Question
The value of $$\mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \frac{{{x^2}}}{4}\log \left( {1 + 3x} \right)}}$$ is :
A.
$$\frac{4}{3}{\left( {{{\log }_e}4} \right)^2}$$
B.
$$\frac{4}{3}{\left( {{{\log }_e}4} \right)^3}$$
C.
$$\frac{3}{2}{\left( {{{\log }_e}4} \right)^2}$$
D.
$$\frac{3}{2}{\left( {{{\log }_e}4} \right)^3}$$
Answer :
$$\frac{4}{3}{\left( {{{\log }_e}4} \right)^3}$$
Solution :
$$\eqalign{
& \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{\sin \frac{{{x^2}}}{4}\log \left( {1 + 3x} \right)}} \cr
& = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {{4^x} - 1} \right)}^3}}}{{{x^3}}}.\frac{{{{\left( {\frac{x}{2}} \right)}^2}}}{{\sin \frac{{{x^2}}}{4}}}.\frac{{3x}}{{\log \left( {1 + 3x} \right)}}.\frac{4}{3} \cr
& = \frac{4}{3}{\left( {{{\log }_e}4} \right)^3}.1.{\log _e}\left( e \right) \cr
& = \frac{4}{3}{\left( {{{\log }_e}4} \right)^3} \cr} $$