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

Question

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.

Answer 1

$\left|\sqrt{x^{2}+1}-\sqrt{y^{2}+1}\right|=\dfrac{|x+y||x-y|}{\sqrt{x^{2}+1}+\sqrt{y^{2}+1}}\leq\dfrac{|x+y||x-y|}{2}\leq\dfrac{2|x-y|}{2}=|x-y|$. Simply choose $\delta=\epsilon$.