4
$\begingroup$

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.

$\endgroup$
4

1 Answer 1

3
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ If it's not clear what was done here, see Claude Leibovici's comment under the question. ${}\qquad{}$ $\endgroup$ Sep 3, 2015 at 11:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .