Question

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

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

Answer 1

Answer: (D)

${\log }_{3}e$

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

This is of the from $\frac{0}{0}$ . Hence

Applying L 'Hospital's rule we get

${lim}_{x\to 0}\frac{\frac{1}{1+x}}{3^{x}lo{g}^3}$

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

$=\frac{1}{lo{g}_e^3}=lo{g}_3^e$