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A question professor couldnt solve!

  1. Mar 6, 2007 #1
    The professor said he didnt know the ans of this qn, so, can u help?

    Question: Can you have lim_x->a f '(x) = D , f ' (a) exists, but not equal to D?

    any one can give an example? thx!
  2. jcsd
  3. Mar 6, 2007 #2
    How about f(x) = (x+1)/(x+1)

    The limit of f'(x) as x-> -1 is 0, however f(-1) is undefined, and therefore I believe f'(-1) is undefined as well.

    I'm no mathematician, so can someone comment on whether this reasoning is correct?
  4. Mar 6, 2007 #3
    i think there's a problem with this ans since the limit does not exist if f(x)=(x+1)/(x+1)....hmm... i am not sure..
  5. Mar 6, 2007 #4
    Like I said, it is best to have someone else verify my claim, but I believe the limit does exist. Approach from either side and f'(x) as x-> -1 is just 0. So why do you feel the limit does not exist? f'(-1) may be undefined, but I believe the limit as x-> -1 is.

    Can someone verify this reasoning?
  6. Mar 6, 2007 #5


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    I think the OP is asking if there's an example where both the limit as f'(x) approaches a and f'(a) exist, but are not equal. I don't think this is possible, but I'm having trouble thinking of a simple proof. For example, if you could justify switching the limits, it'd just be:

    [tex] \lim_{x \rightarrow a} f'(x) = \lim_{x \rightarrow a} \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} [/tex]

    [tex] \lim_{h \rightarrow 0} \lim_{x \rightarrow a} \frac{f(x+h)-f(x)}{h} =\lim_{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}=f'(a)[/tex]

    But how to justify that is escaping me right now.
    Last edited: Mar 6, 2007
  7. Mar 6, 2007 #6
    o i have an ans here, but dont know whether,, let's see

    It seems like you should not, because

    D = lim(x->a) f'(x) {Given}
    = lim(x->a) lim(y->x) (f(y)-f(x))/(y-x) {Definition of derivative}
    = lim(a+h->a) lim(a+h+g->a+h) (f(a+h+g)-f(a+h))/g {Substitution}
    = lim(h->0) lim(g->0) (f(a+h+g)-f(a+h))/g {Arithmetic}
    = lim(h->0) lim(g->0) ((h+g)f'(a) - h f'(a)) /g {Definition of derivative}
    = f'(a) lim(h->0) lim(g->0) g/g {f'(a) is just a number, move it outside}
    = f'(a) lim(h->0) 1 {Easy}
    = f'(a) {Easier}

    What do u guys think???
  8. Mar 6, 2007 #7
    Well, wouldn't it be no by the theorem that says if f is differentiable, f' has the intermediate value property.
  9. Mar 6, 2007 #8


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    It does say that? Where?
  10. Mar 6, 2007 #9
    Yeah, the intermediate value property guarantees no discontinuities of this type. You can have discontinuities where the limit fails to exist though.
  11. Mar 6, 2007 #10


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    my notes in the thread elementary proofs of big theorems, in calculus forum, essentially show all derivatives ahve the intermediate vaklue proeprty.

    in fact it foollows from the rolle theoirem. if f' is positive at a and negative at b, then f is not monotone in the interval [a,b], hence f takes the same value twice, hence by rolle, f has a critical point on that interval, i.e. f' = 0 in there. this is essentially the IV property for f'.

    i.e. f' is always zero between two points where it changes sign. you can easily dress it up to the general statement.
  12. Mar 6, 2007 #11
    Some of you guys are being sloppy, so I should say this for everyone browsing this thread who may get the wrong idea. f' (that is the derivative) has the intermediate value property (Darboux property) if f' is defined on an interval.

    You can easily construct a counterexample to the original claim, just define f as straight lines pieced together so that f is continuous. It is clear that f' only takes on the values of the slopes m1, m2, m3,... etc for each line segment. Hence f' does not have an intermediate value property. This doesn't contradict f' darboux property, as f' wasn't defined in an interval (note how if we restricted f to one line segment we have f' defined and would have the intermediate value property (f'=m) which isn't very interesting).

    Look to see if you can find counterexamples where f' isn't defined on an interval. This is the key to proving your problem.
    Last edited: Mar 6, 2007
  13. Mar 6, 2007 #12


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    Exactly! Darboux's Theorem or Property whichever way you remember it.

    The interval being (a,b) instead of [a,b].
  14. Mar 7, 2007 #13


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    To clarify, the answer to the original question is "no, such a function does not exist". While the derivative of f is not necessarily continuous, every derivative satisfies the "intermediate value theorem" (Darboux's theorem mentioned by Ziox). If f'(a) exists, then it MUST be the lim, as x->a of f'(x).
  15. Mar 7, 2007 #14


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    in the first place it makes little sense to call a function differentiable at a if it is not even defined on an interval containing a.

    secondly "darboux's" theorem is almost trivial as i indicated aboive:

    if f'(a) <0 and f'(b)>0, we want to show f'=0 in between. but clearly f is not monotone on [a,b], hence it takes the same value twice (by IVT), and since assumed differentiable in between a and b, there is a point between where f'=0. bingo.
  16. Mar 7, 2007 #15
    so confused,,,some ppl say it does, some ppl say it does not ..

    anyone can give a conclusion?
  17. Mar 7, 2007 #16


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    the answer is as halls said: no such function exists

    in the questions posed, it only makes sense (because f is differentiable at a) to assume that f is defined and differentiable on an open interval containing a.

    in that situation, if the limit of f'(x) exists as x-->a, then thnat limit must equal f'(a).

    the reaqson is that if not, then there would eventually be an interval (a-e,a+e) such that on the punctured interval (excluding a) f' would be near D, but at a f'(a) would be far from D.

    this would violate the IV propeerty for derivatives, since f' would not take on all values between f'(a) and the values f'(x) near D.

    so the function with f'(a) * D cannot exist, because of the IV property for derivatives. It remains to oprove thaty proeprty, "Darboux's" theorem, which is a trivial corollary of rolles thm. we do this next.
  18. Mar 7, 2007 #17


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    suppose f is differentiable on the interval [a,b] and f'(a) >0 while f'(b) < 0. c;aim f' = 0 in between. \

    proof: by definition of derivative, there must be point to the right of a where f(x) > f(a), and points to the left of b where f(x) > f(b). hence f is not monotone on [a,b].

    hence by the IVT for f, which is continuous, f takes the same value twice between a and b.

    then at a local extremum for f on the closed interval with those two points as endpoints, there is a place where f'=0 between a and b.

    now suppose f'(a) < K and f'(b) > K. then f-Kx has derivative zero between a and b, i.e. f has derivative K between a and b. so the IV proeprty of f' is proved, and the problem is completely solved.

    go show your professor how easy it is, but practice it on your roommate first.

    the fact that this sort of thing is unknown even to some professors, is a result of the subject being taught thoughtlessly year after year, with every book exactly like all the other books, without the authors thinking about what is happening.
  19. Mar 7, 2007 #18
    I think I caused some of the confusion by misreading your question. I gave an example where f'(a) does not equal the limit of f'(x) as x->a and that limit exists.

    This however, as was pointed out, is NOT was you originally asked:

    In the example I gave, f'(a) doesn't exist.
    If f'(a) does exist, everyone here is in agreement that it must be equal to the limit of f'(x) as x->a. And even better, some here have constructed short proofs of this.
  20. Mar 7, 2007 #19
    I'm just practising my LaTeX. Mathwonks proof is completely sufficient.

    The Darboux Property of Derivatives. Let [tex]f[/tex] be differentiable on the interval I. Suppose a and b are in I and there exists an m in I such that m is strictly between [tex]f'(a)[/tex] and [tex]f'(b)[/tex], then there exists a c strictly between a and b such that [tex]f'(c)=m[/tex].

    Proof. Assume WLOG that a is strictly less than b and [tex]f'(a)<m<f'(b)[/tex]. Define [tex]g(x)=f(x)-mx[/tex]. Then [tex]g'(a)=f'(a)-m<0[/tex] and [tex]g'(b)=f'(b)-m>0[/tex]. Since [tex]g[/tex] is continuous it follows that g must have a minimum on [a,b] (continuity on a compact set). Now, c cannot be a as [tex]g(x)\ge g(a)[/tex] so that for all x in (a,b]
    which would imply [tex]g'(a)\ge 0[/tex].

    Similarly c is not b. Therefore, c is in the interior of [a,b]. But then [tex]g'(c)=0[/tex] so that [tex]f'(c)=m[/tex].

    This result is really interesting. It says even though the derivative might not be continuous it still has the intermediate value property.
    Last edited: Mar 7, 2007
  21. Mar 7, 2007 #20


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    I admit that even though I pompously claim it is "trivial" from rolle, it took me about 40 years to notice that fact. BUT I THINK THAT IS WHAT "trivial" usually means in mathematical writing.
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