lim-x-x-2-x-1-x-pls-solve-this-

Question Number 54409 by pooja24 last updated on 03/Feb/19

l i \underset{x \to \infty}{m} \sqrt{(x^{2} + x + 1)} - x = ?

p l s s o l v e t h i s

Commented by maxmathsup by imad last updated on 03/Feb/19

w e h a v e \sqrt{x^{2} + x + 1} =∣ x ∣ \sqrt{1 + \frac{1}{x} + \frac{1}{x^{2}}} \sim 1 + \frac{1}{2} (\frac{1}{x} + \frac{1}{x^{2}}) (x \to \infty) \Rightarrow

\sqrt{x^{2} + x + 1} - x \sim ∣ x ∣ {1 + \frac{1}{2 x} + \frac{1}{2 x^{2}}} - x \Rightarrow

{l i m}_{x \to + \infty} \sqrt{x^{2} + x + 1} - x = {l i m}_{x \to + \infty} x {1 + \frac{1}{2 x} + \frac{1}{2 x^{2}}} - x

= {l i m}_{x \to + \infty} (\frac{1}{2} + \frac{1}{2 x}) = \frac{1}{2}

{l i m}_{x \to - \infty} \sqrt{x^{2} + x + 1} - x = {l i m}_{x \to - \infty} - x {1 + \frac{1}{2 x} + \frac{1}{2 x^{2}}} - x

= {l i m}_{x \to - \infty} - 2 x - \frac{1}{2} - \frac{1}{2 x} = + \infty .

Answered by kaivan.ahmadi last updated on 03/Feb/19

\times \frac{\sqrt{x^{2} + x + 1} + x}{\sqrt{x^{2} + x + 1} + 1} = li \underset{x \to + \infty}{m} \frac{x + 1}{\sqrt{x^{2} + x + 1} + x} \approx li \underset{x \to + \infty}{m} \frac{x}{2 x} = \frac{1}{2}

or

\approx li \underset{x \to \infty}{m} ∣ x + \frac{1}{2} ∣ - x = \frac{1}{2}

Answered by Prithwish sen last updated on 03/Feb/19

let x = \frac{1}{h}

∴ x \to \infty \Rightarrow h \to 0

li \underset{h \to 0}{m} \frac{\sqrt{h^{2} + h + 1} - 1}{h} (∵ form \frac{0}{0}, we apply

L^{'} Hopital rule)

= li \underset{h \to 0}{m} \frac{2 h + 1}{2 \sqrt{h^{2} + h + 1}}

= \frac{1}{2}