Case study (4 Marks)

Question 14 Marks

To find the limits of trigonometric functions, we use the following theorems
Theorem 1: Let $f$ and $g$ be two real valued functions with the same domain such that $f(x) \leq g(x)$ for all $x$ in the domain of definition. For some real number $a$, if both $\lim _{x \rightarrow a} f(x)$ and $\lim _{x \rightarrow a} g(x)$ exist, then
$
\lim _{x \rightarrow a} f(x) \leq \lim _{x \rightarrow a} g(x) .
$
This is shown in the figure

Theorem 2 (Sandwich theorem) : Let $f, g$ and $h$ be real functions such that $f(x) \leq g(x) \leq h(x)$ for all $x$ in the common domain of definition. For some real number $a$, if $\lim _{x \rightarrow a} f(x)=l=\lim _{x \rightarrow a} h(x)$, then $\lim _{x \rightarrow a} g(x)=l$.

This is shown in the figure

Theorem 3 : Three important limits are
(i) $\lim _{x \rightarrow 0} \frac{\sin x}{x}=1$
(ii) $\lim _{x \rightarrow 0} \frac{\frac{x}{1-\cos x}}{x}=0$
(iii) $\lim _{x \rightarrow 0} \frac{\tan ^x x}{x}=1$

Based on above information, answer the following questions.

(i) $\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x}$ is equal to
(a) $\frac{1}{5}$ (b) $\frac{2}{5}$ (c) $\frac{3}{5}$ (d) $\frac{4}{5}$

(ii) $\lim _{\theta \rightarrow b} \frac{\tan (\theta-b)}{\theta-b}$ is equal to
(a) 0 (b) 1 (c) 2 (d) 3

(iii) $\lim _{x \rightarrow 0} \frac{\tan 2 x-\sin 2 x}{x^3}$ is equal to
(a) 4 (b) 3 (c) 2 (d) 1

(iv) $\lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}$ is equal to
(a) 0 (b) 1 (c) 2 (d) 3

(v) $\lim _{x \rightarrow \frac{\pi}{4}} \frac{\sin x-\cos x}{x-\frac{\pi}{4}}$ is equal to
(a) $\sqrt{2}$ (b) 3 (c) 1 (d) $\sqrt{3}$

Answer

(i)
$
\begin{aligned}
\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x} & =\lim _{x \rightarrow 0} \frac{\sin 3 x}{5 x \times \frac{3}{3}} \\
& =\lim _{x \rightarrow 0} \frac{3}{5} \cdot \frac{\sin 3 x}{3 x} \\
& =\frac{3}{5} \cdot \lim _{3 x \rightarrow 0} \frac{\sin 3 x}{3 x} \\
& {[\text { as } x \rightarrow 0, \text { therefore } 3 x \rightarrow 0] } \\
& =\frac{3}{5} \times 1=\frac{3}{5}\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]
\end{aligned}
$

(ii)
We have, $\lim _{\theta \rightarrow b} \frac{\tan (\theta-b)}{\theta-b}$
Put $\theta-b=h \Rightarrow \theta=h+b$
Also, when $\theta \rightarrow b$, then $h \rightarrow b$
$
\therefore \lim _{h \rightarrow b} \frac{\tan (\theta-b)}{\theta-b}=\lim _{h \rightarrow 0} \frac{\tan h}{h}=1
$

(iii)
$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\tan 2 x-\sin 2 x}{x^3} \\
& =\lim _{x \rightarrow 0} \frac{\frac{\sin 2 x}{\cos 2 x}-\sin 2 x}{x^3} \\
& =\lim _{x \rightarrow 0} \frac{\sin 2 x-\sin 2 x \cdot \cos 2 x}{x^3 \cdot \cos 2 x} \\
& =\lim _{x \rightarrow 0} \frac{\sin 2 x(1-\cos 2 x)}{x^3 \cdot \cos 2 x} \\
& =\lim _{x \rightarrow 0} \frac{\tan 2 x}{x} \cdot \lim _{x \rightarrow 0} \frac{2 \sin ^2 x}{x^2}
\end{aligned}
$
[by using product of limits and
$
\begin{aligned}
& \left.\cos 2 \theta=1-2 \sin ^2 \theta\right] \\
& =2 \cdot \lim _{x \rightarrow 0} \frac{\tan 2 x}{2 x} \times 2 \lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)^2 \\
& =2(1) \times 2(1)^2 \\
& =4\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=\lim _{x \rightarrow 0} \frac{\tan x}{x}=1\right]
\end{aligned}
$

(iv)
$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3} \\
= & \lim _{x \rightarrow 0} \frac{2 \sin x-2 \sin x \cos x}{x^3} \\
= & {[\because \sin 2 x=2 \sin x \cos x] } \\
= & \lim _{x \rightarrow 0} \frac{2 \sin x(1-\cos x)}{x^3} \\
= & 2 \lim _{x \rightarrow 0} \frac{\sin x}{x} \cdot \lim _{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right) \\
= & 2 \cdot 1 \lim _{x \rightarrow 0} \frac{1-\cos x}{x^2}\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]
\end{aligned}
$
$\begin{aligned} & =2 \lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{4 \times \frac{x^2}{4}} \\ & =\frac{2 \cdot 2}{4} \lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2=\lim _{x \rightarrow 0}\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2=1\end{aligned}$

(v)
Given,
$
\begin{aligned}
& \lim _{x \rightarrow \pi / 4} \frac{\sqrt{2}\left(\sin x \cdot \frac{1}{\sqrt{2}}-\cos x \cdot \frac{1}{\sqrt{2}}\right)}{\left(x-\frac{\pi}{4}\right)} \\
& =\lim _{x \rightarrow \pi / 4} \frac{\sqrt{2}\left(\sin x \cos \frac{\pi}{4}-\cos x \cdot \sin \frac{\pi}{4}\right)}{\left(x-\frac{\pi}{4}\right)}
\end{aligned}
$
$
\begin{aligned}
& =\lim _{x \rightarrow \pi / 4} \frac{\sqrt{2}\left\{\sin \left(x-\frac{\pi}{4}\right)\right\}}{\left(x-\frac{\pi}{4}\right)} \\
& |\because \sin A \cos B-\cos A \sin B=\sin (A-B)| \\
& =\sqrt{2} \lim _{x-\frac{\pi}{4} \rightarrow 0} \frac{\sin \left(x-\frac{\pi}{4}\right)}{\left(x-\frac{\pi}{4}\right)}=\sqrt{2} \\
& {\left[\because x \rightarrow \frac{\pi}{4} \Rightarrow\left(x-\frac{\pi}{4}\right) \rightarrow 0 \text { and } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]}
\end{aligned}
$

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Question 24 Marks

The logarithmic function expressed as $\log _e R^{+} \rightarrow R$ and given by $\log _e x=y$ iff $e^y=x$. The graph of the function is given below :

(i) Domain of $f(x)=(0, \infty)$ or $R^{+}$
(ii) Range of $f(x)=(-\infty, \infty)$ or $R$

To find the limit of functions involving logarithmic function, we use the following theorem Theorem $\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{x}=1$

Based on above information, answer the following questions.

(i) $\lim _{x \rightarrow 0} \frac{\log _e(1+5 x)}{x}$ is equal to
(a) 5 (b) 4 (c) 3 (d) 1

(ii) $\lim _{x \rightarrow 0} \frac{\log _e(1+6 x)-5 x^2}{x}$ is equal to
(a) 1 (b) 2 (c) 3 (d) 6

(iii) $\quad \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{\log (1+x)}$ is equal to
(a) 1 (b) $\frac{1}{2}$ (c) $\frac{1}{3}$ (d) $\frac{3}{2}$

(iv) $\quad \lim _{x \rightarrow 5} \frac{\log x-\log 5}{x-5}$ is equal to
(a) $\frac{1}{5}$ (b) $\frac{3}{5}$ (c) $\frac{1}{4}$ (d) $\frac{2}{3}$

(v) $\quad \lim _{x \rightarrow 0} \frac{\log (5+x)-\log (5-x)}{x}$ is equal to
(a) $\frac{1}{5}$ (b) $\frac{2}{5}$ (c) $\frac{3}{5}$ (d) $\frac{4}{5}$

Answer

(i)
We have, $\lim _{x \rightarrow 0} \frac{\log _e(1+5 x)}{x}$
$
\begin{aligned}
& =5 \lim _{5 x \rightarrow 0} \frac{\log _e(1+5 x)}{5 x}=5 \times 1=5 \\
& {[\because x \rightarrow 0 \Rightarrow 5 x \rightarrow 0]}
\end{aligned}
$

(ii)
We have, $\lim _{x \rightarrow 0} \frac{\log _{\ell}(1+6 x)-5 x^2}{x}$
$
\begin{aligned}
& =6 \lim _{6 x \rightarrow 0} \frac{\log _e(1+6 x)}{6 x}-5 \lim _{x \rightarrow 0} x \\
& {[\because x \rightarrow 0 \Rightarrow 6 x \rightarrow 0]} \\
& =6 \times(1)-5 \times(0)=6
\end{aligned}
$

(iii)
$
\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1}{\log (1+x)}
$
On multiplying numerator and denominator by $\sqrt{1+x}+1$, we get

$\begin{aligned} \lim _{x \rightarrow 0} & \frac{\sqrt{1+x}-1}{\log (1+x)} \times \frac{\sqrt{1+x}+1}{(\sqrt{1+x}+1)} \\ & =\lim _{x \rightarrow 0} \frac{1+x-1}{(\sqrt{1+x}+1) \log (1+x)} \\ & =\lim _{x \rightarrow 0} \frac{x}{(\sqrt{1+x}+1) \log (1+x)} \\ & =\frac{1}{(\sqrt{1+0}+1)} \lim _{x \rightarrow 0} \frac{1}{\frac{\log (1+x)}{x}} \\ & =\frac{1}{1+1} \times \frac{1}{\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}} \\ & =\frac{1}{1+1} \times 1=\frac{1}{2}\end{aligned}$

(iv)
Put $x-5=h$ and as $x \rightarrow 5$, then $h \rightarrow 0$
$
\begin{aligned}
& \therefore \lim _{h \rightarrow 0} \frac{\log (h+5)-\log 5}{h} \\
& =\lim _{\frac{h}{5} 0} \frac{\log \left(1+\frac{h}{5}\right)}{\frac{h}{5} \times 5}=\frac{1}{5} \\
& {\left[\begin{array}{rl}
\because \log m-\log n & =\log \frac{m}{n}, \\
h \rightarrow 0 & \Rightarrow \frac{h}{5} \rightarrow 0
\end{array}\right]} \\
&
\end{aligned}
$

(v)
$
\begin{gathered}
\lim _{x \rightarrow 0} \frac{\log \left\{5\left(1+\frac{x}{5}\right)\right\}-\log \left\{5\left(1-\frac{x}{5}\right)\right\}}{x} \\
=\lim _{x \rightarrow 0} \frac{\left\{\log 5+\log \left(1+\frac{x}{5}\right)\right\}-\left\{\log 5+\log \left(1-\frac{x}{5}\right)\right\}}{x} \\
=\lim _{\frac{x}{5} \rightarrow 0} \frac{1}{5} \frac{\log \left(1+\frac{x}{5}\right)}{\frac{x}{5}}-\lim _{\frac{x}{5} \rightarrow 0} \frac{\log \left(1-\frac{x}{5}\right)}{-\frac{x}{5}} \cdot \frac{1}{(-5)}
\end{gathered}
$
$\begin{aligned} & {\left[\because x \rightarrow 0 \Rightarrow \frac{x}{5} \rightarrow 0\right]} \\ & =\frac{1}{5} \times(1)+\frac{1}{5} \times(1)=\frac{2}{5}\end{aligned}$

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Question 34 Marks

A function $f$ is said to be a rational function, if $f(x)=\frac{g(x)}{h(x)}$, where $g(x)$ and $h(x)$ are polynomial functions such that $h(x) \neq 0$.
Then, $\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a} \frac{g(x)}{h(x)}$
$
=\frac{\lim _{x \rightarrow a} g(x)}{\lim _{x \rightarrow a} h(x)}=\frac{g(a)}{h(a)}
$
However, if $h(a)=0$, then there are two cases arise,
(i) $g(a) \neq 0$
(ii) $g(a)=0$. In the first case, we say that the limit does not exist.
In the second case, we can find limit.

Based on above information, answer the following questions.

(i) $\lim _{x \rightarrow-1}\left(\frac{x^{10}+x^5+1}{x-1}\right)$ is equal to
(a) $\frac{1}{2}$ (b) $\frac{-1}{2}$ (c) 2 (d) $\frac{3}{2}$

(ii) $\lim _{x \rightarrow-1} \frac{(x-1)^2+3 x^2}{\left(x^4+1\right)^2}$ is equal to
(a) $\frac{7}{4}$ (b) $\frac{6}{5}$ (c) $\frac{4}{7}$ (d) $\frac{3}{4}$

(iii) The value of $\lim _{x \rightarrow 2}\left[\frac{x^2-4}{x^3-4 x^2+4 x}\right]$ is
(a) 0 (b) 1 (c) 2 (d) Does not exist

(iv) $\lim _{x \rightarrow 1} \frac{x^7-2 x^5+1}{x^3-3 x^2+2}$ is equal to
(a) 0 (b) 1 (c) 2 (d) 3

(v) $\lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}$ is equal to
(a) 1 (b) 0 (c) -1 (d) 2

Answer

(i)
$
\begin{aligned}
\lim _{x \rightarrow-1} \frac{x^{10}+x^5+1}{x-1} & =\frac{(-1)^{10}+(-1)^5+1}{-1-1} \\
& =\frac{1-1+1}{-2}=\frac{-1}{2}
\end{aligned}
$

(ii)
$
\begin{aligned}
\lim _{x \rightarrow-1} \frac{(x-1)^2+3 x^2}{\left(x^4+1\right)^2} & =\frac{(-1-1)^2+3(-1)^2}{\left((-1)^4+1\right)^2} \\
& =\frac{(-2)^2+3(1)}{(1+1)^2} \\
& =\frac{4+3}{2^2}=\frac{7}{4}
\end{aligned}
$

(iii) Consider $f(x)=\frac{x^2-4}{x^3-4 x^2+4 x}$
On putting $x=2$, we get
$
f(2)=\frac{4-4}{8-16+8}=\frac{0}{0}
$
i.e., it is the form $\frac{0}{0}$.
So, let us first factorise it.
Consider,
$
\begin{aligned}
& \lim _{x \rightarrow 2} \frac{x^2-4}{x^3-4 x^2+4 x} \\
& =\lim _{x \rightarrow 2} \frac{(x+2)(x-2)}{x(x-2)^2} \\
& =\lim _{x \rightarrow 2} \frac{(x+2)}{x(x-2)} \\
& =\frac{2+2}{2(2-2)}=\frac{4}{0}
\end{aligned}
$
which is not defined.
$\therefore \lim _{x \rightarrow 2}\left[\frac{x^2-4}{x^3-4 x^2+4 x}\right]$ does not exist.
$
\begin{aligned}
\lim _{x \rightarrow 1} & \frac{x^7-2 x^5+1}{x^3-3 x^2+2}\left[\frac{0}{0} \text { form }\right] \\
& =\lim _{x \rightarrow 1} \frac{x^7-x^5-x^5+1}{x^3-x^2-2 x^2+2} \\
& =\lim _{x \rightarrow 1} \frac{x^5\left(x^2-1\right)-1\left(x^5-1\right)}{x^2(x-1)-2\left(x^2-1\right)}
\end{aligned}
$

(iv)
On dividing numerator and denominator by $(x-1)$, then
$
\begin{aligned}
& =\lim _{x \rightarrow 1} \frac{\frac{x^5\left(x^2-1\right)}{(x-1)}-\frac{1\left(x^5-1\right)}{(x-1)}}{\frac{x^2(x-1)}{(x-1)}-\frac{2\left(x^2-1\right)}{(x-1)}} \\
& =\frac{\lim _{x \rightarrow 1} x^5(x+1)-\lim _{x \rightarrow 1}\left(\frac{x^5-1}{x-1}\right)}{\lim _{x \rightarrow 1} x^2-\lim _{x \rightarrow 1} 2(x+1)} \\
& =\frac{1 \times 2-5 \times(1)^4}{1-2 \times 2}=\frac{2-5}{1-4}\left[\because \lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}=n a^{n-1}\right] \\
& =\frac{-3}{-3}=1
\end{aligned}
$

(v) $\begin{aligned} & \lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2} \\ & =\lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2} \\ & \frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}+\sqrt{1-x^3}} \\ & =\lim _{x \rightarrow 0} \frac{\left(1+x^3\right)-\left(1-x^3\right)}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =\lim _{x \rightarrow 0} \frac{1+x^3-1+x^3}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =\lim _{x \rightarrow 0} \frac{2 x^3}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =\lim _{x \rightarrow 0} \frac{2 x}{\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \\ & =0\end{aligned}$

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MATHS Case study (4 Marks)

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