how can I show that this function is uniformly continuous on $(1,0) $ and $(1,\infty)$

Question

How can I show that $\sqrt x$ is uniformly continuous on the intervals: $(0,1)$ and $(1,\infty)?$

Answer 1

Hint : For the first one, show that $\sqrt{x}$ is even uniformly continuous on $[0,1]$. For the second one, show that $\sqrt{x}$ is lipschitz on the given interval.

Answer 2

For the first one, we just need to prove that $\lim_{x \to \ 0^ +} \sqrt(x)$ and $\lim_{x \to \ 1^ -} \sqrt(x)$ exists.

For the second one,

$\forall x_0 \in(1,\infty),\forall \epsilon>0 $ we want to find $\delta>0$ s.t.

$\forall x',x'' \in(1,\infty) $when $|x'-x''|<0$ ,

we have$|\sqrt(x')-\sqrt(x'')|<\epsilon$

Let $|\sqrt(x')-\sqrt(x'')|=|\frac{(x')-(x'')}{\sqrt(x')+\sqrt(x'')}|<|\frac{(x')-(x'')}{2}|<\epsilon$

we take $\delta=2\epsilon$