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

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Question

(A) We are given the limit: $\mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 3}}{{x + 1}}} ight)^{x + 1}}$Rewrite the expression inside t

Vedclass · Accepted Answer

$e^2$

Vedclass · Answer

(A) We are given the limit: $\mathop {\lim }\limits_{x 	o \infty } {\left( {\frac{{x + 3}}{{x + 1}}} ight)^{x + 1}}$Rewrite the expression inside the parenthesis: $\frac{x+3}{x+1} = \frac{x+1+2}{x+1} = 1 + \frac{2}{x+1}$Now the limit becomes: $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{2}{x+1}} ight)^{x+1}}$Using the standard limit formula $\mathop {\lim }\limits_{u 	o \infty } (1 + \frac{k}{u})^u = e^k$,where $u = x+1$ and $k = 2$:$\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{2}{x+1}} ight)^{x+1}} = e^2$

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

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