}\int_{a}^{c} f(x)dx=L\]
exists, we say that
\[\mbox{the improper intergral }\int_{a}^{b-}f(x)dx\mbox{ converges to \(L\).}\]
Otherwise, the improper integral is divergent. In a similar manner, if \(f\) is continuous on \((a,b]\) and unbounded at \(a\), then we consider the limit
\[\lim_{c\rightarrow a^{+}}\int_{c}^{b} f(x)dx.\]
When this limit exists and has the value \(L\), we say that
\[\mbox{the improper intergral }\int_{a+}^{b}f(x)dx\mbox{ converges to \(L\).}\]
Otherwise, the improper integral is divergent.
Example. We have some interesting examples.
(i) We have
\begin{align*} \int_{0}^{1-} (1-x)^{-2/3}dx & =\lim_{c\rightarrow 1^{-}}\int_{0}^{c}
(1-x)^{-2/3}dx=\lim_{c\rightarrow 1^{-}}\left [-3(1-x)^{1/3}\right ]_{0}^{c}\\ & =
\lim_{c\rightarrow 1^{-}}[-3(1-c)^{1/3}+3]=3.\end{align*}
(ii) We have
\[\int_{0+}^{2}\frac{dx}{x}=\lim_{c\rightarrow 0^{+}}\int_{c}^{2}\frac{dx}{x}=\lim_{c\rightarrow 0^{+}}[\ln 2-\ln c]=\infty.\]
Now, suppose that \(f\) is continuous on an interval \([a,b]\) except at some point \(c\) in \((a,b)\) where \(f(x)\rightarrow\pm\infty\) as $x\rightarrow c^{-}$ or as \(x\rightarrow c^{+}\). We say that the improper integral
\[\int_{a}^{b} f(x)dx\]
is convergent when both of the integrals
\[\int_{a}^{c-}f(x)dx\mbox{ and }\int_{c+}^{b} f(x)dx\]
are convergent. When we have
\[\int_{a}^{c-}f(x)dx=L\mbox{ and }\int_{c+}^{b} f(x)dx=M.\]
Then, we obtain
\[\int_{a}^{b} f(x)dx=L+M.\]
Example. To evaluate
\[\int_{1}^{4}\frac{dx}{(x-2)^{2}},\]
we need to calculate
\[\lim_{c\rightarrow 2^{-}}\int_{1}^{c}\frac{dx}{(x-2)^{2}}\mbox{ and }
\lim_{c\rightarrow 2^{+}}\int_{c}^{4}\frac{dx}{(x-2)^{2}}.\]
Neither of these limits exists. Therefore, the above improper integral is divergent. Notice that, if we ignore the fact that the above integral is improper, then we are led to the incorrect conclusion given by
\[\int_{1}^{4}\frac{dx}{(x-2)^{2}}=\left [\frac{-1}{x-2}\right ]_{1}^{4}=-\frac{3}{2}.\]
Example. Evaluate
\[\int_{-2}^{1}\frac{dx}{x^{4/5}},\]
we need to calculate
\begin{align*} \lim_{c\rightarrow 0^{-}}\int_{-2}^{c}\frac{dx}{x^{4/5}} & =
\lim_{c\rightarrow 0^{-}}\left [5x^{1/5}\right ]_{-2}^{c}\\ & =
\lim_{c\rightarrow 0^{-}}[5c^{1/5}-5(-2)^{1/5}]=5\cdot 2^{1/5}\end{align*}
and
\begin{align*} \lim_{c\rightarrow 0^{+}}\int_{c}^{1}\frac{dx}{x^{4/5}} & =
\lim_{c\rightarrow 0^{+}}\left [5x^{1/5}\right ]_{c}^{1}\\ & =
\lim_{c\rightarrow 0^{+}}[5-5c^{1/5}]=5.\end{align*}
Therefore, we obtain
\[\int_{-2}^{1}\frac{dx}{x^{4/5}}=5+5\sqrt[5]{2}.\]
