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$\lim\limits_{x\to 0}\large\frac{2^x-1}{\sqrt{1+x}-1}$=

$(a)\;2\qquad(b)\;\log_e2\qquad(c)\;\large\frac{\log_e2}{2}$$\qquad(d)\;2\log_e2$

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$\lim\limits_{x\to 0}\big(\large\frac{2^x-1}{\sqrt{1+x}-1}\big)$
$\Rightarrow \lim\limits_{x\to 0}\big(\large\frac{2^x-1}{x}.\frac{x}{\sqrt{1+x-1}}\big)$
$\Rightarrow \lim\limits_{x\to 0}\bigg[\large\frac{2^x-1}{x}.\frac{x(\sqrt{1+x}+1)}{(1+x-1)}\bigg]$
$\Rightarrow \big(\lim\limits_{x\to 0}\large\frac{2^x-1}{x}\big)$$\lim\limits_{x\to 0}(\sqrt{1+x}+1)$
$\Rightarrow (\log 2)2$
$\Rightarrow 2\log_e2$
Hence (d) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

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