#### Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 point 12 Question 31

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

Hint:

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

Given:

$f(x)=\left\{\begin{array}{c} \frac{2^{x+2}-16}{4^{x}-16}, \text { if } x \neq 2 \\\\ k, \text { if } x=2 \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{c} \frac{2^{x+2}-16}{4^{x}-16}, \text { if } x \neq 2 \\\\ k, \text { if } x=2 \end{array}\right.$

If $f\left ( x \right )$  is continuous at $x=2$ ,

\begin{aligned} &\lim _{x \rightarrow 2} f(x)=f(2) \\ &\lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=f(2) \end{aligned}

$\lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}=k$   [Taking $4$ as common]                       $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

\begin{aligned} &\lim _{x \rightarrow 2} \frac{4}{\left(2^{x}+4\right)}=k \\ &\frac{4}{\left(2^{2}+4\right)}=k \\ &k=\frac{4}{8} \\ &k=\frac{1}{2} \end{aligned}