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If $\lim\limits_{x\to a}\large\frac{a^x-x^a}{x^x-a^a}$$=-1$ and $a > 0$ then $a$=?

$(a)\;0\qquad(b)\;1\qquad(c)\;e\qquad(d)\;2e$

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$\lim\limits_{\large x\to a}\large\frac{a^x-x^a}{x^x-a^a}\big(\large\frac{0}{0}\big)$
$\Rightarrow \lim\limits_{\large x\to a}\large\frac{a^x\log a-ax^{a-1}}{x(1+\log x)}$ (By L Hospital rule)
$\Rightarrow \large\frac{a^a\log a-a.a^{a-1}}{a^a(1+\log a)}$
$\Rightarrow \large\frac{a^a(\log a-1)}{a^a(1+\log a)}$
$a=1$
Hence (b) is the correct answer.
answered Jan 6, 2014 by sreemathi.v
 

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