Question

Find k if $f\left(x\right)=\{\begin{array}{c}\frac{2^{x+2}-16}{4^x-16},ifx\ne 2\\ k,ifx=2\end{array}$ is continuous at $x=2$

Answer 1

For $f\left(x\right)$ to be continuous at $x=2$, limit must be finite at $x=2$

So, using L'Hopital's rule:

$k=\underset{x\to 2}{lim}\frac{2^{x+2}-16}{4^{x}16}=\underset{x\to 2}{lim}\frac{2^{x+2}ln2}{4^{x}ln4}=\frac{2^{2+2}ln2}{4^{2}2ln2}=\frac{1}{2}$