Question

$f\left(x\right)=\frac{2^{x+2}-16}{4^x-16}$ is continuous at $x=2$, then $f\left(2\right)=$

Answer 1

$RHL=\underset{h\to 0}{lim}f(2+h)=\underset{h\to 0}{lim}\left(\frac{2^{(2+h+2)}-16}{4^{2+h}-16}\right)=\underset{h\to 0}{lim}\left(\frac{16\times 2^h-16}{16\times 4^h-16}\right)=\underset{h\to 0}{lim}\left(\frac{2^h-1}{4^h-1}\right)=\underset{h\to 0}{lim}\left(\frac{\left(\frac{2^h-1}{h}\right)}{\left(\frac{4^h-1}{h}\right)}\right)=\left(\frac{log2}{log4}\right)=\left(\frac{1}{2}\right)\therefore f\left(2\right)=\left(\frac{1}{2}\right)$

as for function to be continuous

$thereforeLHL=RHL=f\left(2\right)=1/2$