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Use the formula $\lim\limits_{x\to 0}\large\frac{a^x-1}{x}=$$ln\; a$ to find $\lim\limits_{x\to 0}\large\frac{2^x-1}{(1+x)^{1/2}-1}$

$(a)\;2\qquad(b)\;2ln \;2\qquad(c)\;ln \;2\qquad(d)\;0$

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$\lim\limits_{x\to 0}\large\frac{2^x-1}{\sqrt{1+x}-1}=$$\lim\limits_{x\to 0}\large\frac{2^x-1}{\sqrt{1+x}-1}\times \frac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{(2^x-1)(\sqrt{1+x}+1)}{1+x-1}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{(2^x-1)(\sqrt{1+x}+1)}{x}$
$\Rightarrow \lim\limits_{x\to 0}\large\frac{2^x-1}{x}$$\lim\limits_{x\to 0}(\sqrt{1+x}+1)$
$\Rightarrow ln \;2.(1+1)$
$\Rightarrow 2ln \;2$
Hence (b) is the correct answer.
answered Dec 31, 2013 by sreemathi.v
 
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