<s>[INST] Why is $\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = 1/2$ I've been doing some calculus problems lately out of an old Russian book, and I came across something I didn't fully understand: One of the problems said that $$\lim_{x\to \infty} x(\sqrt{x^2+1} - x) = \frac{1}{2}$$ 
Could someone please explain to my why this is the case?
Thanks a lot.
 [/INST] Another approach based on Taylor expansions.
Consider $$y=x(\sqrt{x^2+1} - x)=x^2\left(\sqrt{1+\frac{1}{x^2}}-1\right)$$ and remember that, for small $y$, $\sqrt{1+y}=1+\frac{y}{2}-\frac{y^2}{8}+O\left(y^3\right)$. Replace $y$ by $\frac{1}{x^2}$ to get $$\sqrt{1+\frac{1}{x^2}}-1=\frac{1}{2 x^2}-\frac{1}{8 x^4}+O\left(\frac{1}{x^5}\right)$$ which makes $$y=\frac{1}{2}-\frac{1}{8 x^2}+O\left(\frac{1}{x^3}\right)$$ showing the limit and also how it is approched
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