#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 27

$k=\pm 1$

Hint:

$f\left ( x \right )$ must be defined. The limit of the $f\left ( x \right )$ approaches the value $x$ must exist.

Given:

$f(x)=\left\{\begin{array}{cl} \frac{1-\cos k x}{x \sin x} & , x \neq 0 \\\\ \frac{1}{2} & , x=0 \end{array}\right.$

Solution:

$f\left ( x \right )$  is continuous at$x=0$  , then

$\lim _{x \rightarrow 0} f(x)=f(0)$

Consider,

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{1-\cos k x}{x \sin x}\right)=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} k \frac{x}{2}}{x \sin x}\right)$                      $\left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]$

$\Rightarrow$         $\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(\frac{2 \sin ^{2} k \frac{x}{2}}{x^{2}\left(\frac{\sin x}{x}\right)}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{2 \frac{k^{2}}{4}\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}\left(\frac{\sin x}{x}\right)}\right)$         [Multiplying and dividing by $\left ( \frac{k}{2} \right )^{2}$ ]

$=\frac{2 k^{2}}{4} \lim _{x \rightarrow 0}\left(\frac{\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}\left(\frac{\sin x}{x}\right)}\right)$

$\lim _{x \rightarrow 0} f(x)=\frac{2 k^{2}}{4}\left(\frac{\lim _{x \rightarrow 0} \frac{\left(\sin k \frac{x}{2}\right)^{2}}{\left(\frac{k x}{2}\right)^{2}}}{\lim _{x \rightarrow 0}\left(\frac{\sin x}{x}\right)} \right)$                                                      $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$\lim _{x \rightarrow 0} f(x)=\frac{2 k^{2}}{4} \times 1=\frac{k^{2}}{2}$                                                                               … (i)

From (i)

\begin{aligned} &\frac{k^{2}}{2}=f(0) \\\\ &\frac{k^{2}}{2}=\frac{1}{2}, k=\pm 1 \end{aligned}