<s>[INST] 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.
 [/INST] the trouble with your approach stems from $0 \times \infty$ being indeterminate.
here is an explanation. 
(a) what you are doing is approximating $\sqrt{x^2+1} = x + \cdots$
(b) $\sqrt{x^2 + 1} - x = 0 + \cdots$ 
(c) $x(\sqrt{x^2+1} - x) = x(0 + \cdots) = 0?$
you think the answer is 0. but, perhaps $x \times \cdots \neq 0?$
now, i will go back and elaborate on that $\cdots$ by using binomial thorem
$$(BIG + small)^{1/2} = BIG^{1/2} + {1 \over 2}BIG^{-1/2}small + \cdots$$
$(x^2+1)^{1/2} = x + {1 \over 2x} + \cdots$ 
 now, if you go thru the steps (a)-(c) again, you see that $x(\sqrt{x^2+1} - x) = x({1 \over 2x}+ \cdots) = 1/2$
the reason is that $\cdots = const . 1/x^2 + \cdots$ and $x \times \cdots = 0$ as $x \to \infty$
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