#### Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 22 Maths Textbook Solution.

$k=\frac{2}{5}$

Hint:

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

Given:

$f(x)=\left\{\begin{array}{l} \frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.$  is continuous at  $x=0$

Solution:

$f(x)=\left\{\begin{array}{l} \frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.$

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{\sin 2 x}{5 x}=k \end{aligned}

$\lim _{x \rightarrow 0} \frac{2 \sin 2 x}{5 \times 2 x}=k$    [Multiplying and dividing by $2$ ]

$\frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=k$                                                                                                             $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$

$\frac{2}{5}\times 1=k$

$k=\frac{2}{5}$