If mathop {lim }limits_{x to a} frac{{{a^x} - | vedclass

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(A) Applying $L$-Hospital's rule to the expression $\mathop {\lim }\limits_{x 	o a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}}$,we differentiate the num

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$a = 1$

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(A) Applying $L$-Hospital's rule to the expression $\mathop {\lim }\limits_{x 	o a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}}$,we differentiate the numerator and denominator with respect to $x$:Numerator: $\frac{d}{dx}(a^x - x^a) = a^x \ln a - a x^{a-1}$Denominator: $\frac{d}{dx}(x^x - a^a) = x^x(1 + \ln x) - 0 = x^x(1 + \ln x)$Evaluating the limit as $x 	o a$:$-1 = \frac{a^a \ln a - a(a^{a-1})}{a^a(1 + \ln a)} = \frac{a^a(\ln a - 1)}{a^a(1 + \ln a)} = \frac{\ln a - 1}{\ln a + 1}$$-1(\ln a + 1) = \ln a - 1$$-\ln a - 1 = \ln a - 1$$2 \ln a = 0$ $\Rightarrow \ln a = 0$ $\Rightarrow a = 1$.

If $\mathop {\lim }\limits_{x \to a} \frac{{{a^x} - {x^a}}}{{{x^x} - {a^a}}} = - 1$,then

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