Question

If $f\left(x\right)=\frac{\sin 2x-\tan 2x}{(e^x-1)x^2},$ then $\underset{x\to 0}{lim}f\left(x\right)$ is equal to

• A. $-\frac{1}{2}$
• B. $-4$
• C. $\frac{1}{2}$
• D. $4$

Answer 1

The correct option is B $-4$

$\underset{x\to 0}{lim}f\left(x\right)$
$=\underset{x\to 0}{lim}\frac{\sin 2x-\tan 2x}{(e^x-1)x^2}$
$=\underset{x\to 0}{lim}\frac{\sin 2x-\frac{\sin 2x}{\cos 2x}}{(e^x-1)x^2}=\underset{x\to 0}{lim}\frac{\sin 2x(\cos 2x-1)}{\cos 2x(e^x-1)x^2}=-\underset{x\to 0}{lim}[\left(\frac{\sin 2x}{2x}\right)\left(\frac{2{\sin }^{2}x}{x^2}\right)\left(\frac{1}{\cos 2x}\right)\left(\frac{x}{e^x-1}\right)\cdot 2]=-1\cdot 2\cdot 1\cdot 1\cdot 2=-4$