Question

$\underset{x\to 0}{lim}\frac{x(e^x-1)}{1-\cos x}$ is equal to

• A. $0$
• B. $\infty$
• C. $-2$
• D. $2$

Answer 1

The correct option is D $2$

$\underset{x\to 0}{lim}\frac{x(e^x-1)}{1-\cos x}$

$\underset{x\to 0}{lim}\frac{x(1+x-1)}{2{\sin }^2\left(x/2\right)}$

$\left\{\begin{array}{ccc}{e}^x=1+x& for& x\to 0\\ 1-\cos x=2{\sin }^2\frac{x}{2}& & \end{array}\right\}$

$\Rightarrow \underset{x\to 0}{lim}\frac{x\cdot x}{2\cdot {\left(\frac{x}{2}\right)}^2}$

$\{\sin \frac{x}{2}\to \frac{x}{2}for\frac{x}{2}\to 0\}$

$\Rightarrow \underset{x\to 0}{lim}2\frac{x^2}{x^2}$

$=2$.