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?
 A: 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*}
A: HINT
$$\sqrt{x^2+c} - x = \dfrac{c}{\sqrt{x^2+c}+x}$$
A: Note that for positive $x$ we have 
$$x\le \sqrt{x^2+c}\le x+\frac{c}{2x}.$$
This is because 
$$\left(x+\frac{c}{2x}\right)^2=x^2+c+\frac{c^2}{4x^2}\ge x^2+c.$$
Thus 
$$0\le \sqrt{x^2+c}-x\le \frac{c}{2x}.$$
Now Squeeze.
A: Hints: 
$$\sqrt{x^2+c}=|x|\sqrt{1+\frac{c}{x^2}} = |x|(1+\frac{c}{2x^2} + o(\frac{c}{x^2}))$$
A: something you might notice that $\lim_{x\to\infty}\sqrt{x^2 + c} = x$ is trivially true for positive and negative finite values of c.
$$\lim_{x\to\infty}\sqrt{x^2 + c} -\lim_{x\to\infty}x = U$$
$$\lim_{x\to\infty}x-\lim_{x\to\infty}x = U$$
$$\lim_{x\to\infty}x -\lim_{x\to\infty}x = 0$$
perhaps not the most mathematically rigorous argument, but it works.
A: \begin{align*}
\lim_{ x \to \infty}\sqrt{x^2+c}-x & =\lim_{ x \to \infty}(\sqrt{x^2+c}-x)\frac{\sqrt{x^2+c}+x}{\sqrt{x^2+c}+x}\\
&=\lim_{ x \to \infty}\frac{x^2-x^2+c}{\sqrt{x^2+c}+x}\\
&=\lim_{ x \to \infty}\frac{c}{\sqrt{x^2+c}+x} \\ & =0.
\end{align*}
