Question

$\underset{x\to 0}{lim}\frac{lo{g}_e(1+x)}{3^x-1}=$

• A. ${\log }_{e}3$
• B. $0$
• C. $1$
• D. ${\log }_{3}e$

Answer 1

Answer: (D)

${\log }_{3}e$

Clearly form of the limit is $\frac{0}{0}$,
Thus using L-hospital rule,
$\underset{x\to 0}{lim}\frac{lo{g}_e(1+x)}{3^x-1}=\underset{x\to 0}{lim}\frac{\frac{1}{1+x}}{3^x\ln 3}$
$=\frac{1}{{\log }_{e}3}={\log }_{3}e[\because {\log }_{a}b=\frac{1}{{\log }_{b}a}]$