Question

Find the limit :-
$lim{}_{x\to 0}\frac{2^{3x}-3^x}{\sin 3x}$

Answer 1

$\underset{x\to 0}{lim}\frac{8^x-3^x}{\sin 3^x}$

As numerator & denominator both tend to zero, we can use $L$ Hospitals rule

$\underset{x\to 0}{lim}\frac{8^x-3^x}{\sin 3^x}=\underset{x\to 0}{lim}\frac{8^x{\log }_{e}8-3^x{\log }_{e}3}{3\cos 3^x}$

$=\frac{{\log }_e\frac{9}{3}}{3}$

$=ln{\left(\frac{8}{3}\right)}^{1/3}$