Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am having difficulty determining what value I should assign to $\delta$ for the following problem. How do I determine what it should be?

Define $f:[3.4,5] \rightarrow \mathbb{R}$ by $f(x)=2/(x-3)$. Show that $f$ is uniformly continuous on [3.4,5].

This is what I have so far.

Choose $\varepsilon >0,$ and $\delta=???$. If $|x-y|< \delta$ and $x,y \in [3.4,5]$, then

$|f(x)-f(y)|=|2/(x-3)-2/(y-3)| = |2(y-x)/((x-3)(y-3))|<???$

share|cite|improve this question

Here We have a very important result that ensures the following:

"If $f : K \rightarrow \mathbb{R}$ is a continuous function and K is compact, then $f$ is uniformly continuous."

In our case [3.4 , 5] is closed and bounded in the line hence is compact by Heine-Borel's Theorem, $f$ is continuous (note that denominator of the fraction never is zero so we have a quotient of continuous functions hence continuous) Hence the result above apllies and $f$ is uniformly continuous.

share|cite|improve this answer

You are off to a good start. Whenever we have epsilon-delta proofs, we need to use a two step process. First we work to figure out what delta should be. Then we show it works.

Step 1 is to determine what delta should be. This means we start as you did by saying that we WANT: $$ \epsilon > |f(x) - f(y)| = \left| \frac{2}{x-3} - \frac{2}{y-3} \right| = \left| \frac{2(y-x)}{(x-3)(y-3)} \right| $$

Using this we can multiply and rewrite the inequality as $$ \epsilon \frac{\left|(x-3)(y-3)\right|}{2} > \left| y-x \right|$$

As $f$ is defined on the interval $[3.4, 5]$, we can see that $|(x-3)| < 5-3=2$ for all $x \in [3.4, 5]$. Therefore, we have $$ \epsilon \frac{\left|(x-3)(y-3)\right|}{2} < \epsilon \frac{2\cdot2}{2} = 2\epsilon.$$

Now we proceed to step 2, where we choose $\delta = 2 \epsilon.$ We then proceed to show that this choice of delta will make the limit uniformly continuous.

(Proof) Let $\epsilon > 0$ then choose $\delta = 2\epsilon$ and let $x,y \in [3.4, 5]$ and $|x-y| < \delta.$ Then we have $$|x-y| < 2 \epsilon \implies \frac{|x-y|}{2} < \epsilon $$

Now we note that $$ \frac{|x-y|}{2} \geq \frac{2\cdot2}{2} \geq \frac{\left|(x-3)(y-3)\right|}{2} = |f(x) - f(y)|$$ as $x,y \in [3.4, 5]$. We can then rewrite the right side as in step 1 until we have $|f(x) - f(y)|$.

Putting all of the inequalities together yields $|f(x)-f(y)| < \epsilon$.

share|cite|improve this answer
First of all is $|x-y|=|y-x|$? Is it true that now we note that $$ \frac{|x-y|}{2} \geq \frac{2\cdot2}{2} \geq \frac{\left|(x-3)(y-3)\right|}{2} = |f(x) - f(y)|$$(*) as $x,y \in [3.4, 5]$? Especially can you tell me why last equivalence of (*) is true? – alvoutila Oct 23 '13 at 15:49

Your function is continuous over the interval, with continuous derivative there. Weierstrass' theorem ensures that $f'$ attains its maximum, say $M$, over $[3.4\;,\; 5]$. Then, by the Mean Value Theorem, we have that $$|f(x)-f(y)|\leq M|x-y|$$ over $[3.4, \; 5]$. Thus, given $\epsilon>0$, take $\delta =\frac{\epsilon}{M}$ and conclude.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.