# How do I prove $f(x) = \frac{1}{x-x_0}$ is not uniformly continuous?

Let $E\subset \mathbb{R}$ and $E$ be a noncompact bounded set.

Then, there exists a limit point $x_0$ of $E$ such that $x_0 \notin E$.

Thus, $f(x) = \frac{1}{x-x_0}$ is continuous on $E$.

I can't figure out how to make $d(f(x),f(y))$ arbitrarily large, for some $x,y$ such that $d(x,y) < \delta$ for a given $\delta$.

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You should make a drawing. –  Lierre Oct 15 '12 at 11:50
@Lierre I know it has to be arbitrarily large. I'm asking how do i prove it precisely. –  Katlus Oct 15 '12 at 11:52

Let wlog. $E=(0,1)$ and $x_0=0$. Take for example $x<\frac{1}{n}$ and $y<\frac{1}{2n}$. Then by the inverse triangle inequality, $$|f(x)-f(y)|\ge \left||f(x)|-|f(y)|\right|=n.$$
I meant $E$ to be an arbitrary set. This should not be a proof –  Katlus Oct 15 '12 at 12:00
Well, the above is a proof and anyway you now have a rather good hint to prove it with a general $\,E\,$, +1 –  DonAntonio Oct 15 '12 at 12:06