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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

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

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3 Answers 3

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.

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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 $|f(x)-f(y)| < \epsilon$.

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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? –  laovultai 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.

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