lim-x-x-1-x-2-1-1-

Question Number 141356 by physicstutes last updated on 17/May/21

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

Answered by bemath last updated on 18/May/21

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

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

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

Answered by Mathspace last updated on 18/May/21

f (x) = \frac{x + 1}{\sqrt{x^{2} + 1}} - 1 \Rightarrow f o r x > 0

f (x) = \frac{1 + \frac{1}{x}}{\sqrt{1 + \frac{1}{x^{2}}}} - 1

\Rightarrow f (x) \sim \frac{1 + \frac{1}{x}}{1 + \frac{1}{2 x^{2}}} - 1

f (x) \sim (1 + \frac{1}{x}) (1 - \frac{1}{2 x^{2}}) - 1

= 1 - \frac{1}{2 x^{2}} + \frac{1}{x} - \frac{1}{2 x^{3}} - 1 \Rightarrow

f (x) \sim \frac{1}{x} - \frac{1}{2 x^{2}} \to 0 (x \to + \infty) \Rightarrow

{l i m}_{x \to + \infty} f (x) = 0