# Coming up with an example, a function that is continuous but not uniformly continuous

What would be a example of a function that is continuous, but not uniformly continous? will f(x)=1/x be a example? Give domain from (0,2), why? Strictly, in definitions

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Clearly $\,\displaystyle{\frac{1}{x}}\,$ is continuous in $\,(0,2)\,$ as it is the quotient of two polynomials and the denominator doesn't vanish there.

Now, if the function was uniformly continuous there then

$$\forall\,\epsilon>0\,\,\exists\,\delta>0\,\,s.t.\,\,|x-y|<\delta\Longrightarrow \left|\frac{1}{x}-\frac{1}{y}\right|<\epsilon$$

But taking $\,\epsilon=1\,$ , then for any $\,\delta>0\,$ we take

$$x:=\min(\delta,1)\,\,,\,y=\frac{x}{2}\Longrightarrow |x-y|=\frac{x}{2}<\delta, \,\text{but nevertheless}$$

$$\left|\frac{1}{x}-\frac{1}{y}\right|=\left|\frac{1}{x}-\frac{2}{x}\right|=\left|\frac{1}{x}\right|\geq 1=\epsilon$$

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+1 for this crystal way. –  Babak S. Jan 30 '13 at 12:17

$f(x)=\frac{1}{x}$ is an example, since its derivative is unbounded on $(0,2)$. This easily follows from definition.

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So what? $\sqrt{x}$ defined on $(0,2)$ has unbounded derivative, yet is uniformly continuous. –  Chris Eagle Jan 30 '13 at 10:52

Here is a very simple graphical example, where the domain of the function is the interval of the $x$ axis between zero and the last doted line, minus the white points.

The function fails to be uniformly continuous at the two first white points. It is however continuous.

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Its domain is not $(0,2)$, though. –  Henning Makholm Dec 15 '13 at 3:07
Right, this only answers the first part of the question. –  Martin Van der Linden Dec 15 '13 at 9:25
I could be wrong, but I'm pretty sure this function isn't continuous at the white dot points because it doesn't satisfy Weierstrass for all $\epsilon$. –  René G Jan 5 '14 at 16:29