Prove that $ \lim_\limits{x \to \infty} x(\sqrt{x^2+1}-x) $ = $ \frac{1}{2}$?

Question

How can you prove that $$ \lim_\limits{x \to \infty} x(\sqrt{x^2+1}-x) = \frac{1}{2} \text{ ?}$$

I can not find a way to calculate this.

This is one idea: $$ \lim_{x \to \infty} x(\sqrt{x^2+1}-x) \approx \lim_\limits{x \to \infty} x(\sqrt{x^2}-x) = 0 $$ but that is wrong.

Hint: use the formula $x^2 - y^2 = (x+y)(x-y)$ on the expression in brackets. — ajr, Sep 03 '15 at 11:29
See also: Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? and Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$ — Martin Sleziak, Oct 15 '19 at 15:58

score 3 · Accepted Answer · answered Sep 03 '15 at 11:31

3

Hint: Here is another idea $$\lim_{x \to \infty} \frac{x (x^2 +1 - x^2)}{\sqrt{x^2 +1} + x} =\lim_{x \to \infty} \frac{x}{\sqrt{x^2 +1} + x} = \lim_{x \to \infty} \frac{1}{\sqrt{1 +\frac{1}{x^2}} + 1} = \ldots$$

answered Sep 03 '15 at 11:31

Aaron Maroja

17,571

If it's not clear what was done here, see Claude Leibovici's comment under the question. ${}\qquad{}$ – Michael Hardy Sep 03 '15 at 11:39

Prove that $ \lim_\limits{x \to \infty} x(\sqrt{x^2+1}-x) $ = $ \frac{1}{2}$?

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