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.
 A: When you get questions like this (i.e. a one-line limit with a radical), always consider multiplying both numerator and denominator by "something" to make it look nicer. 
In this case, consider what would happen if you multiply both numerator and denominator by $\sqrt{x^2+1} + x$. 
A: Mutiplying the numerator and denominator by $\sqrt{x^2+1}+x$ we get that
$$\begin{align}\lim_{x\to \infty} x(\sqrt{x^2+1} - x) 
&=\lim_{x\to \infty} \frac{x(x^2+1-x^2)}{\sqrt{x^2+1}+x}\\
&=\lim_{x\to \infty} \frac{x}{|x|\sqrt{1+x^{-2}}+x}\\
&=\lim_{x\to \infty} \frac{1}{\sqrt{1+x^{-2}}+1}
\end{align}$$
where we have used the fact that $|x|=x$ because the limit is as $x$ goes to infinity. From here the limit is easy to compute.
A: Hint: Let $t=1/x^2$ and recall the definition of the derivative as a difference quotient.
A: One interesting thing to note is that: 
$k=x\sqrt{x^2+1}-x^2$
Implies that 
$x=\frac{k}{\sqrt{1-2k}}$
And you can then see how $k=\frac{1}{2}$ is the asymptote quite easily. 
A: $\lim_{x\to\infty} x(\sqrt{x^2+1}-x)=\lim_{x\to\infty} \sqrt{x^4+x^2}-x^2$
$=\lim_{x\to\infty}\frac{(\sqrt{x^4+x^2}-x^2)(\sqrt{x^4+x^2}+x^2)}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{x^4+x^2-x^4}{\sqrt{x^4+x^2}+x^2}$
$=\lim_{x\to\infty}\frac{x^2}{\sqrt{x^4+x^2}+x^2}=\lim_{x\to\infty}\frac{1}{\sqrt{1+\frac{1}{x^2}}+1}\rightarrow \frac{1}{2}$
A: 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
