Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty}  x\left( (x^2+1)^{\frac{1}{2}} - x\right)  = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.

Let $x=\tan\theta$.
Then $L=\displaystyle\lim_{x\to \infty}  x( (x^2+1)^{\frac{1}{2}} - x )  \\= \displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}}  \tan\theta( \sec\theta - \tan\theta )\\=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}}  \frac{\sin\theta-\sin^2\theta}{\cos^2\theta}\\
=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}}  \frac{\sin\theta-\sin^2\theta}{1-\sin^2\theta}\\
=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}}  \frac{(1-\sin\theta)\sin\theta}{(1-\sin\theta)(1+\sin\theta)}\\
=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}}\frac{\sin\theta}{1+\sin\theta}\\
=\dfrac{1}{2}.$
