Question

Evaluate the limit:
$\underset{x\to 0}{lim}\frac{\log (1+x)}{3^x-1}$

Answer 1

Given:
$\underset{x\to 0}{lim}\frac{\log (1+x)}{3^x-1}$
Dividing the Numerator and Denominator by $x$

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

$=\frac{1}{\log 3}$
$[\because \underset{x\to 0}{lim}\frac{\log (1+x)}{x}=1,\underset{x\to 0}{lim}\left(\frac{a^x-1}{x}\right)=\log a]$

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