mathop {lim }limits_{x to infty } {left( {fra | vedclass

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Question

(B) Let $A = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 2}}{{x + 1}}} ight)^{x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } {\lef

Vedclass · Accepted Answer

$e$

Vedclass · Answer

(B) Let $A = \mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 2}}{{x + 1}}} ight)^{x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{1}{{x + 1}}} ight)^{x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } {\left[ {{{\left( {1 + \frac{1}{{x + 1}}} ight)}^{x + 1}}} ight]^{\frac{{x + 3}}{{x + 1}}}}$Since $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{1}{{x + 1}}} ight)^{x + 1}} = e$ and $\mathop {\lim }\limits_{x 	o \infty } \frac{{x + 3}}{{x + 1}} = 1$,$A = e^1 = e$.

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$ is

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