#### Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 45

$k=1$

Hint:

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

Given:

$f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & , \text { if } x=0 \\\\ \frac{x}{|x|} & , \text { if } x>0\end{cases}$

Solution:

$f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & , \text { if } x=0 \\\\ \frac{x}{|x|} & , \text { if } x>0\end{cases}$

$f(x)= \begin{cases}\frac{1-\cos 2 x}{2 x^{2}} & , \text { if } x<0 \\\\ k & \text { , if } x=0 \\\\ 1 & , \text { if } x>0\end{cases}$

We have

(LHL at $x=0$  )

$\! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \! \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2(-h)}{2(-h)^{2}}\right)=\lim _{h \rightarrow 0}\left(\frac{1-\cos 2 h}{2 h^{2}}\right)=\frac{1}{2} \lim _{h \rightarrow 0}\left(\frac{2 \sin ^{2} h}{h^{2}}\right)$

$\left[\because 1-\cos 2 x=2 \sin ^{2} x\right]$

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

(RHL at  $x=0$  )

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)=\lim _{h \rightarrow 0}(1)=1$

Also

$f(0)=k$

If $f\left ( x \right )$  is continuous at $x=0$ .

\begin{aligned} &\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{-}} f(x)=f(0) \\\\ &1=1=k \\\\ &k=1 \end{aligned}

Hence, the required value of  $k$  is $1$  .