#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 35 Maths Textbook Solution.

$k=1$

Hint:

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

Given:

$f(x)=\left(\begin{array}{c} \frac{1-\cos 4 x}{8 x^{2}}, \text { when } x \neq 0 \\\\ k, w h e n\: x=0 \end{array}\right)$

Solution:

$f(x)=\left(\begin{array}{c} \frac{1-\cos 4 x}{8 x^{2}}, \text { when } x \neq 0 \\\\ k, w h e n\: x=0 \end{array}\right)$

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{1-\cos 4 x}{8 x^{2}}=f(0) \end{aligned}

$\lim _{x \rightarrow 0} \frac{2 \sin ^{2} 2 x}{8 x^{2}}=f(0)$                                                               $\left[\because 1-\cos 2 x=2 \sin ^{2} x\right]$

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

\begin{aligned} &1 \times 1=f(0) \\ &k=1 \end{aligned}                                                                            $[\because f(0)=k]$