Question

If $f\left(x\right)=\frac{{\log }_e(1+x^2\tan x)}{\sin x^3},x\ne 0$ is continuous at $x=0$, then the value of $f\left(0\right)$ is

• A. $-1$
• B. $0$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is D $1$

For $f\left(x\right)$ to be continuous at $x=0$, we must have
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}\frac{{\log }_e(1+x^2\tan x)}{\sin x^3}$
$=\underset{x\to 0}{lim}\frac{{\log }_e(1+x^2\tan x)}{x^2\tan x}\cdot \frac{x^2\tan x}{x^3}\cdot \frac{x^3}{\sin x^3}$
$=\underset{x\to 0}{lim}\left\{\frac{{\log }_e(1+x^2\tan x)}{x^2\tan x}\right\}\cdot \underset{x\to 0}{lim}\left(\frac{\tan x}{x}\right)\cdot \frac{1}{\underset{x\to 0}{lim}\left(\frac{\sin x^3}{x^3}\right)}$
$=1\times 1\times \frac{1}{1}=1$