Question

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

Answer 1

Answer: 4.00

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