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Higher Engineering Mathematics

John Bird

Chapter 8

Maclaurin's series - all with Video Answers

Educators

Chapter Questions

04:27

Problem 1

Determine the first four terms of the power series for $\cos x$
The values of $f(0), f^{\prime}(0), f^{\prime \prime}(0), \ldots$ in Maclaurin's series are obtained as follows:
$$
\begin{array}{cc}
f(x)=\cos x & f(0)=\cos 0=1 \\
f^{\prime}(x)=-\sin x & f^{\prime}(0)=-\sin 0=0 \\
f^{\prime \prime}(x)=-\cos x & f^{\prime \prime}(0)=-\cos 0=-1 \\
f^{\prime \prime \prime}(x)=\sin x & f^{\prime \prime \prime}(0)=\sin 0=0 \\
f^{i v}(x)=\cos x & f^{i v}(0)=\cos 0=1 \\
f^{v}(x)=-\sin x & f^{\mathrm{v}}(0)=-\sin 0=0 \\
f^{\mathrm{vi}}(x)=-\cos x & f^{\mathrm{vi}}(0)=-\cos 0=-1
\end{array}
$$
Substituting these values into equation (5) gives:
$$
\begin{aligned}
f(x)=\cos x=& 1+x(0)+\frac{x^{2}}{2 !}(-1)+\frac{x^{3}}{3 !}(0) \\
&+\frac{x^{4}}{4 !}(1)+\frac{x^{5}}{5 !}(0)+\frac{x^{6}}{6 !}(-1)+\cdots
\end{aligned}
$$
i.e. $\quad \cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\cdots$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
02:59

Problem 2

Determine the power series for $\cos 2 \theta$
Replacing $x$ with $2 \theta$ in the series obtained in Problem 1 gives:
$$
\begin{aligned}
\cos 2 \theta &=1-\frac{(2 \theta)^{2}}{2 !}+\frac{(2 \theta)^{4}}{4 !}-\frac{(2 \theta)^{6}}{6 !}+\cdots \\
&=1-\frac{4 \theta^{2}}{2}+\frac{16 \theta^{4}}{24}-\frac{64 \theta^{6}}{720}+\cdots
\end{aligned}
$$
i.e. $\cos 2 \theta=1-2 \theta^{2}+\frac{2}{3} \theta^{4}-\frac{4}{45} \theta^{6}+\cdots$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
03:56

Problem 3

Using Maclaurin's series, find the first 4 (non zero) terms for the function $f(x)=\sin x$
$$
\begin{array}{cc}
f(x)=\sin x & f(0)=\sin 0=0 \\
f^{\prime \prime}(x)=-\sin x \quad f^{\prime \prime}(0)=-\sin 0=0 \\
f^{\prime \prime \prime}(x)=-\cos x \quad f^{\prime \prime \prime}(0)=-\cos 0=-1 \\
f^{\mathrm{iv}}(x)=\sin x \quad & f^{\mathrm{i}}(0)=\sin 0=0 \\
f^{\mathrm{v}}(x)=\cos x & f^{\mathrm{v}}(0)=\cos 0=1 \\
f^{\mathrm{vi}}(x)=-\sin x & f^{\mathrm{vi}}(0)=-\sin 0=0 \\
f^{\mathrm{vii}}(x)=-\cos x & f^{\mathrm{vii}}(0)=-\cos 0=-1
\end{array}
$$
Substituting the above values into Maclaurin's series of equation (5) gives:
$$
\begin{aligned}
\sin x=0+x(1)+\frac{x^{2}}{2 !}(0)+\frac{x^{3}}{3 !}(-1)+\frac{x^{4}}{4 !}(0) \\
+\frac{x^{5}}{5 !}(1)+\frac{x^{6}}{6 !}(0)+\frac{x^{7}}{7 !}(-1)+\cdots \\
\text { i.e. } \sin x=x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\frac{x^{7}}{7 !}+\cdots
\end{aligned}
$$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
03:38

Problem 4

Using Maclaurin's series, find the first five terms for the expansion of the function $f(x)=\mathrm{e}^{3 x}$
$$
\begin{array}{rlr}
f(x) & =\mathrm{e}^{3 x} & f(0)=\mathrm{e}^{0}=1 \\
f^{\prime}(x) & =3 \mathrm{e}^{3 x} & f^{\prime}(0)=3 \mathrm{e}^{0}=3 \\
f^{\prime \prime}(x) & =9 \mathrm{e}^{3 x} & f^{\prime \prime}(0)=9 \mathrm{e}^{0}=9 \\
f^{\prime \prime \prime}(x) & =27 \mathrm{e}^{3 x} & f^{\prime \prime \prime}(0)=27 \mathrm{e}^{0}=27 \\
f^{\mathrm{iv}}(x) & =81 \mathrm{e}^{3 x} & f^{\mathrm{iv}}(0)=81 \mathrm{e}^{0}=81
\end{array}
$$
Substituting the above values into Maclaurin's series of equation (5) gives:
$$
\begin{aligned}
&\mathrm{e}^{3 x}=1+x(3)+\frac{x^{2}}{2 !}(9)+\frac{x^{3}}{3 !}(27) \\
&+\frac{x^{4}}{4 !}(81)+\cdots \\
&\mathrm{e}^{3 x}=1+3 x+\frac{9 x^{2}}{2 !}+\frac{27 x^{3}}{3 !}+\frac{81 x^{4}}{4 !}+\cdots
\end{aligned}
$$ \text { i.e. } \mathrm{e}^{3 x}=1+3 x+\frac{9 x^{2}}{2}+\frac{9 x^{3}}{2}+\frac{27 x^{4}}{8}+\cdots

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
04:05

Problem 5

Determine the power series for $\tan x$ as far as the term in $x^{3}$
$$
\begin{aligned}
f(x) &=\tan x \\
f(0) &=\tan 0=0 \\
f^{\prime}(x) &=\sec ^{2} x \\
f^{\prime}(0) &=\sec ^{2} 0=\frac{1}{\cos ^{2} 0}=1 \\
f^{\prime \prime}(x)=&(2 \sec x)(\sec x \tan x) \\
=& 2 \sec ^{2} x \tan x \\
f^{\prime \prime}(0)=& 2 \sec ^{2} 0 \tan 0=0 \\
f^{\prime \prime \prime}(x)=&\left(2 \sec ^{2} x\right)\left(\sec ^{2} x\right) \\
&+(\tan x)(4 \sec x \sec x \tan x), \text { by the } \\
\text { product rule, } \\
=& 2 \sec ^{4} x+4 \sec ^{2} x \tan ^{2} x \\
f^{\prime \prime \prime}(0) &=2 \sec ^{4} 0+4 \sec ^{2} 0 \tan ^{2} 0=2
\end{aligned}
$$
Substituting these values into equation (5) gives:
$$
f(x)=\tan x=0+(x)(1)+\frac{x^{2}}{2 !}(0)+\frac{x^{3}}{3 !}(2)
$$
i.e. $\tan x=x+\frac{1}{3} x^{3}$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
04:54

Problem 6

Expand $\ln (1+x)$ to five terms.
$$
\begin{array}{cc}
f(x)=\ln (1+x) & f(0)=\ln (1+0)=0 \\
f^{\prime}(x)=\frac{1}{(1+x)} \quad f^{\prime}(0)=\frac{1}{1+0}=1 \\
f^{\prime \prime}(x)=\frac{-1}{(1+x)^{2}} & f^{\prime \prime}(0)=\frac{-1}{(1+0)^{2}}=-1 \\
f^{\prime \prime \prime}(x)=\frac{2}{(1+x)^{3}} & f^{\prime \prime \prime}(0)=\frac{2}{(1+0)^{3}}=2 \\
f^{\text {iv }}(x)=\frac{-6}{(1+x)^{4}} \quad f^{\mathrm{iv}}(0)=\frac{-6}{(1+0)^{4}}=-6 \\
f^{\mathrm{v}}(x)=\frac{24}{(1+x)^{5}} \quad f^{\mathrm{v}}(0)=\frac{24}{(1+0)^{5}}=24
\end{array}
$$ Substituting these values into equation (5) gives:
$$
\begin{aligned}
f(x)=\ln (1+x)=& 0+x(1)+\frac{x^{2}}{2 !}(-1) \\
&+\frac{x^{3}}{3 !}(2)+\frac{x^{4}}{4 !}(-6)+\frac{x^{5}}{5 !}(24)
\end{aligned}
$$
1.e. $\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
01:30

Problem 7

Expand $\ln (1-x)$ to five terms.
Replacing $x$ by $-x$ in the series for $\ln (1+x)$ in Problem 6 gives:
$\ln (1-x)=(-x)-\frac{(-x)^{2}}{2}+\frac{(-x)^{3}}{3}$ $-\frac{(-x)^{4}}{4}+\frac{(-x)^{5}}{5}-\cdots$ i.e. $\ln (1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\frac{x^{5}}{5}-\cdots$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
04:12

Problem 8

Determine the power series for $\ln \left(\frac{1+x}{1-x}\right)$
$\ln \left(\frac{1+x}{1-x}\right)=\ln (1+x)-\ln (1-x)$ by the laws of logarithms, and from Problems 6 and 7,
$$
\begin{aligned}
\ln \left(\frac{1+x}{1-x}\right)=&\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots\right) \\
&-\left(-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-\frac{x^{5}}{5}-\cdots\right) \\
=& 2 x+\frac{2}{3} x^{3}+\frac{2}{5} x^{5}+\cdots
\end{aligned}
$$
i.e. $\ln \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right)$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
03:13

Problem 9

Use Maclaurin's series to find the expansion of $(2+x)^{4}$
$$
\begin{array}{ll}
f(x)=(2+x)^{4} & f(0)=2^{4}=16 \\
f^{\prime}(x)=4(2+x)^{3} & f^{\prime}(0)=4(2)^{3}=32
\end{array}
$$ $$
\begin{array}{ll}
f^{\prime \prime}(x)=12(2+x)^{2} & f^{\prime \prime}(0)=12(2)^{2}=48 \\
f^{\prime \prime \prime}(x)=24(2+x)^{1} & f^{\prime \prime \prime}(0)=24(2)=48 \\
f^{\mathrm{iv}}(x)=24 & f^{\mathrm{iv}}(0)=24
\end{array}
$$
Substituting in equation (5) gives:
$$
\begin{aligned}
&(2+x)^{4} \\
&=f(0)+x f^{\prime}(0)+\frac{x^{2}}{2 !} f^{\prime \prime}(0)+\frac{x^{3}}{3 !} f^{\prime \prime \prime}(0)+\frac{x^{4}}{4 !} f^{\mathrm{iv}}(0) \\
&=16+(x)(32)+\frac{x^{2}}{2 !}(48)+\frac{x^{3}}{3 !}(48)+\frac{x^{4}}{4 !}(24) \\
&=16+32 x+24 x^{2}+8 x^{3}+x^{4}
\end{aligned}
$$
(This expression could have been obtained by applying the binomial theorem.)

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
02:53

Problem 10

Expand $\mathrm{e}^{\frac{x}{2}}$ as far as the term in $x^{4}$
$$
\begin{array}{cc}
f(x)=\mathrm{e}^{\frac{x}{2}} \quad f(0)=\mathrm{e}^{0}=1 \\
f^{\prime}(x)=\frac{1}{2} \mathrm{e}^{\frac{x}{2}} & f^{\prime}(0)=\frac{1}{2} \mathrm{e}^{0}=\frac{1}{2} \\
f^{\prime \prime}(x)=\frac{1}{4} \mathrm{e}^{\frac{x}{2}} & f^{\prime \prime}(0)=\frac{1}{4} \mathrm{e}^{0}=\frac{1}{4} \\
f^{\prime \prime \prime}(x)=\frac{1}{8} \mathrm{e}^{\frac{x}{2}} & f^{\prime \prime \prime}(0)=\frac{1}{8} \mathrm{e}^{0}=\frac{1}{8} \\
f^{\mathrm{iv}}(x)=\frac{1}{16} \mathrm{e}^{\frac{x}{2}} & f^{\mathrm{iv}}(0)=\frac{1}{16} \mathrm{e}^{0}=\frac{1}{16}
\end{array}
$$
Substituting in equation (5) gives:
$$
\begin{gathered}
\mathrm{e}^{\frac{x}{2}}=f(0)+x f^{\prime}(0)+\frac{x^{2}}{2 !} f^{\prime \prime}(0) \\
+\frac{x^{3}}{3 !} f^{\prime \prime \prime}(0)+\frac{x^{4}}{4 !} f^{\mathrm{iv}}(0)+\cdots \\
=1+(x)\left(\frac{1}{2}\right)+\frac{x^{2}}{2 !}\left(\frac{1}{4}\right)+\frac{x^{3}}{3 !}\left(\frac{1}{8}\right) \\
+\frac{x^{4}}{4 !}\left(\frac{1}{16}\right)+\cdots
\end{gathered}
$$
i.e. $e^{y}=1+\frac{1}{2} x+\frac{1}{8} x^{2}+\frac{1}{48} x^{3}+\frac{1}{384} x^{4}+\cdots$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
02:39

Problem 11

Develop a series for $\sinh x$ using Maclaurin's series.
$$
\begin{array}{rl}
f(x) & =\sinh x & f(0)=\sinh 0=\frac{e^{0}-e^{-0}}{2}=0 \\
f^{\prime}(x) & =\cosh x & f^{\prime}(0)=\cosh 0=\frac{e^{0}+e^{-0}}{2}=1 \\
f^{\prime \prime}(x) & =\sinh x & f^{\prime \prime}(0)=\sinh 0=0 \\
f^{\prime \prime \prime}(x) & =\cosh x & f^{\prime \prime \prime}(0)=\cosh 0=1 \\
f^{\text {iv }}(x) & =\sinh x & f^{\text {iv }}(0)=\sinh 0=0 \\
f^{v}(x) & =\cosh x & f^{v}(0)=\cosh 0=1
\end{array}
$$
Substituting in equation (5) gives:
$$
\begin{gathered}
\sinh x=f(0)+x f^{\prime}(0)+\frac{x^{2}}{2 !} f^{\prime \prime}(0)+\frac{x^{3}}{3 !} f^{\prime \prime \prime}(0) \\
+\frac{x^{4}}{4 !} f^{\mathrm{iv}}(0)+\frac{x^{5}}{5 !} f^{v}(0)+\cdots \\
=0+(x)(1)+\frac{x^{2}}{2 !}(0)+\frac{x^{3}}{3 !}(1)+\frac{x^{4}}{4 !}(0) \\
+\frac{x^{5}}{5 !}(1)+\cdots
\end{gathered}
$$
i.e. $\sinh x=x+\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}+\cdots$
(as shown in Chapter 16 )

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
01:53

Problem 12

Produce a power series for $\cos ^{2} 2 x$ as far as the term in $x^{6}$
From double angle formulae, $\cos 2 A=2 \cos ^{2} A-1$ (see Chapter 19).
from which, $\quad \cos ^{2} A=\frac{1}{2}(1+\cos 2 A)$ and $\quad \cos ^{2} 2 x=\frac{1}{2}(1+\cos 4 x)$
From Problem 1 ,
$$
\begin{aligned}
&\cos x=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !}-\frac{x^{6}}{6 !}+\cdots \\
&\text { hence } \quad \cos 4 x=1-\frac{(4 x)^{2}}{2 !}+\frac{(4 x)^{4}}{4 !}-\frac{(4 x)^{6}}{6 !}+\cdots \\
&=1-8 x^{2}+\frac{32}{3} x^{4}-\frac{256}{45} x^{6}+\cdots
\end{aligned}
$$
$$
=\frac{1}{2}\left(1+1-8 x^{2}+\frac{32}{3} x^{4}-\frac{256}{45} x^{6}+\cdots\right)
$$
i.e. $\cos ^{2} 2 x=1-4 x^{2}+\frac{16}{3} x^{4}-\frac{128}{45} x^{6}+\cdots$

Mirza Aslam Beig
Mirza Aslam Beig
Numerade Educator
04:22

Problem 13

Evaluate $\int_{0.1}^{0.4} 2 \mathrm{e}^{\sin \theta} \mathrm{d} \theta$, conrect to 3 significant figures.
A power series for $\mathrm{e}^{\sin \theta}$ is firstly obtained using Maclaurin's series.
$$
\begin{aligned}
&f(\theta)=\mathrm{e}^{\sin \theta} \quad f(0)=\mathrm{e}^{\sin 0}=\mathrm{e}^{0}=1 \\
&f^{\prime}(\theta)=\cos \theta \mathrm{e}^{\sin \theta} \quad f^{\prime}(0)=\cos 0 \mathrm{e}^{\sin 0}=(1) \mathrm{e}^{0}=1 \\
&f^{\prime \prime}(\theta)=(\cos \theta)\left(\cos \theta \mathrm{e}^{\sin \theta}\right)+\left(\mathrm{e}^{\sin \theta}\right)(-\sin \theta)
\end{aligned}
$$
by the product rule
$$
\begin{aligned}
=& \mathrm{e}^{\sin \theta}\left(\cos ^{2} \theta-\sin \theta\right) \\
f^{\prime \prime}(0)=& \mathrm{e}^{0}\left(\cos ^{2} 0-\sin 0\right)=1 \\
f^{\prime \prime \prime}(\theta)=&\left(\mathrm{e}^{\sin \theta}\right)[(2 \cos \theta(-\sin \theta)-\cos \theta)] \\
& \quad+\left(\cos ^{2} \theta-\sin \theta\right)\left(\cos \theta \mathrm{e}^{\sin \theta}\right) \\
&=\mathrm{e}^{\sin \theta} \cos \theta\left[-2 \sin \theta-1+\cos ^{2} \theta-\sin \theta\right] \\
f^{\prime \prime \prime}(0) &=\mathrm{e}^{0} \cos 0[(0-1+1-0)]=0
\end{aligned}
$$
Hence from equation (5):
$$
\begin{aligned}
\mathrm{e}^{\sin \theta} &=f(0)+\theta f^{\prime}(0)+\frac{\theta^{2}}{2 !} f^{\prime \prime}(0)+\frac{\theta^{3}}{3 !} f^{\prime \prime \prime}(0)+\cdots \\
&=1+\theta+\frac{\theta^{2}}{2}+0
\end{aligned}
$$
Thus $\int_{0.1}^{0.4} 2 \mathrm{e}^{\sin \theta} \mathrm{d} \theta=\int_{0.1}^{0.4} 2\left(1+\theta+\frac{\theta^{2}}{2}\right) \mathrm{d} \theta$
$$
\begin{aligned}
=& \int_{0.1}^{0.4}\left(2+2 \theta+\theta^{2}\right) \mathrm{d} \theta \\
=&\left[2 \theta+\frac{2 \theta^{2}}{2}+\frac{\theta^{3}}{3}\right]_{0.1}^{0.4} \\
=&\left(0.8+(0.4)^{2}+\frac{(0.4)^{3}}{3}\right) \\
& \quad-\left(0.2+(0.1)^{2}+\frac{(0.1)^{3}}{3}\right) \\
=& 0.98133-0.21033
\end{aligned}
$$
$=0.771$, correct to 3 significant figures.

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
04:04

Problem 14

Evaluate $\int_{0}^{1} \frac{\sin \theta}{\theta} \mathrm{d} \theta$ using Maclaurin's series, correct to 3 significant figures.
Let $f(\theta)=\sin \theta \quad f(0)=0$
$$
\begin{aligned}
f^{\prime}(\theta) &=\cos \theta & f^{\prime}(0)=1 \\
f^{\prime \prime}(\theta) &=-\sin \theta & f^{\prime \prime}(0)=0 \\
f^{m \prime}(\theta) &=-\cos \theta & f^{\prime \prime \prime}(0)=-1 \\
f^{i v}(\theta) &=\sin \theta & f^{\mathrm{iv}}(0)=0 \\
f^{\mathrm{v}}(\theta) &=\cos \theta & f^{\mathrm{v}}(0)=1
\end{aligned}
$$
Hence from equation $(5)$ :
$$
\begin{gathered}
\sin \theta=f(0)+\theta f^{\prime}(0)+\frac{\theta^{2}}{2 !} f^{\prime \prime}(0)+\frac{\theta^{3}}{3 !} f^{\prime \prime \prime}(0) \\
+\frac{\theta^{4}}{4 !} f^{\mathrm{iv}}(0)+\frac{\theta^{5}}{5 !} f^{\mathrm{v}}(0)+\cdots \\
=0+\theta(1)+\frac{\theta^{2}}{2 !}(0)+\frac{\theta^{3}}{3 !}(-1) \\
+\frac{\theta^{4}}{4 !}(0)+\frac{\theta^{5}}{5 !}(1)+\cdots
\end{gathered}
$$
i.e. $\sin \theta=\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}-\cdots$
Hence
$$
\begin{aligned}
&\int_{0}^{1} \frac{\sin \theta}{\theta} \mathrm{d} \theta \\
&\quad=\int_{0}^{1} \frac{\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}-\frac{\theta^{7}}{7 !}+\cdots\right)}{\theta} \mathrm{d} \theta \\
&\quad=\int_{0}^{1}\left(1-\frac{\theta^{2}}{6}+\frac{\theta^{4}}{120}-\frac{\theta^{6}}{5040}+\cdots\right) \mathrm{d} \theta \\
&\quad=\left[\theta-\frac{\theta^{3}}{18}+\frac{\theta^{5}}{600}-\frac{\theta^{7}}{7(5040)}+\cdots\right]_{0}^{1} \\
&=1-\frac{1}{18}+\frac{1}{600}-\frac{1}{7(5040)}+\cdots
\end{aligned}
$$
$=0.946$, correct to 3 significant figures.

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
02:16

Problem 15

Evaluate $\int_{0}^{0.4} x \ln (1+x) \mathrm{d} x$ using Maclaurin's theorem, correct to 3 decimal places.
From Problem 6 ,
$$
\ln (1+x)=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots
$$
Hence $\int_{0}^{0.4} x \ln (1+x) \mathrm{d} x$
$$
\begin{aligned}
&=\int_{0}^{0.4} x\left(x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\frac{x^{5}}{5}-\cdots\right) \mathrm{d} x \\
&=\int_{0}^{0.4}\left(x^{2}-\frac{x^{3}}{2}+\frac{x^{4}}{3}-\frac{x^{5}}{4}+\frac{x^{6}}{5}-\cdots\right) \mathrm{d} x
\end{aligned}
$$
$$
\begin{aligned}
=&\left[\frac{x^{3}}{3}-\frac{x^{4}}{8}+\frac{x^{5}}{15}-\frac{x^{6}}{24}+\frac{x^{7}}{35}-\cdots\right]_{0}^{0.4} \\
=&\left(\frac{(0.4)^{3}}{3}-\frac{(0.4)^{4}}{8}+\frac{(0.4)^{5}}{15}-\frac{(0.4)^{6}}{24}\right.\\
&\left.\quad+\frac{(0.4)^{7}}{35}-\cdots\right)-(0)
\end{aligned}
$$
$$
=0.02133-0.0032+0.0006827-\cdots
$$
$=\mathbf{0 . 0 1 9}$, correct to 3 decimal places.

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
01:29

Problem 16

Determine $\lim _{x \rightarrow 1}\left\{\frac{x^{2}+3 x-4}{x^{2}-7 x+6}\right\}$
The first step is to substitute $x=1$ into both numerator and denominator. In this case we obtain $\frac{0}{0}$. It is only when we obtain such a result that we then use L'H么pital's rule. Hence applying L'H么pital's rule,
$$
\lim _{x \rightarrow 1}\left\{\frac{x^{2}+3 x-4}{x^{2}-7 x+6}\right\}=\lim _{x \rightarrow 1}\left\{\frac{2 x+3}{2 x-7}\right\}
$$
i.e. both numerator and denominator have been differentiated
$$
=\frac{5}{-5}=-1
$$

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
01:46

Problem 17

Determine $\lim _{x \rightarrow 0}\left\{\frac{\sin x-x}{x^{2}}\right\}$
Substituting $x=0$ gives
$$
\lim _{x \rightarrow 0}\left\{\frac{\sin x-x}{x^{2}}\right\}=\frac{\sin 0-0}{0}=\frac{0}{0}
$$
Applying L'H么pital's rule gives
$$
\lim _{x \rightarrow 0}\left\{\frac{\sin x-x}{x^{2}}\right\}=\lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{2 x}\right\}
$$
Substituting $x=0$ gives
$$
\frac{\cos 0-1}{0}=\frac{1-1}{0}=\frac{0}{0} \text { again }
$$
Applying L'H么pital's rule again gives
$$
\lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{2 x}\right\}=\lim _{x \rightarrow 0}\left\{\frac{-\sin x}{2}\right\}=0
$$

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator
03:29

Problem 18

Determine $\lim _{x \rightarrow 0}\left\{\frac{x-\sin x}{x-\tan x}\right\}$.
Substituting $x=0$ gives
$$
\lim _{x \rightarrow 0}\left\{\frac{x-\sin x}{x-\tan x}\right\}=\frac{0-\sin 0}{0-\tan 0}=\frac{0}{0}
$$
Applying L'H么pital's rule gives
$$
\lim _{x \rightarrow 0}\left\{\frac{x-\sin x}{x-\tan x}\right\}=\lim _{x \rightarrow 0}\left\{\frac{1-\cos x}{1-\sec ^{2} x}\right\}
$$
Substituting $x=0$ gives
$$
\lim _{x \rightarrow 0}\left\{\frac{1-\cos x}{1-\sec ^{2} x}\right\}=\frac{1-\cos 0}{1-\sec ^{2} 0}=\frac{1-1}{1-1}=\frac{0}{0} \text { again }
$$ Applying L'H么pital's rule gives
$$
\begin{aligned}
\lim _{x \rightarrow 0}\left\{\frac{1-\cos x}{1-\sec ^{2} x}\right\} &=\lim _{x \rightarrow 0}\left\{\frac{\sin x}{(-2 \sec x)(\sec x \tan x)}\right\} \\
&=\lim _{x \rightarrow 0}\left\{\frac{\sin x}{-2 \sec ^{2} x \tan x}\right\}
\end{aligned}
$$
Substituting $x=0$ gives
$$
\frac{\sin 0}{-2 \sec ^{2} 0 \tan 0}=\frac{0}{0} \text { again }
$$
Applying L'H么pital's rule gives
$$
\begin{aligned}
&\lim _{x \rightarrow 0}\left\{\frac{\sin x}{-2 \sec ^{2} x \tan x}\right\} \\
&\quad=\lim _{x \rightarrow 0}\left\{\begin{array}{c}
\left(-2 \sec ^{2} x\right)\left(\sec ^{2} x\right) \\
+(\tan x)\left(-4 \sec ^{2} x \tan x\right)
\end{array}\right\}
\end{aligned}
$$
using the product rule
Substituting $x=0$ gives
$$
\begin{aligned}
\frac{\cos 0}{-2 \sec ^{4} 0-4 \sec ^{2} 0 \tan ^{2} 0} &=\frac{1}{-2-0} \\
&=-\frac{1}{2}
\end{aligned}
$$
Hence $\lim _{x \rightarrow 0}\left\{\frac{x-\sin x}{x-\tan x}\right\}=-\frac{1}{2}$

Varsha Aggarwal
Varsha Aggarwal
Numerade Educator