Question

Find the value of $p$, for which
$f(x)= \begin{cases}\frac{\left(4^{x}-1\right)^{3}}{\sin \left(\frac{x}{p}\right) \log _{e}\left\{1+\left(\frac{x^{2}}{3}\right)\right\}}, & x \neq 0 \\ 12\left(\log _{e} 4\right)^{3}, & x=0\end{cases}$
is continuous at $x=0$.

Answer 1

$\displaystyle\lim _{x \rightarrow 0} \frac{\left(4^{x}-1\right)^{3}}{\sin (x / p) \log _{e}\left(1+\frac{x^{2}}{3}\right)}=12\left(\log _{e} 4\right)^{3}$
$\displaystyle\lim _{x \rightarrow 0} \frac{\left(\frac{4^{x}-1}{x}\right)^{3} x^{3}}{\left(\frac{\sin (x / p)}{(x / p)}\right)\left(\frac{x}{p}\right)\left(\frac{\log \left(1+\left(x^{2} / 3\right)\right)}{x^{2} / 3}\right) x^{2} / 3}$
$=12\left(\log _{e} 4\right)^{3}$
$3 p \left(\log _{e} 4\right)^{3}=12\left(\log _{e} 4\right)^{3}$
$p=4$