If $a > 0, \underset{x \to a}{L t} \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$ then $' a' =$

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$e$

$2 e$

answer is B.

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Detailed Solution

If $a > 0, \underset{x \to a}{L t} \frac{a^{x} - x^{a}}{x^{x} - a^{a}} = - 1$

$\Rightarrow \underset{x \to a}{L t} \frac{a^{x} \log a - a x^{a - 1}}{x^{x} (1 + \log x)} = - 1$

$\Rightarrow \frac{a^{a} \log a - a a^{a - 1}}{a^{a} (1 + \log a)} = - 1$

$\Rightarrow \frac{a^{a} (\log a - 1)}{a^{a} (\log a + 1)} = - 1$

$\Rightarrow \log a - 1 = - \log a - 1$

$\Rightarrow 2 \log a = 0$

$\Rightarrow \log a = 0$

$\Rightarrow a = 1$

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