Question

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

Answer 1

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

Apply L-hospital

$=\underset{x\to 0}{lim}\frac{x{e}^x+e^x-1}{\sin x}$

Again apply L-hospital

$=\underset{x\to 0}{lim}\frac{x{e}^x+e^x+ex}{\cos x}$

$=2$