If a 0 and lim _{x rightarrow a} frac{a^x - | vedclass

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(B) Given the limit: $\lim _{x ightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$. Applying $L'	ext{Hospital's rule}$ since the form is $\frac{0}{0}$:

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$1$

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(B) Given the limit: $\lim _{x ightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$. Applying $L'	ext{Hospital's rule}$ since the form is $\frac{0}{0}$: $\lim _{x ightarrow a} \frac{\frac{d}{dx}(a^x - x^a)}{\frac{d}{dx}(x^x - a^a)} = -1$ $\Rightarrow \lim _{x}$ ${ightarrow a} \frac{a^x \ln a - a x^{a-1}}{x^x(1 + \ln x)} = -1$ Substituting $x = a$: $\frac{a^a \ln a - a \cdot a^{a-1}}{a^a(1 + \ln a)} = -1$ $\Rightarrow \frac{a^a \ln a - a^a}{a^a(1 + \ln a)} = -1$ Dividing numerator and denominator by $a^a$: $\frac{\ln a - 1}{1 + \ln a} = -1$ $\Rightarrow \ln a - 1 = -1 - \ln a$ $\Rightarrow 2 \ln a = 0$ $\Rightarrow \ln a = 0$ $	herefore a = e^0 = 1$.

If $a > 0$ and $\lim _{x \rightarrow a} \frac{a^x - x^a}{x^x - a^a} = -1$,then $a$ is equal to

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