Question

If $f\left(x\right)=\frac{4^x-2^{x+1}+1}{1-\cos x},\text{for}x\ne 0$ is continuous at $x=0$, find $f\left(0\right)$

Answer 1

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\left(\frac{4^x-2^{x+1}+1}{1-\cos x}\right)$

Using L Hospitals rule

$=\underset{x\to 0}{lim}\left(\frac{4^x\ln 4-2^{x+1}\ln 2}{1-(-\sin x)}\right)$

$=\frac{4^o\ln 4-2\ln 2}{1+0}$

$=\ln 4-\ln 4=0$.