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

Question

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

Answer 1

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

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

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

Answer 2

• $(1)$ Is $f(x)=1/x$ continuous on $(0,2)$? Yes. Why?
  
  Justify your answer by showing $f(x)$ satisfies the definition of continuity: e.g., since both $1$ and $x$ are polynomials, $x \ne 0$ on $(0, 2)$.

• $(2)$ Now, turn to the definition of uniform continuity of a function on an interval $(a, b)$. In your case $(a, b) = (0, 2)$:
  
  Is $f'(x)$ bounded on $(0, 2)$?
  
  So, by definition, can $f(x)$ be uniformly continuous?

• If you can answer yes to $(1)$ and justify "why", and answer "no" to $(2)$ explaining why not, then you have a fine example of a continuous function $f(x)$ on $(0, 2)$ which is not uniformly continuous there.

Answer 3

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