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If $a>0, \lim _{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=-1$, then $a$ is equal to
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We have,
$\lim _{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=-1$
Using $L^{\prime}$ hospital's rule
$\Rightarrow \quad \lim _{x \rightarrow a} \frac{a^x \log a-a x^{a-1}}{x^x(1+\log x)}=-1$
$\Rightarrow \quad \frac{a^a \log a-a \cdot a^{a-1}}{a^a(1+\log a)}=-1$
$\Rightarrow \quad \log a-1=-1-\log a$
$\therefore \quad 2 \log _e a=0 \Rightarrow a=e^0=1$
$\lim _{x \rightarrow a} \frac{a^x-x^a}{x^x-a^a}=-1$
Using $L^{\prime}$ hospital's rule
$\Rightarrow \quad \lim _{x \rightarrow a} \frac{a^x \log a-a x^{a-1}}{x^x(1+\log x)}=-1$
$\Rightarrow \quad \frac{a^a \log a-a \cdot a^{a-1}}{a^a(1+\log a)}=-1$
$\Rightarrow \quad \log a-1=-1-\log a$
$\therefore \quad 2 \log _e a=0 \Rightarrow a=e^0=1$
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