Is the function $f(x) = |x|$ uniformly continuous on the interval $ (0,1] $?

Question

Is the function $f(x) = |x|$ uniformly continuous on the interval $ (0,1] $ ? is it suffices to show that $ f(x) = x $ is uniformly continuous on $(0,1]$ or not ? Since $f(x) = |x| = x$ on $(0,1]$

Answer 1

Let $\epsilon>0$. if $|x-y|<\delta$ then $|f(x)-f(y)|=||x|-|y||\le{}|x-y|<\delta$. thus, taking $\delta=\epsilon$, we see that $$|x-y|<\delta$$ implies $$|f(x)-f(y)|<\epsilon$$ for all $x,y\in{}(0,1]$ (actually, on all $\mathbb{R}$). so f is uniformly continuous on $(0,1]$.

Answer 2

Yes. Let $\delta=\epsilon$ then for all $x,x'\in(0,1]$ such that $\vert x-x'\vert<\delta$ we have by triangle inequality

$$\vert f(x)-f(x')\vert\le \vert x-x'\vert<\epsilon$$