<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] Let
$$a=\lim_{x\to\infty}x\left(\sqrt{x^2+1}-x\right)$$
Now rationalize it:
$$a=\lim_{x\to\infty}\dfrac{x\left(x^2+1-x^2\right)}{\sqrt{x^2+1}+x}$$
Then simplify it:
$$a=\lim_{x\to\infty}\dfrac{x}{\sqrt{x^2+1}+x}$$
$$a=\lim_{x\to0}\dfrac{\dfrac1x}{\sqrt{\left(\dfrac1x\right)^2+1}+\dfrac1x}$$
$$a=\lim_{x\to0}\dfrac{1}{\sqrt{x^2+1}+1}$$
$$a=\dfrac{1}{\sqrt{0^2+1}+1}=\dfrac{1}{1+1}=\dfrac12$$
</s>