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.

Rewrite the given term as
$$x\frac{x^2+1-x^2}{\sqrt{x^2+1}+x}=\frac{x}{\sqrt{x^2+1}+x}$$
Assuming $x>0$, we have
$$\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}$$
This clearly tends to $\frac{1}{2}$ as $x\rightarrow\infty$.
