#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 32

$k=-4$

Hint:

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

Given:

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

Solution:

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

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

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\lim _{x \rightarrow 0} \frac{\cos ^{2} x-\sin ^{2} x-1}{\sqrt{x^{2}+1}-1}=k \end{aligned}

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

$\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x}{\sqrt{x^{2}+1}-1}=k$

$\lim _{x \rightarrow 0} \frac{-2 \sin ^{2} x\left(\sqrt{x^{2}+1}+1\right)}{\left(\sqrt{x^{2}+1}-1\right)\left(\sqrt{x^{2}+1}+1\right)}=k$  [Multiplying and dividing by $\sqrt{x^{2}+1}+1$  ]

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

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

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

$k=-4$