Question

If $f\left(x\right)=\frac{2x-3\sin x}{3x+4\tan x},x\ne 0$ is continuous at $x=0$, then the value of $|14f\left(0\right)|$ is

Answer 1

For continuity of $f\left(x\right)$ at $x=0$, we must have,
$f\left(0\right)=\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\frac{2x-3\sin x}{3x+4\tan x}$ $\left[\frac{0}{0}\text{form}\right]$
Dividing by $x$ on both numerator and denominator
$=\underset{x\to 0}{lim}\frac{2-\frac{3\sin x}{x}}{3+\frac{4\tan x}{x}}=\frac{2-3}{3+4}=\frac{-1}{7}$
$\therefore |14f\left(0\right)|=2$