Is $f(x) = 1/x$ over $[0.1 , 1]$ uniformly continuous but $1/x$ on $(0,1)$ is not?

Solution 1:

Uniform Continuous function

To understand it clearly, you should think what uniform continuity looks like. Let me illustrate it.

Given function is:
$f(x) = \frac{1}{x}$

Claim: In the interval (0, 1), you cannot find $\delta$ corresponding to any $\epsilon$.

To understand this, fix $\epsilon = 1$, and choose whatever $\delta$ you think will satisfy the definition. You will find that this will not work. Reduce $\delta$ by half. You will find this will also not work. Keep on dividing it by half. You will find that no such $\delta$ works.

This happens because $f$ grows too fast in the neighbourhood of 0 so that no such $\delta$ sized sub-interval of (0, 1) can capture the growth of $f$.

In the case of interval [0.1, 1], there is no such neighborhood where $f$ grows rapidly (which is true for continuous functions on closed and bounded intervals), therefore you can find a $\delta>0$ corresponding to every $\epsilon>0$.

How to find $\delta$ corresponding to $\epsilon$ in the interval [0.1, 1]:

Choose an arbitrary $\epsilon > 0$.
$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{|xy|}$

Using the fact that $x, y \geq 0.1$,
$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{|xy|} \leq 100 |x-y| < \epsilon$

Clearly, $\delta = \frac{\epsilon}{100}$

Solution 2:

$$\left|\frac1{x-\delta}-\frac1x\right|=\left|\frac{\delta}{(x-\delta)x}\right|<\epsilon$$ requires

$$\delta<\frac{\epsilon x^2}{\epsilon x+1}<\epsilon x^2$$ so that no finite bound can hold for all $x$.