mathop {lim }limits_{x to infty } {left( {1 - | vedclass

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(B) 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

$e^{-12}$

Vedclass · Answer

(B) We use the standard limit formula: $\mathop {\lim }\limits_{x 	o \infty } (1 + \frac{a}{x})^x = e^a$. Given expression: $\mathop {\lim }\limits_{x 	o \infty } (1 - \frac{4}{x - 1})^{3x - 1}$. Let $t = x - 1$,then as $x 	o \infty$,$t 	o \infty$. Substituting $x = t + 1$: $\mathop {\lim }\limits_{t 	o \infty } (1 - \frac{4}{t})^{3(t + 1) - 1} = \mathop {\lim }\limits_{t 	o \infty } (1 - \frac{4}{t})^{3t + 2}$. This can be written as: $\mathop {\lim }\limits_{t 	o \infty } [(1 - \frac{4}{t})^t]^3 	imes \mathop {\lim }\limits_{t 	o \infty } (1 - \frac{4}{t})^2$. $= (e^{-4})^3 	imes (1 - 0)^2 = e^{-12} 	imes 1 = e^{-12}$.

$\mathop {\lim }\limits_{x \to \infty } {\left( {1 - \frac{4}{{x - 1}}} \right)^{3x - 1}} = $

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