Prove a function is uniformly continuous

Question

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

Answer 1

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.

Answer 2

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 $\mathbb R.$