The value of mathop {lim }limits_{x to a} fra | vedclass

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(A) We are given the limit $L = \mathop {\lim }\limits_{x 	o a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$.Since the expression is in the form $\

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$1$

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(A) We are given the limit $L = \mathop {\lim }\limits_{x 	o a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$.Since the expression is in the form $\frac{-\infty}{-\infty}$ as $x 	o a^+$,we apply $L$'$H$ôpital's Rule:$L = \mathop {\lim }\limits_{x 	o a} \frac{\frac{d}{dx}[\log(x-a)]}{\frac{d}{dx}[\log(e^x - e^a)]} = \mathop {\lim }\limits_{x 	o a} \frac{\frac{1}{x-a}}{\frac{e^x}{e^x - e^a}} = \mathop {\lim }\limits_{x 	o a} \frac{e^x - e^a}{(x-a)e^x}$.This is now in the $\frac{0}{0}$ form. Applying $L$'$H$ôpital's Rule again:$L = \mathop {\lim }\limits_{x 	o a} \frac{e^x}{e^x + (x-a)e^x} = \frac{e^a}{e^a + (a-a)e^a} = \frac{e^a}{e^a} = 1$.

The value of $\mathop {\lim }\limits_{x \to a} \frac{{\log (x - a)}}{{\log ({e^x} - {e^a})}}$ is

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