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

Note that $f'(x)= x/\sqrt {x^2+1}$ for all $x.$ We thus have $|f'(x)|<1$ everywhere. Suppose $x,y\in \mathbb R.$ Then the mean value theorem gives
$$f(y)-f(x) = f'(c)(y-x) \implies |f(y)-f(x)| \le |y-x|.$$
This easily gives the uniform continuity of $f$ on $\mathbb R.$
