Question

$li{m}_{x\to 0}\frac{lo{g}_e(1+4x)}{e^{3x}-1}$ is equal to

• A. 0
• B. $\frac{1}{12}$
• C. $\frac{4}{3}$
• D. 1

Answer 1

The correct option is C $\frac{4}{3}$

Given limit is
${lim}_{x\to 0}\frac{lo{g}_e(1+4x)}{e^{3x}-1}\left(\frac{0}{0}form\right)$

Applying L ' Hospital's rule

=${lim}_{x\to 0}\frac{\frac{1}{1+4x}.\left(4\right)}{e^{3x}.3-0}=\frac{4}{3}$