Question

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

Answer 1

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

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

$\therefore L=\frac{(\log 3{)}^2\times (1{)}^2}{1\times 2}$
$L=\frac{(\log 3{)}^2}{2}$