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Question

(C) We have the limit $\mathop {\lim }\limits_{x 	o \infty } \,{\left( {\frac{{x + a}}{{x + b}}} ight)^{x + b}}$.Rewrite the expression inside the

Vedclass · Accepted Answer

$e^{a - b}$

Vedclass · Answer

(C) We have the limit $\mathop {\lim }\limits_{x 	o \infty } \,{\left( {\frac{{x + a}}{{x + b}}} ight)^{x + b}}$.Rewrite the expression inside the limit as $\mathop {\lim }\limits_{x 	o \infty } \,{\left( {1 + \frac{{a - b}}{{x + b}}} ight)^{x + b}}$.Using the standard limit formula $\mathop {\lim }\limits_{u 	o \infty } (1 + \frac{k}{u})^u = e^k$,where $u = x + b$ and $k = a - b$:$\mathop {\lim }\limits_{x 	o \infty } \,{\left( {1 + \frac{{a - b}}{{x + b}}} ight)^{x + b}} = e^{a - b}$.Thus,the correct option is $C$.

$\mathop {\lim }\limits_{x \to \infty } \,{\left( {\frac{{x + a}}{{x + b}}} \right)^{x + b}} = $

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