For x in R,mathop {lim }limits_{x to infty } | vedclass

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(C) We use the standard limit formula $\mathop {\lim }\limits_{x 	o \infty } {(1 + \frac{a}{x})^x} = e^a$.Given expression: $\mathop {\lim }\limits_{

Vedclass · Accepted Answer

$e^{-5}$

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(C) We use the standard limit formula $\mathop {\lim }\limits_{x 	o \infty } {(1 + \frac{a}{x})^x} = e^a$.Given expression: $\mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x - 3}}{{x + 2}}} ight)^x} = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 2 - 5}}{{x + 2}}} ight)^x} = \mathop {\lim }\limits_{x 	o \infty } {\left( {1 - \frac{5}{{x + 2}}} ight)^x}$.Rewrite the expression as: $\mathop {\lim }\limits_{x 	o \infty } {\left[ {{{\left( {1 - \frac{5}{{x + 2}}} ight)}^{\frac{{x + 2}}{{-5}}}}} ight]^{\frac{-5x}{x + 2}}}$.Since $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 - \frac{5}{{x + 2}}} ight)^{\frac{x + 2}{-5}}} = e$ and $\mathop {\lim }\limits_{x 	o \infty } \frac{-5x}{x + 2} = \mathop {\lim }\limits_{x 	o \infty } \frac{-5}{1 + \frac{2}{x}} = -5$.The limit is $e^{-5}$.

For $x \in R$,$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}$ is equal to

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