Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$ I've been doing some calculus problems lately out of an old Russian book, and I came across something I didn't fully understand: One of the problems said that $$\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = \frac{1}{2}$$ 
Could someone please explain to my why this is the case?
Thanks a lot.

$\lim_{x\to\infty} x(\sqrt{x^2+1}-x)=\lim_{x\to\infty} \sqrt{x^4+x^2}-x^2$
$=\lim_{x\to\infty}\frac{(\sqrt{x^4+x^2}-x^2)(\sqrt{x^4+x^2}+x^2)}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{x^4+x^2-x^4}{\sqrt{x^4+x^2}+x^2}$
$=\lim_{x\to\infty}\frac{x^2}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}\rightarrow \frac{1}{2}$
