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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

The value of $f(0)$ so that the function $f(x) = \frac{(256 - 8x)^{\frac{1}{4}} - 4}{16 - 4(64 + 3x)^{\frac{1}{3}}}$, $x \neq 0$ is continuous at $x = 0$, is

For a certain value of $c$, $\mathop {Lim}\limits_{x \to - \infty } [(x^5 + 7x^4 + 2)^c - x]$ is finite and non-zero. The value of $c$ and the value of the limit are:

$\lim _{x \rightarrow 0} \frac{\sqrt{1+\sqrt{1+x^4}}-\sqrt{2+x^5+x^6}}{x^4} = $

Let $L = \lim_{x \rightarrow 0} \frac{a - \sqrt{a^2 - x^2} - \frac{x^2}{4}}{x^4}$,where $a > 0$. If $L$ is finite,then which of the following is true?

$\mathop {\lim }\limits_{x \to 0} \frac{{{{(1 + x)}^n} - 1}}{x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module