# Thread: Trouble with Uniform continuity calculation

1. ## Trouble with Uniform continuity calculation

So i have to prove that $f(x)=\frac{1}{\sqrt{x}}$ is uniformly continuous using the definition of uniform continuity...

aka I have to start with $|\frac{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{c}}}{x-c} + \frac{1}{2\sqrt{c^3}}|$

and knowing that $|x-c|<\delta$, I have to make the first inequality less than $\epsilon$.

I am getting so lost in the calculation and this problem is due tomorrow, someone please help....

2. Uniformly continuous on what domain?

3. $x>0$

4. Does anyone have any idea??

5. Well, it is NOT uniformly continuous on $(0,\infty)$, and it's easy to show this as follows: Consider the sequence $s_{n} = \frac{1}{n^{2}}$, which is Cauchy, but $f(s_{n}) = n$ is not Cauchy, and so $f$ cannot be uniformly continuous. Not exactly sure how to show this using the $\epsilon - \delta$ definition though.

6. yea sorry the problem says for x>0

7. Yes, and for that interval it is NOT uniformly continuous and to somehow use the epsilon-delta definition, you have to show that $\forall \, \delta > 0, \exists \, \epsilon > 0$ such that for some $x,y \in (0,\infty ), |x-y| < \delta$ and $|f(x) - f(y)| \ge \epsilon$.