When the integrand contains $\sqrt{a^2+x^2}$, $\sqrt{a^2-x^2}$ or $\sqrt{x^2-a^2}$ and the integral cannot be evaluated by a simple substitution, the following trigonometric substitutions may come in handy:

1. For $\sqrt{a^2+x^2}$, use $x=a\tan\theta$ with $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$. Accordingly $dx=a\sec^2\theta d\theta$ and $a^2+x^2=a^2\sec^2\theta$.
2. For $\sqrt{a^2-x^2}$, use $x=a\sin\theta$ with $-\frac{\pi}{2}\leq x\leq\frac{\pi}{2}$. Accordingly $dx=a\cos\theta d\theta$ and $a^2-x^2=a^2\cos^2\theta$. Note: You can also use the substitution $x=a\cos\theta$ with $0\leq\theta\leq\pi$. Accordingly $dx=-a\sin\theta d\theta$ and $a^2-x^2=a^2\sin^2\theta$.
3. For $\sqrt{x^2-a^2}$, use $x=a\sec\theta$ with $0\leq\theta<\frac{\pi}{2}$ or $\pi\leq\theta<\frac{3\pi}{2}$. Accordingly $dx=a\sec\theta\tan\theta d\theta$ and $x^2-a^2=a^2\tan^2\theta$.

Example. Evaluate $\int\frac{dx}{\sqrt{4+x^2}}$.

Solution. Let $x=2\tan\theta$. Then $dx=2\sec^2\theta d\theta$ and $4+x^2=4\sec^2\theta$. So \begin{align*}\int\frac{dx}{\sqrt{4+x^2}}&=\int\frac{2\sec^2\theta d\theta}{2|\sec\theta|}\\&=\int\sec\theta d\theta\ \left(\sec\theta>0\ \mbox{for}\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)\\&=\ln|\sec\theta+\tan\theta|+C’\end{align*} SInce $\tan\theta=\frac{x}{2}$ one may consider the triangle

Hence $$\int\frac{dx}{\sqrt{4+x^2}}=\ln|\sqrt{4+x^2}+x|+C$$ where $C=C’-\ln 2$.

Example. Evaluate $\int\frac{x^2}{\sqrt{9-x^2}}dx$.

Solution. Let $x=3\sin\theta$. Then $dx=3\cos\theta d\theta$ and $9-x^2=9\cos^2\theta$. \begin{align*}\int\frac{x^2}{\sqrt{9-x^2}}dx&=\int\frac{9\sin^2\theta(3\cos\theta d\theta)}{3|\cos\theta|}\\&=9\int\sin^2\theta d\theta\ \left(\cos\theta>0\ \mbox{for}\ -\frac{\pi}{2}<\theta<\frac{\pi}{2}\right)\\&=9\int\frac{1-\cos 2\theta}{2}d\theta\\&=\frac{9}{2}(\theta-\sin\theta\cos\theta)+C\\&=\frac{9}{2}\left(\sin^{-1}\frac{x}{3}-\frac{x}{3}\frac{\sqrt{9-x^2}}{3}\right)+C\\&=\frac{9}{2}\sin^{-1}\frac{x}{3}-\frac{x}{2}\sqrt{9-x^2}+C\end{align*}

Example. Evaluate $\int\frac{dx}{\sqrt{25x^2-4}}$.

Solution. $\int\frac{dx}{\sqrt{25x^2-4}}=\frac{1}{5}\int\frac{dx}{\sqrt{x^2-\left(\frac{2}{5}\right)^2}}$. Let $x=\frac{2}{5}\sec\theta$. Then $dx=\frac{2}{5}\sec\theta\tan\theta d\theta$ and $x^2-\left(\frac{2}{5}\right)^2=\frac{4}{25}\tan^2\theta$. \begin{align*}\int\frac{dx}{\sqrt{25x^2-4}}&=\frac{1}{5}\int\frac{dx}{\sqrt{x^2-\left(\frac{2}{5}\right)^2}}\\&=\frac{1}{5}\int\frac{\frac{2}{5}\sec\theta\tan\theta d\theta}{5\left(\frac{2}{5}\right)\tan\theta}\ \left(\tan\theta>0\ \mbox{for}\ 0<\theta<\frac{\pi}{2}\right)\\&=\frac{1}{5}\int\sec\theta d\theta\\&=\frac{1}{5}\ln|\sec\theta +\tan\theta|+C’\\&=\frac{1}{5}\ln|5x+\sqrt{25x^2-4}|+C\end{align*} where $C=C’-\frac{\ln 2}{5}$.

Example. Evaluate $\int\sqrt{1+x^2}dx$.

Solution. Let $x=\tan\theta$. Then $dx=\sec^2\theta d\theta$ and $1+x^2=\sec^2\theta$. \begin{align*}\int\sqrt{1+x^2}dx&=\int\sec^3\theta d\theta\\&=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C\\&=\frac{1}{2}x\sqrt{1+x^2}+\frac{1}{2}\ln|x+\sqrt{1+x^2}|+C\end{align*} The integration $$\int\sec^3\theta d\theta=\frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln|\sec\theta+\tan\theta|+C$$ was discussed here.

Example. Evaluate $\int\frac{\sqrt{1+x^2}}{x}dx$.

Solution. Let $x=\tan\theta$. Then $dx=\sec^2\theta d\theta$ and $1+x^2=\sec^2\theta$. \begin{align*}\int\frac{\sqrt{1+x^2}}{x}dx&=\int\frac{\sec^3\theta}{\tan\theta}d\theta\\&=\int\frac{\sec\theta\sec^2\theta}{\tan\theta}d\theta\\&=\int\frac{\sec\theta(1+\tan^2\theta)}{\tan\theta}d\theta\\&=\int\frac{\sec\theta}{\tan\theta}d\theta+\int\sec\theta\tan\theta d\theta\\&=\int\csc\theta d\theta+\int\sec\theta\tan\theta d\theta\\&=-\ln|\csc\theta+\cot\theta|+\sec\theta+C\\&=-\ln\left|\frac{\sqrt{1+x^2}}{x}+\frac{1}{x}\right|+\sqrt{1+x^2}+C\end{align*}

Example. Find the area enclosed by the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

Solution. The ellipse in the first quadrant is given by the function $y=\frac{b}{a}\sqrt{a^2-x^2}$, $0\leq x\leq a$. So the area is \begin{align*}A&=4\int_0^a\frac{b}{a}\sqrt{a^2-x^2}dx\\&=4\frac{b}{a}\int_0^{\frac{\pi}{2}}a\cos\theta(a\cos\theta d\theta)\ (x=a\sin\theta$)\\&=4ab\int_0^{\frac{\pi}{2}}\cos^2\theta d\theta\\&=4ab\int_0^{\frac{\pi}{2}}\frac{1+\cos 2\theta}{2}d\theta\\&=\pi ab\end{align*}

If two functions $f(x)$ and $g(x)$ both approach zero as $x\to a$, the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{0}{0}$ at $x=a$. Regardless of the term, $\frac{f(x)}{g(x)}$ may approach a limit as $x$ approaches $a$. The following theorem named after the French mathematician G. F. A. de L’Hôpital (1661-1704) is useful for the process of determining this limit if it exists.

Theorem (L’Hôpital’s Rule). If the functions $f(x)$ and $g(x)$ are continuous in an interval containing $a$ and if $f'(x)$ and $g'(x)$ exist such that $g'(x)\ne 0$ in this interval (except possibly at $x=a$), then when $\lim_{x\to a}f(x)=\lim_{x\to 0}g(x)=0$ (or equivalently $f(a)=g(a)=0$), we have $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$$ provided the limit on the right exists.

Example. Evaluate $\lim_{x\to 0}\frac{\tan x}{x}$.

Solution. The limit is an indeterminate form $\frac{0}{0}$. By applying L’Hôpital’s Rule, we obtain $$\lim_{x\to 0}\frac{\tan x}{x}=\lim_{x\to 0}\frac{\sec^2x}{1}=1$$

The Indeterminate form $\frac{\infty}{\infty}$

If $f(x)\to\infty$ and  $g(x)\to \infty$ as $x\to a$ (or $x\to\pm\infty$), the function $\frac{f(x)}{g(x)}$ is said to assume the indeterminate form $\frac{\infty}{\infty}$ at $x=a$ (or at $x=\pm\infty$). The limit of $\frac{\infty}{\infty}$ as $x\to a$ (or $x\to\pm\infty$) may still be found by L’Hôpital’s Rule if it exists.

Example. $$\lim_{x\to\infty}\frac{x^2}{e^x}=\lim_{x\to\infty}\frac{2x}{e^2}=\lim_{x\to\infty}\frac{2}{e^x}=0$$

Example. Evaluate $\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}$.

Solution. Applying L’Hôpital’s Rule, we have $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x^2}$$ While L’Hôpital’s rule can still be applied, we would accomplish nothing by doing so. This example shows that L’Hôpital’s rule may not necessarily leads to a desirable result. For the above example, the limit can be found by $$\lim_{x\to 0+}\frac{e^{-\frac{1}{x}}}{x}=\lim_{x\to 0+}\frac{\frac{1}{x}}{e^{\frac{1}{x}}}=\lim_{z\to\infty}\frac{z}{e^z}=\lim_{z\to\infty}\frac{1}{e^z}=0$$

The indeterminate form $0\cdot\infty$

If $f(x)\to 0$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the function $f(x)g(x)$ is said to assume the indeterminate form $0\cdot\infty$ at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)g(x)$ as $x\to a$ (or $x\to\pm\infty$) exists, it may be found by writing $f(x)g(x)$ as $$\frac{f(x)}{\frac{1}{g(x)}}\ \mbox{or}\ \frac{g(x)}{\frac{1}{f(x)}}$$ and applying L’Hôpital’s rule.

Example.

1. $\lim_{x\to\infty}xe^{-x}=\lim_{x\to\infty}\frac{x}{e^x}=\lim_{x\to\infty}\frac{1}{e^x}=0$.
2. $\lim_{x\to 0}\sin 3x\cot 2x=\lim_{x\to 0}\frac{\sin 3x}{\tan 2x}=\lim_{x\to 0}\frac{3\cos 3x}{2\sec^22x}=\frac{3}{2}$.

The Indeterminate Form $\infty-\infty$

If $f(x)\to\infty$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), the difference $f(x)-g(x)$ is said to assume the indeterminate form $\infty-\infty$  at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)-g(x)$ as $x\to a$ (or as $x\to\pm\infty$) exists, it may be found by transforming the difference into a fraction by algebraic means and applying L’Hôpital’s rule.

Example. $\lim_{x\to 0}(\csc x-\cot x)=\lim_{x\to 0}\frac{1-\cos x}{\sin x}=\lim_{x\to 0}\frac{\sin x}{\cos x}=0$

Example. Evaluate $\lim_{x\to\infty}(x-\ln x)$.

Solution. While $x-\ln x$ assumes the indeterminate form $\infty-\infty$ at $x=\infty$, there is no algebraic means to transform it to a fraction. An indeterminate of the form $\infty-\infty$ such as the one in consideration may be evaluated by finding the limit of its exponential. Let $y=x-\ln x$. Then $$e^y=e^{x-\ln x}=\frac{e^x}{e^{\ln x}}=\frac{e^x}{x}$$ Hence, $$\lim_{x\to\infty}e^y=\lim_{x\to\infty}\frac{e^x}{x}=\lim_{x\to\infty}\frac{e^x}{1}=\infty$$ Since $y\to\infty$ when $e^y\to\infty$, $$\lim_{x\to\infty}(x-\ln x)=\infty$$

The Indeterminate Forms $0^0$, $\infty^0$, $1^\infty$

If $f(x)\to 0$ and $g(x)\to 0$, or $f(x)\to\infty$ and $g(x)\to 0$, or $f(x)\to 1$ and $g(x)\to\infty$ as $x\to a$ (or $x\to\pm\infty$), $f(x)^{g(x)}$ is said to assume the indeterminate form $0^0$, $\infty^0$, or $1^\infty$, respectively at $x=a$ (or at $x=\pm\infty$). If the limit of $f(x)^{g(x)}$ exists as $x\to a$ (or $x\to\pm\infty$), it may be found by denoting $f(x)^{g(x)}$ by $y$ and investigating the limit approached by the logarithm $$\ln y=g(x)\ln f(x)$$ If $\lim_{x\to a}\ln y=k$, then $\lim_{x\to a}y=e^k$.

Example. Evaluate $\lim_{x\to 0}x^x$.

Solution. $x^x$ assumes the indeterminate form $0^0$ at $x=0$. Let $y=x^x$. Then $\ln y=x\ln x$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln x}{\frac{1}{x}}=\lim_{x\to 0}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\to 0}(-x)=0$$ Hence $\lim_{x\to 0}y=\lim_{x\to 0}x^x=e^0=1$.

Example. Evaluate $\lim_{x\to 0}(1-\sin x)^{\frac{1}{x}}$.

Solution. The function assumes the indeterminate form $1^\infty$ at $x=0$. Let $y=(1-\sin x)^{\frac{1}{x}}$. Then $\ln y=\frac{\ln(1-\sin x)}{x}$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln(1-\sin x)}{x}=\lim_{x\to 0}\frac{\frac{-\cos x}{1-\sin x}}{1}=-1$$ Hence, $\lim_{x\to 0}(1-\sin x)^{\frac{1}{x}}=e^{-1}=\frac{1}{e}$.

Example. Use L’Hôpital’s rule to show the limit $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Proof. Let $y=(1+x)^{\frac{1}{x}}$. Then $y$ assumes the indeterminate form $1^\infty$ at $x=0$. $\ln y=\frac{\ln(1+x)}{x}$ and $$\lim_{x\to 0}\ln y=\lim_{x\to 0}\frac{\ln(1+x)}{x}=\lim_{x\to 0}\frac{1}{1+x}=1$$ Hence, $$\lim_{x\to 0}(1+x)^{\frac{1}{x}}=e$$

Suppose that $f$ is differentiable infinitely many times on an open interval containing $a$. Then $$\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f^{\prime\prime}(a)}{2!}(x-a)^2+\frac{f^{\prime\prime\prime}(a)}{3!}(x-a)^3+\cdots$$ is called the Taylor series of $f$ centered at $a$. A Taylor series centered at 0 is called a Maclaurin series.

Example. Find the Maclaurin series of $f(x)=e^x$ and its radius of convergence $R$.

Solution. The Maclaurin series is $\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}x^n=\sum_{n=0}^\infty\frac{x^n}{n!}$. By the ratio test $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\lim_{n\to\infty}\frac{|x|}{n+1}=0<1$$ for all $x$ so the series converges for all $x$ i.e. the radius of convergence is $R=\infty$.

The following theorem tells when a function $f(x)$ can be represented by its Taylor series. Recall that $f(x)=T_n(x)+R_n(x)$ where $T_n(x)$ is the $n$-th degree Taylor polynomial of $f$ at $a$ and $R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$ where $a<\xi<x$ or $x<\xi<a$.

Theorem. $f=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$ for $|x-a|<R$ if and only if $\lim_{n\to\infty}R_n(x)=0$ for $|x-a|<R$.

Since $\sum_{n=0}^\infty\frac{x^n}{n!}$ converges for all $x$, \begin{equation}\label{eq:explim}\lim_{n\to\infty}\frac{x^n}{n!}=0\end{equation} Now we show that \begin{equation}\label{eq:expx}e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\end{equation} for all $x$. $R_n(x)=\frac{e^\xi}{(n+1)!}x^{n+1}$ where $0<\xi<x$ or $x<\xi<0$. If $0<\xi<x$ then $0<R_n(x)<\frac{e^x}{(n+1)!}x^{n+1}\to 0$ as $n\to\infty$ by \eqref{eq:explim}. If $x<\xi<0$ then $0\leq |R_n(x)|<\frac{|x|^{n+1}}{(n+1)!}\to 0$ as $n\to\infty$ again by \eqref{eq:explim}. This completes the proof of \eqref{eq:expx}. For $x=1$ we have a definition of the Euler number $e$ in terms of a series as \begin{equation}\label{eq:eulernum}e=\sum_{n=0}^\infty\frac{1}{n!}=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots\end{equation}

Example. Find the Taylor series of $f(x)=e^x$ at $a=2$.

Solution. By the same manner, one can show that $$e^x=\sum_{n=0}^\infty\frac{e^2}{n!}(x-2)^n$$ for all $x$.

Example. Find the Maclaurin series of $\sin x$ and show that it represents $\sin x$ for all $x$.

Solution. The Maclaurin series is $\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}$ and by the ratio test one can show that it converges for all $x$. Since $|f^{(n+1)}(\xi)|\leq 1$, $0\leq |R_n(x)|\leq\frac{|x|^{n+1}}{(n+1)!}\to 0$ as $n\to\infty$. Hence, \begin{equation}\label{eq:sinx}\sin x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\end{equation} for all $x$.

Similarly one can also show that \begin{equation}\label{eq:cosx}\cos x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}\end{equation} for all $x$.

Euler’s Formula

If $x$ is replaced by $ix$ in \eqref{eq:expx}, we obtain \begin{align*}e^{ix}&=\sum_{n=0}^\infty\frac{(ix)^n}{n!}\\&=\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}+i\sum_{n=0}^\infty\frac{x^{2n+1}}{(2n+1)!}\\&=\cos x+i\sin x\end{align*}\begin{equation}\label{eq:eulerformula}e^{ix}=\cos x+i\sin x\end{equation} is called the Euler’s Formula and it represents the point on the unit circle centered at the origin corresponding to the angle $x$. The formula in \eqref{eq:eulerformula} comes in handy in so many places of mathematics from trigonometry, calculus, differential equations to abstract algebra, topology, geometry to name a few. It is also a useful tool in physics on many occasions. For example, one can derive the sine sum and the cosine sum formulas using \eqref{eq:eulerformula}: $e^{i(\theta_1+\theta_2)}=\cos(\theta_1+\theta_2)+i\sin(\theta_1+\theta_2)$.  On the other hand, \begin{align*}e^{i(\theta_1+\theta_2)}&=e^{i\theta_1}e^{i\theta_2}\\&=(\cos\theta_1+i\sin\theta_1)(\cos\theta_2+i\sin\theta_2)\\&=(\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2)+i(\cos\theta_1\sin\theta_2+\sin\theta_1\cos\theta_2)\end{align*} Hence we obtain the sine and the cosine sum formulas \begin{align*}\sin(\theta_1+\theta_2)&=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\\\cos(\theta_1+\theta_2)&=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\end{align*} When $\theta=\pi$, we obtain the so-called the Euler’s identity \begin{equation}\label{eq:euleridentity}e^{\pi i}+1=0\end{equation} which contains the most fundamental constants of mathematics $\pi$, $e$, $i$, 0, and 1.

The Binomial Series

Let us find the Maclaurin series of $f(x)=(1+x)^k$ where $k$ is any real number. First, we find $$f^{(n)}(x)=k(k-1)\cdots(k-n+1)(1+x)^{k-n}$$ and so $$f^{(n)}(0)=k(k-1)\cdots(k-n+1)$$ Hence the Maclaurin series is given by $$\sum_{n=0}^\infty\frac{k(k-1)\cdots(k-n+1)}{n!}x^n$$ The coefficients $\frac{k(k-1)\cdots(k-n+1)}{n!}$, $n=0,1,2,\cdots$ are denoted by ${}_k\mathrm{C}_n$ or $\begin{pmatrix}k\\n\end{pmatrix}$ and are called binomial coefficients. $$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{k-n}{n+1}\right||x|=\lim_{n\to\infty} |x|$$ So by the ratio test if $|x|<1$ then the series converges and if $|x|>1$ then it diverges. Furthermore it can be shown (we omit the proof) that \begin{equation}\label{eq:binomialseries}(1+x)^k=\sum_{n=0}^\infty\frac{k(k-1)\cdots(k-n+1)}{n!}x^n\end{equation} for $|x|<1$. The series in \eqref{eq:binomialseries} is called the binomial series.

Example. FInd the Maclaurin series of $f(x)=\frac{1}{\sqrt{4-x}}$.

Solution. \begin{align*}\frac{1}{\sqrt{4-x}}&=\frac{1}{2}\frac{1}{\sqrt{1-\frac{x}{4}}}\\&=\frac{1}{2}\left(1-\frac{x}{4}\right)^{-\frac{1}{2}}\\&=\frac{1}{2}\sum_{n=0}^\infty\begin{pmatrix}-\frac{1}{2}\\n\end{pmatrix}\left(-\frac{x}{4}\right)^n\\&=\frac{1}{2}\left\{1+\left(-\frac{1}{2}\right)\left(-\frac{x}{4}\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}\left(-\frac{x}{4}\right)^2+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}\left(-\frac{x}{4}\right)^3+\cdots\\+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{2n-1}{2}\right)}{n!}\left(-\frac{x}{4}\right)^n+\cdots\right\}\\&=\frac{1}{2}\left\{1+\frac{x}{8}+\frac{1\cdot 3}{2!8^2}x^2+\frac{1\cdot\cdot 5}{3!8^3}x^3+\cdots+\frac{1\cdot 3\cdot 5\cdots(2n-1)}{n!8^n}x^n+\cdots\right\}\end{align*} This series converges if $\left|-\frac{x}{4}\right|<1$ i.e. $|x|<4$. The radius of convergence is 4.

Working with Taylor Series

Example. Use Taylor series to evaluate $\lim_{x\to 0}\frac{x^2+2\cos x-2}{3x^4}$.

Solution. \begin{align*}\lim_{x\to 0}\frac{x^2+2\cos x-2}{3x^4}&=\lim_{x\to 0}\frac{x^2+2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)-2}{3x^4}\\&=\lim_{x\to 0}\frac{2\left(\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)}{3x^4}\\&=\frac{2}{4!3}=\frac{1}{36}\end{align*}

Remark. The above limit can be also calculated using the L’Hôpital’s rule as it is an indeterminate form of type $\frac{0}{0}$.

Example. Approximate $\int_0^1 e^{-x^2}dx$ with an error no greater than $5\times 10^{-4}$.

Solution. \begin{align*}\int_0^1 e^{-x^2}dx&=\int_0^1\left(1-x^2+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots+\frac{(-1)^nx^{2n}}{n!}+\cdots\right)dx\\&=\left[\left(x-\frac{x^3}{3}+\frac{x^5}{2!5}-\frac{x^7}{3!7}+\cdots+\frac{(-1)^nx^{2n+1}}{n!(2n+1)}+\cdots\right)\right]_0^1\\&=1-\frac{1}{3}+\frac{1}{2!5}-\frac{1}{3!7}+\cdots+\frac{(-1)^n}{n!(2n+1)}+\cdots\end{align*} Recall that for an alternating series $\sum_{n=0}^\infty(-1)^n a_n$, $|R_n|\leq a_{n+1}$. For $n=4$, $|R_4|\leq a_5=\frac{1}{5!11}=7.6\times 10^{-3}>5\times 10^{-4}$. For $n=5$, $|R_5|\leq a_6=\frac{1}{6!13}=1.07\times 10^{-4}<5\times 10^{-4}$. Thus the error is less than $5\times 10^{-4}$ if $n\geq 5$. The approximation with $n=5$ is given by \begin{align*}\int_0^1 e^{-x^2}dx&\approx 1-\frac{1}{3}+\frac{1}{2!5}-\frac{1}{3!7}+\frac{1}{5!11}\\&\approx 0.747\end{align*}