Question

$f\left(x\right)=\frac{{(3^x-1)}^2}{\sin x\log (1+x)},x\ne 0$ is continuous at $x=0$ then $f\left(0\right)={\left(\log b\right)}^2$. Find $b$.

Answer 1

$f\left(x\right)=\frac{{(3^x-1)}^2}{\sin x\log (1+x)},x\ne 0$
For $f\left(x\right)$ to be continuous at $x=0$
$f\left(0\right)=\underset{x\to 0}{lim}\frac{{(3^x-1)}^2}{\sin x\log (1+x)}$
$=\underset{x\to 0}{lim}\frac{{\left(\frac{3^x-1}{x}\right)}^2}{\frac{\sin x}{x}\frac{\log (1+x)}{x}}=$

applying $\frac{3^x-1}{x}=$ $\left(log3\right)$

$\frac{(\log 3{)}^2}{1.1}=(\log 3{)}^2$