MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{2^x} - 1}}{{{{(1 + x)}^{1/2}} - 1}} = $
  • A
    $\log 2$
  • $\log 4$
  • C
    $\log \sqrt 2 $
  • D
    None of these

Answer

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

$\left\{ \because \,\,\,\mathop {\lim }\limits_{x \to a} \,\,\frac{f(x)}{g(x)}=\mathop {\lim }\limits_{x \to a} \,\,\frac{{f}'(x)}{{g}'(x)} \right\}$

$ = 2\,\log 2 = \log 4.$

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