Question

The value of $\underset{x\to 0}{\lim }\frac{{\left(4^x-1\right)}^3}{sin\frac{x^2}{4}log\left(1+3x\right)}$, is

• A. $\frac{4}{3}{\left(ln4\right)}^2$
• B. $\frac{4}{3}{\left(ln4\right)}^3$
• C. $\frac{3}{2}{\left(ln4\right)}^2$
• D. $\frac{3}{2}{\left(ln4\right)}^3$

Answer 1

Answer: (B)

$\underset{x\to 0}{\lim }\frac{{\left(4^x-1\right)}^3}{sin\frac{x^2}{4}log\left(1+3x\right)}$
$=\underset{x\to 0}{\lim }\frac{\left(4^x-1^x\right)}{x^3}.\frac{{\left(x/2\right)}^2}{sin{x}^2/4}.\frac{3x}{log\left(1+3x\right)}.\frac{4}{3}$
$=\frac{4}{3}{\left(lo{g}_{e}4\right)}^3.1.lo{g}_e\left(e\right)=\frac{4}{3}{\left(lo{g}_{e}4\right)}^3$.