# Trouble with Uniform continuity calculation

• Mar 24th 2010, 03:30 PM
frenchguy87
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....
• Mar 24th 2010, 04:35 PM
Pinkk
Uniformly continuous on what domain?
• Mar 24th 2010, 04:43 PM
frenchguy87
$x>0$
• Mar 25th 2010, 12:26 PM
frenchguy87
Does anyone have any idea?? :(
• Mar 25th 2010, 12:41 PM
Pinkk
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.
• Mar 25th 2010, 02:39 PM
frenchguy87
yea sorry the problem says for x>0
• Mar 25th 2010, 02:44 PM
Pinkk
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 $|f(x) - f(y)| \ge \epsilon$.