Prove a function is uniformly continuous 
Prove the function $f(x)=\sqrt{x^2+1}$     $ (x\in\mathbb{R})$ is uniformly continuous.

Now I understand the definition, I am just struggling on what to assign $x$ and $x_0$
Let $\epsilon>0$ we want $|x-x_0|<\delta$ so that $|f(x)-f(x_0)|<\epsilon$
Could anyone help fill in the missing bits? Thanks

You have $$\begin{aligned}\vert f(x)-f(y) \vert &= \left\vert \sqrt{x^2+1}-\sqrt{y^2+1} \right\vert \\
&= \left\vert (\sqrt{x^2+1}-\sqrt{y^2+1}) \frac{\sqrt{x^2+1}+\sqrt{y^2+1}}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\
&= \left\vert \frac{x^2-y^2}{\sqrt{x^2+1}+\sqrt{y^2+1}} \right\vert \\
&\le \frac{\vert x-y \vert (\vert x \vert + \vert y \vert )}{ \sqrt{x^2+1}+\sqrt{y^2+1}} \\
&\le \vert x-y \vert
\end{aligned}$$ hence choosing $\delta = \epsilon$ will work.
