Epsilon/Delta.

**pearlyc** · August 11th 2008, 11:47 PM

Just needed a double check on my working to see if I am on the right track!

Prove that $f(x) = x^4$ is continuous at $x = 0$ .

$| f(x) - f(0) | = | x^4 |$

Given $\epsilon > 0$ , we have $| f(x) - f(0)| < \epsilon$ , provided that $| x^4 | < \epsilon$

Therefore, $\delta = \epsilon$ .

Is this correct?

And I have another question about continuity!

Let ${Q}$ denotes the set of rational numbers, consider the function

$g(x)=\left\{\begin{array}{cc}sin x, &\mbox{ } x \in {Q}\\x, &\mbox{ } x \notin {Q}\end{array}\right.$

(i) Prove that g(x) is continuous at x=0.
(ii) State the least upper bound for g on the interval [0,1] and state whether g attains a maximum on [0,1]. Give clear explanations to justify your statements.

**Isomorphism** · August 12th 2008, 12:31 AM

Originally Posted by pearlyc

Just needed a double check on my working to see if I am on the right track!

Prove that $f(x) = x^4$ is continuous at $x = 0$ .

$| f(x) - f(0) | = | x^4 |$

Given [math\epsilon > 0[/tex], we have $| f(x) - f(0)| < \epsilon$ , provided that $| x^4 | < \epsilon$

Therefore, $\delta = \epsilon$ .

Is this correct?

For a given $\epsilon > 0$ , you have to obtain a $\delta$ such that whenever $|x - 0| < \delta, |f(x) - f(0)| < \epsilon$

And you have claimed that $\delta = \epsilon$ will work for all epsilon...

So check if your choice works for $\epsilon = 3$ .

Hints:

Hint 1: So does $|x| < 3 \implies |x^4| < 3$ ?

Hint 2: What about x=2?

Think about it again. You have to break your argument into two parts. For one part choosing $\delta = \epsilon$ will work. Figure out the other part and we can continue.

**pearlyc** · August 12th 2008, 02:34 AM

I guess it does not satisfy all epsilons :\ But I don't know what went wrong. I am really confused with this whole epsilon/delta business but I really want to know how to nail it as I know it's a very fundamental concept to understand if I wanna further my degree in Math.

Is there like a guide or something :\?

**Plato** · August 12th 2008, 03:54 AM

Because $\left| x \right| < 1\quad \Rightarrow \quad x^4 < 1$
so you want to make sure that $\delta = \min \left\{ {1,\varepsilon } \right\}$ .

**Isomorphism** · August 12th 2008, 04:02 AM

To elaborate on:

Originally Posted by Isomorphism

For a given $\epsilon > 0$ , you have to obtain a $\delta$ such that whenever $|x - 0| < \delta, |f(x) - f(0)| < \epsilon$

And you have claimed that $\delta = \epsilon$ will work for all epsilon...

So check if your choice works for $\epsilon = 3$ .

Hints:

Hint 1: So does $|x| < 3 \implies |x^4| < 3$ ?

Hint 2: What about x=2?

Think about it again. You have to break your argument into two parts. For one part choosing $\delta = \epsilon$ will work. Figure out the other part and we can continue.

Observe that for $\epsilon < 1, \epsilon ^4 < 1$ and thus choosing $\delta = \epsilon$ will work.

However when $\epsilon \ge 1$ , the above argument fails.

To understand what we need to do with this specific case, lets work with a concrete example. How will you choose $\delta$ if i give $\epsilon = 3$ ?

I want a $\delta$ such that $|x| < \delta \implies |x^4| < 3$ .

Note that it suffices if I find x's such that $|x^4| < 1$ , since 1 < 3.

But for all $|x| < 1, |x^4| < 1$ . Thus if

a) $\epsilon < 1$ then choose $\delta = \epsilon$

b) $\epsilon \ge 1$ then choose $\delta = 1$

You are actually done with the problem. But generally there is a tricky way of combining both parts and writing $\delta = \text{min}\{1,\epsilon\}$ .

NOTE:If fourth root is defined to you, you dont need all this circus. The problem is straightforward then.

**pearlyc** · August 12th 2008, 04:44 AM

therefore i just have to change my final part of the working to $\delta = \text{min}\{1,\epsilon\}$ ?

how about the 2nd question? i don't know where to start. i reckon the approach is almost similar. but what is L in this case?

**pearlyc** · August 13th 2008, 03:04 AM

wait, should i let $\delta = \epsilon^{\frac{1}{4}}$ instead?

**kalagota** · August 13th 2008, 03:14 AM

Originally Posted by pearlyc

wait, should i let $\delta = \epsilon^{\frac{1}{4}}$ instead?

yes, that is better..

**pearlyc** · August 13th 2008, 04:37 AM

would the answer still be the same ?

or would it be $\delta = min \{1,\epsilon^{\frac{1}{4}}\}$ ?

**kalagota** · August 13th 2008, 05:22 AM

Originally Posted by pearlyc

would the answer still be the same ?

or would it be $\delta = min \{1,\epsilon^{\frac{1}{4}}\}$ ?

no, it is just the same, it can be $\delta = \epsilon^{\frac{1}{4}}$ since you can take any positive epsilon.

for example, if you take $\epsilon = 16$ , taking $\delta = \epsilon^{\frac{1}{4}} = 2$ such that if $|x| < \delta=2$ , then $|x^4| < \delta^4 =16=\epsilon$

the more important thing is, you should be able to come up with $|f(x) - f(a)| < \epsilon$

unlike if you only take $\delta = \epsilon = 16$ , if $|x| < \delta=16$ , then $|x^4| < \delta^4 =16^4$ and after that, you are not sure whether it will still be less that $16$ which is your epsilon.

**pearlyc** · August 13th 2008, 09:56 PM

okay. so my working above is sufficient? all i have to do is just change the delta value right? how about the 2nd question? how do i start with it?

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