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Is it correct/ true to say that since $f=1/\sqrt{x}$ is continuous on the interval $(0,1)$, so f is uniformly continuous?

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No unbounded function on a bounded region is uniformly continuous. –  Thomas Andrews Dec 3 '12 at 5:08
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No. A function that is continuous on a compact set is uniformly continuous (cf Theorem 4.19 in Rudin's Principles).

The idea of uniform continuity is that for every $\varepsilon$ you can pick a $\delta$ so that independent of where you go in the domain \begin{equation} |x_0 - x_1| < \delta \Rightarrow |f(x_0) - f(x_1)| < \varepsilon. \end{equation}

Now, because for your function you won't be able to do this. For any fixed $\delta$ if you go close enough to zero then $|f(x_0) - f(x_1)| > \varepsilon$.

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Simply having the derivative get arbitrarily steep near $0$ doesn't preclude uniform continuity on $(0,1)$. Consider the function $g(x)=\sqrt[3]{x}$, for example. –  Cameron Buie Dec 3 '12 at 5:17
    
You're right. I've corrected the answer. Thanks! –  Alex Dec 3 '12 at 5:22
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$1/\sqrt x$ is not uniformly continuous on the interval $(0,1)$.

If $f$ is continuous on the closed interval $[a,b]$ then it is uniformly continuous on the interval $[a,b]$

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