Question

If $f\left(x\right)=\left\{\begin{array}{ll}\frac{2^{x+2}-16}{4^x-16},\mathrm{if}& x\ne 2\\ k,\mathrm{if}& x=2\end{array}\right.$is continuous at x = 2, find k.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}\frac{2^{x+2}-16}{4^x-16},\mathrm{if}x\ne 2\\ k,\mathrm{if}x=2\end{array}\right.$

If f(x) is continuous at x = 2, then

​$$\begin{gathered} \underset{x\to 2}{\lim }f\left(x\right)=f\left(2\right) \\ \Rightarrow \underset{x\to 2}{\lim }\frac{2^{x+2}-16}{4^x-16}=f\left(2\right) \\ \Rightarrow \underset{x\to 2}{\lim }\frac{4\left(2^x-4\right)}{\left(2^x-4\right)\left(2^x+4\right)}=k \\ \Rightarrow \underset{x\to 2}{\lim }\frac{4}{\left(2^x+4\right)}=k \\ \Rightarrow \frac{4}{\left(2^2+4\right)}=k \\ \Rightarrow \frac{4}{8}=k \\ \Rightarrow k=\frac{1}{2} \end{gathered}$$