Question

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

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

Answer 1

Answer: (B)

$ \displaystyle \lim_{x \to 0} \frac{\left(4^{x}-1\right)^{3}}{sin \frac{x^{2}}{4}log\left(1+3x\right)}$
$= \displaystyle \lim _{x \to 0} \frac{\left(4^{x}-1^{x}\right)}{x^{3}} . \frac{\left(x / 2\right)^{2}}{sin\,x^{2}/4}. \frac{3x}{log\left(1+3x\right)}. \frac{4}{3}$
$= \frac{4}{3} \left(log_{e}\, 4\right)^{3}.1.log_{e} \left( e\right) = \frac{4}{3} \left(log_{e}\,4\right)^{3}$.