# Proving uniform continuity of $g(x)=\sqrt{x^2 + 1}$ on $(0, 1)$

I'm not quite sure how to do this. Here is the definition of uniform continuity (which my professor stipulated we must use to prove this):

$$\forall\epsilon >0, \exists\delta > 0, \forall x,y\in (0, 1), |x-y|<\delta \Rightarrow |f(x)-f(y)| <\epsilon$$

So, I know I must fix an arbitrary epsilon... how do I choose the delta? Forgive me for not having much to start with, my book (Springer) doesn't have concrete examples of uniform continuity proofs to go off of.

• Note that a continuous function on a compact is uniformly continuous (thus the $2$ that appear in user284331's answer, we bound the expression relatively to the lower and upper bounds of the compact). Here $(0,1)$ is not compact, but the $\sqrt{x^2+1}$ is trivially extendable by continuity on $[0,1]$.
– zwim
Nov 10, 2017 at 4:23

$\left|\sqrt{x^{2}+1}-\sqrt{y^{2}+1}\right|=\dfrac{|x+y||x-y|}{\sqrt{x^{2}+1}+\sqrt{y^{2}+1}}\leq\dfrac{|x+y||x-y|}{2}\leq\dfrac{2|x-y|}{2}=|x-y|$. Simply choose $\delta=\epsilon$.
• Hmm, in your second expression, did you determine that you could maximize that expression by making $x^2$ and $y^2$ both equal to $0$? Clever. So for someone new to these types of proofs (epsilon delta) can you please explain why we always start with the epsilon expression and transform it into the delta? Not sure why we always do that Nov 10, 2017 at 3:29
• Yes, take $x=y=0$ to maximize that. Because you want $...\delta$ implies $...\epsilon$, if you start with $\delta$, it seems that you will probably go too further away to have something to do with $\epsilon$. Start with $\epsilon$, then a sequence of inequalities of the form $\leq...\leq...\leq$ lead to $|x-y|$, then $|x-y|<\delta$ implies $...\epsilon$ will go through. Nov 10, 2017 at 3:36