Find-limit-lim-x-x-x-2-1-x-and-lim-x-x-x-2-1-x-

Question Number 100920 by ajfour last updated on 29/Jun/20

F i n d l i m i t

\lim_{x \to + \infty} x (\sqrt{x^{2} + 1} - x) a n d

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

Commented by bramlex last updated on 29/Jun/20

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

s e t \frac{1}{x} = z, z \to 0

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

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

= \frac{1}{2} ★

Commented by bramlex last updated on 29/Jun/20

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

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

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

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

s e t \frac{1}{- x} = q, q \to 0

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

- \lim_{q \to 0} \frac{\sqrt{1 + q^{2}} + 1}{q^{2}} = - \infty . ★

Answered by mathmax by abdo last updated on 29/Jun/20

for x > 0 we have x (\sqrt{x^{2} + 1} - x) = x (x \sqrt{1 + \frac{1}{x^{2}}} - x) \sim x (x (1 + \frac{1}{{2 x}^{2}}) - x)

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

for x < 0 we have x (\sqrt{x^{2} + 1} - x) = x {- x \sqrt{1 + \frac{1}{x^{2}}} - x}

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

Commented by ajfour last updated on 29/Jun/20

c o r r e c t S i r, i m e a n \lim_{x \to + \infty} f (x) = \frac{1}{2}

a n d \lim_{x \to - \infty} f (x) = - \infty . T h a n k s .

Commented by mathmax by abdo last updated on 29/Jun/20

you are welcome