Question 3 in Calculus Proof Questions

[Rule]

See here for an introduction. It is a guide as to the prerequisite knowledge that you need, but if you feel comfortable with single variable calculus you should be fine.

This question is easier than the question in the above link.

Let f be a continuous function on the reals.  Suppose that f'(x) exists whenever x doesn't equal 0, and


Does it follow that f'(0) exists?

[Rule]

Do people want hints or something?  These problems aren't that hard and I thought they were fun in the sense that they pull a lot of basic calculus together to prove things that have some useful meaning.  If you post something that's wrong or only partly right, I'm not going to change my view of your intelligence and then be like "5/10, and ps.you're an idiot".

[Sidoh]

If you post something that's wrong or only partly right, I'm not going to change my view of your intelligence and then be like "5/10, and ps.you're an idiot".

lmao.

These are interesting problems, Rule, and I'd love to work on them.  However, this is the last week of classes with finals next week, so I'm pretty busy.  I know the answers are probably pretty easy to get to if you're clever, but I really don't have the time right now.  Will you still be willing to run this in a week an a half?

[Rule]

Ah, makes sense.  I have just finished my work for the year and have about a week off, so absent-mindedly assumed everyone else was in the same position.

And of course, we can run this indefinitely.   

[Sidoh]

Excellent.

Have a good week off.

[Newby]

Ah, makes sense.  I have just finished my work for the year and have about a week off, so absent-mindedly assumed everyone else was in the same position.

Yeah. I've got a literary criticism paper due Friday, (into next week now) a book talk (synthesizing the information from a book with Vietnam) Monday, AP Spanish test Tuesday, AP Calc BC test Wednesday (!!!), AP US History test Friday.

[Warrior]

5.

Give me something harder next time.

[Ender]

I'm currently very busy but I definitely plan to pour some energy into these questions later in the week.

EDIT: I forgot to mention that I'm really very happy about you posting all these problems ^_^

[Rule]

No problem, I like working through problems like this.  I've written a bunch more, but lets see what comes of these first.  

[d&q]

I'll try to "answer" these..

[Rule]

In 1 week I will post this problem on Yoni's forum.  When he posts a solution, I will link it, or I will post a solution myself in 2 weeks, whichever is faster. 

[Ender]

Yes, it does follow that f'(0) exists.

Hmm, you never said whether this has to be proven or just shown, so my explanation will not be all that rigorous.

Basically, as f is continuous on all the reals, its derivative cannot have any removable discontinuities, i.e. discontinuities where the lefthand and righthand limits agree as x->c but f(c) does not exist. This is because all removable discontinuities are removed in the limit.

To clarify this point, consider



If the lefthand limit agrees with the righthand limit, and f and g are both continuous, then the limit is defined as a number.

Making use of this point, for f'(0) not to exist, there must be a removable discontinuity at x = 0, yet we know there cannot be, since all removable discontinuities are removed in the process of differentiation, i.e.



Thus f'(0) exists.

[Rule]

Your instinct is right, but |x| appears to be a counterexample to what you are saying.  Although maybe not, depending on if you mean something other than the lim x--> c, f/g clarification with this:

This is because all removable discontinuities are removed in the limit.

[Ender]

Your instinct is right, but |x| appears to be a counterexample to what you are saying.  Although maybe not, depending on if you mean something other than the lim x--> c, f/g clarification with this:

This is because all removable discontinuities are removed in the limit.

|x| has a jump discontinuity, and so the lefthand limts and righthands limits don't agree. lim x -> c f(x) / g(x) where the lefthand and righthand limits agree but f(c) / g(c) is undefined is just a poor representation of a removable discontinuity.

Yeah, basically what I meant was that all removable discontinuities are removed in the limit included in the differentiation process.

[Rule]

Your instinct is right, but |x| appears to be a counterexample to what you are saying.  Although maybe not, depending on if you mean something other than the lim x--> c, f/g clarification with this:

This is because all removable discontinuities are removed in the limit.

|x| has a jump discontinuity, and so the lefthand limts and righthands limits don't agree. lim x -> c f(x) / g(x) where the lefthand and righthand limits agree but f(c) / g(c) is undefined is just a poor representation of a removable discontinuity.

Yeah, basically what I meant was that all removable discontinuities are removed in the limit included in the differentiation process.

Supposing that the derivative of a continuous function has no removable discontinuities doesn't rule out |x| as a counterexample.  Of course |x| doesn't show that f'(0) doesn't have to exist, as it says in the question that lim x--> 0 f'(x) exists.  I think I see what you're saying, but it's not totally convincing .

For example,

Basically, as f is continuous on all the reals, its derivative cannot have any removable discontinuities

does not rule out other cases where f'(0) might not exist.

[Ender]

But since it says in the question that lim x -> 0 f'(x) exists, doesn't that imply the lefthand and righthand limits agree? This rules out jump discontinuities. Also, as infinity as a limit does not make the limit exist -- since limits only exist as numbers -- this rules out infinite discontinuities as well. So we've ruled out removable discontinuities, jump discontinuities, and infinite discontinuities,  thus there can be no discontinuity at x = 0 since there are no other types of discontinuities left for all that single var. calc is concerned.

I guess I should include this in my proof to complete it, but I'll wait for you to respond.

By the way, this question seems to be rooted in analysis, albeit basic analysis.

[Rule]

Yes, this falls under analysis.  I'm being screamed at to get out the door, so I will have to reply properly later. 

[Rule]

since all removable discontinuities are removed in the process of differentiation

Your reasoning is more clear in your most recent post.  Though I think you need to further substantiate the claim in the quotation here.  It is intuitive, but I don't think you've "proven" it. 

Good, though!

I will write three proofs and post them here when the vL ftp comes back up! (I store the images there so they will be semi-permanent)

[Ender]

Okay, here's the latest "proof" that I blogged. I'm interested to see whether it passes =p

http://www.rolyata.com/?p=31

[Rule]

There is a problem with your definition of removable discontinuity:

As f is continuous on all the reals, its derivative cannot have any removable discontinuities, i.e. discontinuities where the lefthand and righthand limits agree as x->c but f(c) does not exist

The function
f(x) =  { x^2  for x!= 0
            9      for x = 0
contains a removable discontinuity at x = 0, but f(0) exists.

With this in mind, if f'(x) were to have a removable discontinuity at x=0, it is possible that f'(0) may still exist.

So for this particular question, infinite removable discontinuities should be the only concern.  I don't think you've shown that finite removable discontinuities are ruled out, but they won't cause problems for f'(0) not existing.

Despite this, your proof is fairly convincing.  It's not very rigorously stated, but that's ok.

You can find a more immediate way of showing f'(0) exists using the Mean Value Theorem or L'Hospital's rule in combination with the definition of the derivative.

[Ender]

We first fix h, with h > 0. We know that f is continuous on [0, h] and differentiable on (0, h) so we can use the Mean Value Theorem. By MVT we know there exists an x = x(h) in (0, h) such that

f'(x) = [f(h) - f(0)]/h

Since 0 < x < h, observe that as h->0 it is necessary that x->0. Thus lim h->0 f'(x) = lim x->0 f'(x) which we know exists, and it follows that

f'(0) = lim h->0 [f(h) - f(0)]/h = lim h->0 f'(x) = lim x->0 f'(x).

We conclude that f'(0) exists and f'(0) = lim x->0 f'(x).

[Rule]

We first fix h, with h > 0. We know that f is continuous on [0, h] and differentiable on (0, h) so we can use the Mean Value Theorem. By MVT we know there exists an x = x(h) in (0, h) such that

f'(x) = [f(h) - f(0)]/h

Since 0 < x < h, observe that as h->0 it is necessary that x->0. Thus lim h->0 f'(x) = lim x->0 f'(x) which we know exists, and it follows that

f'(0) = lim h->0 [f(h) - f(0)]/h = lim h->0 f'(x) = lim x->0 f'(x).

We conclude that f'(0) exists and f'(0) = lim x->0 f'(x).

Good job

[Rule]

In a binge of procrastination, I am going to write an alternate proof:

Given: lim x-->0  f'(x) = 9.   f is continuous on the reals. f is differentiable except possibly at x=0.

Proof:

Since f is continuous on the reals, we know f(0) exists.

Let g(x) = f(x) - f(0)

Consider: lim x--> 0 g(x)/x
lim x-->0 g(x) = 0, lim x-->0 x = 0.
This, together with the differentiability of f(x), except perhaps where x=0, allows us to invoke L'Hospital's Rule.

e.g.  lim x-->0  [d/dx g(x)]/[d/dx x] = lim x-->0 g(x)/x

The limit on the left hand side is lim x-->0 f'(x) = 9.
The limit on the right hand side is lim x-->0 [f(x)-f(0)]/[x-0] = f'(0).
Therefore f'(0) exists and equals 9.