MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}} = $
- A$0$
- B$\infty $
- C$-2$
- ✓$2$
$ = 2\mathop {\lim }\limits_{x \to 0} \,\left[ {\frac{{{{(x/2)}^2}}}{{{{\sin }^2}\frac{x}{2}}}} \right]\,\left( {\frac{{{e^x} - 1}}{x}} \right) = 2.$
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