Showing that $\lim_{x\to\infty}\left(\sqrt{x^2+c}-x\right)=0$ A limit I had to compute recently boiled down to the following limit:
$$\lim_{x\to\infty}\left(\sqrt{x^2+c}-x\right)=0\quad\mbox{for $c\ge0$}$$
How can I show that this limit is correct?

Note that \begin{gather*}
\sqrt{x^2+c}-x=(\sqrt{x^2+c}-x)\frac{\sqrt{x^2+c}+x}{\sqrt{x^2+c}+x}\\
=\frac{x^2-x^2+c}{\sqrt{x^2+c}+x}\\
=\frac{c}{\sqrt{x^2+c}+x}
\end{gather*}
