# Thread: Trouble with Uniform continuity calculation

1. ## Trouble with Uniform continuity calculation

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

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

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

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

2. Uniformly continuous on what domain?

3. $\displaystyle x>0$

4. Does anyone have any idea??

5. Well, it is NOT uniformly continuous on $\displaystyle (0,\infty)$, and it's easy to show this as follows: Consider the sequence $\displaystyle s_{n} = \frac{1}{n^{2}}$, which is Cauchy, but $\displaystyle f(s_{n}) = n$ is not Cauchy, and so $\displaystyle f$ cannot be uniformly continuous. Not exactly sure how to show this using the $\displaystyle \epsilon - \delta$ 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 $\displaystyle \forall \, \delta > 0, \exists \, \epsilon > 0$ such that for some $\displaystyle x,y \in (0,\infty ), |x-y| < \delta$ and $\displaystyle |f(x) - f(y)| \ge \epsilon$.