Question

Let $f\left(x\right)=\frac{\ln (1+xtanx)}{4x}$, $x\ne 0$ is continuous at $x=0$,then $f\left(0\right)$ must be equal to

• A. $1$
• B. $0$
• C. $3$
• D. $6$

Answer 1

The correct option is B $0$

At $x=0$ the function $f\left(x\right)$ is of form $\frac{0}{0}$. So using L'Hopital's rule:

$f\left(0\right)=\underset{x\to 0}{lim}\frac{ln(1+xtanx)}{4x}=\underset{x\to 0}{lim}\frac{xse{c}^{2}x+tanx}{4(1+xtanx)}=0$