Question

Solve : $\underset{x\to 0}{lim}\frac{(3^x-1{)}^3}{(2^x-1)\sin x.\log (1+x)}$

Answer 1

$\underset{x\to 0}{lim}\frac{{(3^x-1)}^3}{(2^x-1)\sin x.\log (1+x)}$

$=\underset{x\to 0}{lim}\frac{\frac{{(3^x-1)}^3}{x^3}}{\frac{(2^x-1)}{x}.\frac{\sin x}{x}.\frac{\log (1+x)}{x}}$

$=\frac{\underset{x\to 0}{lim}\frac{{(3^x-1)}^3}{x^3}}{\underset{x\to 0}{lim}(\frac{(2^x-1)}{x}.\frac{\sin x}{x}.\frac{\log (1+x)}{x})}$

$=\frac{{\left(\ln 3\right)}^3}{\left(\left(\ln 2\right).1.1\right)}=\frac{{\left(\ln 3\right)}^3}{\ln 2}$