# Showing a function is continuous but not uniformly continuous

Let $f(x)=\frac{1}{x-3}$ on $(3,4]$. I need to show that f is continuous in this interval, but not uniformly continuous.

Idea: Since $f'(x)=\frac{-1}{(x-3)^{2}}$, the derivative exists at all points in $(3,4]$ then $f$ is continuous in $(3,4]$ But if we look at uniform continuity, $\lim_{x \to a^{+}}f(x)$ is infinity while $f(3)$ is simply undefined. Is this accurate to say? More worried about the second part.

-
For continuity, all you really need is that it is a rational function, and those are continuous wherever they are defined. –  Francis Adams Nov 28 '12 at 3:38
The derivative exists at all points in (3,4], so as you note, $f$ is continuous there. But is the derivative $f'$ bounded on (3,4]? What is the implication for uniform continuity? –  hardmath Nov 28 '12 at 3:40

To show that $f$ is not uniformly continuous:

Let $\varepsilon=1$. Consider two sequences $(x_n)=3+\frac{1}{n}$, $(y_n)=3+\frac{1}{n+2}$ which are contained in $(3,4]$ and $\lim x_n-y_n=0$, then for all $\delta>0$ exists $N$ such that $|x_n-y_n|<\delta$ for every $n > N$, but note that:

$$|f(x_n)-f(y_n)|=\left|\frac{1}{3+\frac{1}{n}-3}-\frac{1}{3+\frac{1}{n+2}-3}\right|=|n-(n+2)|=2 > \varepsilon$$

Therefore, $f$ is not uniformly continuous in $(3,4]$.

-

It is not too hard to show that any uniformly continuous function defined on a bounded set has to be bounded, e.g. using sequences and the Bolzano Weierstrass theorem.

-