Evaluating $\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $ The question is $$\lim_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $$ I know the answer is $\frac{1}{2}$ and I found it using this equality :
$$(\sqrt{x^2+1} - x)(x+1) = \frac{x+1}{\sqrt{x^2+1} + x}$$
But is there any other way to solve this? Any hints would be appreciated.

Use binomial series:
$\sqrt{1+\dfrac1{x^2}}=1+\dfrac1{2x^2}\cdots$
Addendum with further explanation:
$\lim\limits_{x\to\infty}(\sqrt{x^2+1} - x)(x+1) $
$=\lim\limits_{x\to\infty}\left(\sqrt{1+\dfrac1{x^2}}-1\right)x(x+1)$
$=\lim\limits_{x\to\infty}\left(1+\dfrac1{2x^2}\cdots-1\right)(x^2+x)$
$=\lim\limits_{x\to\infty}\left(\dfrac1{2x^2}\cdots\right)(x^2+x).$
Can you take it from here?
