MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{x({e^x} - 1)}}{{1 - \cos x}} = $
  • A
    $0$
  • B
    $\infty $
  • C
    $-2$
  • $2$

Answer

Correct option: D.
$2$
d
(d) $\mathop {\lim }\limits_{x \to 0} \,\,\,\frac{{x\,({e^x} - 1)}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \,\frac{{2x\,({e^x} - 1)}}{{4.{{\sin }^2}\frac{x}{2}}}$

$ = 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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