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$\lim\limits_{x\to 1}\large\frac{\sin(e^x-1)}{\log x}$ is

$(a)\;0\qquad(b)\;1\qquad(c)\;e\qquad(d)\;\text{None of these}$

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Put $x=1+h$ so that as $x\to 1,h\to 0$
Given limit=$\lim\limits_{h\to 0}\large\frac{\sin(e^h-1)}{\log(1+h)}$
$\Rightarrow \lim\limits_{h\to 0}\large\frac{\sin(e^h-1)}{e^h-1}\times \frac{e^h-1}{h}\times \frac{h}{\log(1+h)}$
$\Rightarrow 1\times 1\times 1$
$\Rightarrow 1$
Hence (b) is the correct answer.
answered Jan 6, 2014 by sreemathi.v

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