Question

The function $f\left(x\right)=\frac{(3^x-1{)}^2}{\sin x\ln (1+x)},x\ne 0$, is continuous at $x=0$. Then the value of $f\left(0\right)$ is

• A. $2{\log }_{e}3$
• B. $({\log }_{e}3{)}^2$
• C. ${\log }_{e}6$
• D. none of these

Answer 1

Answer: (B)

$({\log }_{e}3{)}^2$

For $f\left(x\right)$ to be continuous at $x=0$

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

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

$=\underset{x\to 0}{lim}\frac{3^x-1}{x}\times \underset{x\to 0}{lim}\frac{3^x-1}{x}\times \underset{x\to 0}{lim}\frac{x}{\sin x}\times \underset{x\to 0}{lim}\frac{x}{\cos x}$

$={\log }_{e}3\times {\log }_{e}3\times 1\times 1$

$={\left({\log }_{e}3\right)}^2$