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

$(a)\;1\qquad(b)\;e^{\large a}\qquad(c)\;e^{\large -a}\qquad(d)\;-1$

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1 Answer

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$\lim\limits_{x\to a}\large\frac{\log(x-a)}{\log(e^x-e^a)}=$$\lim\limits_{x\to a}\big(\large\frac{1}{x-a}\big)/\big(\large\frac{e^x}{e^x-e^a}\big)$ (Using L Hospital's rule)
$\Rightarrow \lim\limits_{x\to a}\large\frac{\Large\frac{e^x-e^a}{e^x}}{x-a}$
$\Rightarrow \lim\limits_{x\to a}\large\frac{1-e^{\large a-x}}{x-a}$
$\Rightarrow \lim\limits_{x\to a}\large\frac{e^{\large a-x}-1}{a-x}$
$\Rightarrow 1$
Hence (a) is the correct answer.
answered Dec 24, 2013 by sreemathi.v
 
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