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

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Question

(B) We use the standard limit formula $\mathop {\lim }\limits_{x 	o \infty } (1 + \frac{a}{x})^{bx} = e^{ab}$.First,rewrite the expression as $\matho

Vedclass · Accepted Answer

$e^{-2/3}$

Vedclass · Answer

(B) We use the standard limit formula $\mathop {\lim }\limits_{x 	o \infty } (1 + \frac{a}{x})^{bx} = e^{ab}$.First,rewrite the expression as $\mathop {\lim }\limits_{x 	o \infty } {\left( {1 + \frac{-6}{3x + 2}} ight)^{\frac{x + 1}{3}}}$.This is of the form $1^{\infty}$,so the limit is $e^{\mathop {\lim }\limits_{x 	o \infty } (\frac{-6}{3x + 2} 	imes \frac{x + 1}{3})}$.Calculate the exponent limit: $\mathop {\lim }\limits_{x 	o \infty } \frac{-6(x + 1)}{3(3x + 2)} = \mathop {\lim }\limits_{x 	o \infty } \frac{-6x - 6}{9x + 6} = -\frac{6}{9} = -\frac{2}{3}$.Thus,the final value is $e^{-2/3}$.

The value of $\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{3x - 4}}{{3x + 2}}} \right)^{\frac{{x + 1}}{3}}}$ is equal to

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