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.
 A: Just for fun, try letting $x=\tan\theta=\sin\theta/\cos\theta$ with $\theta\to\pi/2^-$, and use
$$\sqrt{\tan^2\theta+1}=\sqrt{\sec^2\theta}=\sec\theta={1\over\cos\theta}$$
so that
$$\begin{align}
(\sqrt{x^2+1}-x)(x+1)
&=\left({1\over\cos\theta}-{\sin\theta\over\cos\theta}\right)\left({\sin\theta\over\cos\theta}+1\right)\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over\cos^2\theta}\\
&={(1-\sin\theta)(\sin\theta+\cos\theta)\over1-\sin^2\theta}\\
&={\sin\theta+\cos\theta\over1+\sin\theta}
\end{align}$$
We get
$$\lim_{x\to\infty}(\sqrt{x^2+1}-x)(x+1)=\lim_{\theta\to\pi/2^-}{\sin\theta+\cos\theta\over1+\sin\theta}={1+0\over1+1}={1\over2}$$
A: 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?
