Question

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

Answer 1

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

Dividing the numerator and the denominator by x²:

$$\begin{gathered} =\underset{x\to 0}{\lim }\left[\frac{\left({\displaystyle \frac{e^x-1}{x}}\right)}{{\displaystyle \left(\frac{1-\cos x}{x^2}\right)}}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{e^x-1}{x}\times \frac{1}{\frac{2{\sin }^2\left({\displaystyle \frac{x}{2}}\right)}{4\times {\displaystyle \frac{x^2}{4}}}}\right] \\ =1\times \frac{2}{1^2} \\ =2 \end{gathered}$$