MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{x{{.2}^x} - x}}{{1 - \cos x}} = $
  • A
    $0$
  • $\log 4$
  • C
    $\log 2$
  • D
    None of these

Answer

Correct option: B.
$\log 4$
b
(b) $\mathop {\lim }\limits_{x \to 0} \,\frac{{x.({2^x} - 1)}}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{x}.\frac{{{x^2}}}{{1 - \cos x}}$

$ = \log \,\,2\,.\mathop {\lim }\limits_{x \to 0} \,\,\frac{{{x^2}}}{{2\,\,{{\sin }^2}\frac{x}{2}}} = (\log \,\,2)\,.\,2 = 2\log 2 = \log 4$.

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