evaluate-the-limit-if-it-exists-lim-n-n-4-2n-3-n-2-n-2-3-

Question Number 169527 by Giantyusuf last updated on 01/May/22

e v a l u a t e t h e l i m i t (i f i t e x i s t s)

{l i m}_{n \to \infty} {[\frac{\sqrt{n^{4} - 2 n^{3}} - n^{2}}{n + 2}]}^{3}

Commented by infinityaction last updated on 02/May/22

p = \lim_{n \to \infty} {[\frac{n^{2} (\sqrt{1 - \frac{2}{n}} - 1)}{n + 2}]}^{3}

use binomail theorem for rational index

{(1 + x)}^{n} = 1 + nx + \frac{n (n + 1)}{2!} x^{2} + \dots . . where - 1 ⪇ x ⪇ 1

p = \lim_{n \to \infty} {[\frac{1 - \frac{1}{n} - 1}{\frac{1}{n} (1 + \frac{2}{n})}]}^{3}

p = \lim_{n \to \infty} {[\frac{- 1}{(1 + \frac{2}{n})}]}^{3} = - 1

Answered by cortano1 last updated on 02/May/22

\lim_{x \to \infty} {(\frac{\sqrt{x^{4} - 2 x^{3}} - x^{2}}{x + 2})}^{3} = {(\lim_{x \to \infty} \frac{x^{2} \sqrt{1 - \frac{2}{x}} - x^{2}}{x + 2})}^{3}

= {(\lim_{x \to \infty} \frac{\sqrt{1 - \frac{2}{x}} - 1}{\frac{1}{x} + \frac{2}{x^{2}}})}^{3}

= {(\lim_{x \to 0} \frac{\sqrt{1 - 2 x} - 1}{x + 2 x^{2}})}^{3}

= {(\lim_{x \to 0} \frac{- 2 x}{x (1 + 2 x) (\sqrt{1 - 2 x} + 1)})}^{3}

= {(\lim_{x \to 0} \frac{- 2}{(1 + 2 x) (\sqrt{1 - 2 x} + 1)})}^{3}

= {(\frac{- 2}{1 (1 + 1)})}^{3} = - 1

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