Question

If the function $f\left(x\right)=\frac{(4^{\sin x}-1{)}^2}{x\log (1+2x)}$ for $x\ne 0$ is continuous at $x=0$, Find $f\left(0\right)$

Answer 1

$RHL={lim}_{h\to 0}f(0+h)={lim}_{h\to 0}f\left(h\right)={lim}_{h\to 0}\left(\frac{{(4^{sinh}-1)}^2}{hlog(1+2h)}\right)$

$={lim}_{h\to 0}\left(\frac{(4^{sinh}-1{)}^2}{(sinh{)}^2}\right)\times \left(\frac{(sinh{)}^2}{h\left(\frac{log(1+2h)}{2h}\right)\times 2h}\right)$

$={lim}_{h\to 0}{\left(\right(\frac{4^{sinh}-1}{sinh}\left)\right)}^2\times \left(\frac{1}{\left(\frac{log(1+2h)}{2h}\right)}\right)\times \left(\frac{(sinh{)}^2}{h^2}\right)\times \left(\frac{1}{2}\right)$

$=(log4{)}^2\times 1\times 1\times (\frac{1}{2})=(log2{)}^2\times 2$

$forf\left(x\right)tobecontinuousRHL=LHL=f\left(0\right)\therefore f\left(0\right)=2(log2{)}^2$