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

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Question

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

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) Let $L = \mathop {\lim }\limits_{x 	o \infty } {x^{\frac{1}{3}}}\left( {{{\left( {x + 1} ight)}^{\frac{2}{3}}} - {{\left( {x - 1} ight)}^{\frac{2}{3}}}} ight)$.We can rewrite the expression as $L = \mathop {\lim }\limits_{x 	o \infty } x \left( {\left( {1 + \frac{1}{x}} ight)^{\frac{2}{3}} - \left( {1 - \frac{1}{x}} ight)^{\frac{2}{3}}} ight)$.Using the binomial expansion $(1+u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2$ for small $u$,where $u = \pm \frac{1}{x}$ and $n = \frac{2}{3}$:$\left( {1 + \frac{1}{x}} ight)^{\frac{2}{3}} \approx 1 + \frac{2}{3x} + \frac{\frac{2}{3}(\frac{2}{3}-1)}{2x^2} = 1 + \frac{2}{3x} - \frac{1}{9x^2}$.$\left( {1 - \frac{1}{x}} ight)^{\frac{2}{3}} \approx 1 - \frac{2}{3x} + \frac{\frac{2}{3}(\frac{2}{3}-1)}{2x^2} = 1 - \frac{2}{3x} - \frac{1}{9x^2}$.Substituting these into the expression:$L = \mathop {\lim }\limits_{x 	o \infty } x \left( (1 + \frac{2}{3x} - \frac{1}{9x^2}) - (1 - \frac{2}{3x} - \frac{1}{9x^2}) ight)$.$L = \mathop {\lim }\limits_{x 	o \infty } x \left( \frac{4}{3x} ight) = \frac{4}{3}$.

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

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