Question

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

• A. $lo{g}_{e}3$
• B. $0$
• C. $lo{g}_{3}e$
• D. $1$

Answer 1

The correct option is C $lo{g}_{3}e$

Given that: $\underset{x\to 0}{lim}\frac{{\log }_e(1+x)}{3^x-1}$

Using L'Hospital rule, we differentiate the numerator and denominator to find the limit.

We thus get $\underset{x\to 0}{lim}\frac{1}{(1+x)\left(3^x\right)\left({\log }_{e}3\right)}$

$=\frac{1}{1\times 1\times {\log }_{e}3}$$=\frac{1}{{\log }_{e}3}$$={\log }_{3}e$