Question:
If $f(x)=\left\{\begin{array}{cc}\frac{2^{x+2}-16}{4^{x}-16}, & \text { if } \quad x \neq 2 \\ k & \text {, if } x=2\end{array}\right.$ is continuous at $x=2$, find $k$.
Solution:
Given:
$f(x)= \begin{cases}\frac{2^{x+2}-16}{4^{x}-16}, & \text { if } x \neq 2 \\ k & , \text { if } x=2\end{cases}$
If $f(x)$ is continuous at $x=2$, then
$\lim _{x \rightarrow 2} f(x)=f(2)$
$\Rightarrow \lim _{x \rightarrow 2} \frac{2^{x+2}-16}{4^{x}-16}=f(2)$
$\Rightarrow \lim _{x \rightarrow 2} \frac{4\left(2^{x}-4\right)}{\left(2^{x}-4\right)\left(2^{x}+4\right)}=k$
$\Rightarrow \lim _{x \rightarrow 2} \frac{4}{\left(2^{x}+4\right)}=k$
$\Rightarrow \frac{4}{\left(2^{2}+4\right)}=k$
$\Rightarrow \frac{4}{8}=k$
$\Rightarrow k=\frac{1}{2}$