Question

$\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{3^x-1}$ is equal to

(a) $\log _e 3$

(b) 0

(c) 1

(d) $\log _3 e$

Answer 1

$\lim _{x \rightarrow 0} \frac{\log _e(1+x)}{3^x-1}$

$\left(\frac{0}{0}\right.$ form $)$

$\begin{array}{l}=\lim _{x \rightarrow 0} \frac{\frac{1}{1+x}}{3^x \log _e 3} \\=\frac{\frac{1}{1+0}}{3^0 \log _e 3} \\=\frac{1}{\log _e 3}=\log _3 e \quad\left(\because \frac{\log _e e}{\log _e 3}=\log _3 e\right)\end{array}$

So, The correct option of this question will be (D).