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Math Help - Need some constructive criticism...(advanced calc proof)

  1. #1
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    Need some constructive criticism...(advanced calc proof)

    I am unsure of this proof, and I would appreciate any input about whether my reasoning is on the right track. Due tomorrow btw.

    The question is:

    Suppose that f is continuous everywhere, f' exists for x\neq0, and \mathop {\lim }\limits_{x \to 0} f'\left( x \right) exists. Prove that f'\left( 0 \right) exists.
    Professor said to use L'H˘pital's rule if we get stuck.

    I started by trying to let g\left( x \right) = f'\left( x \right) and working through an argument using L'H˘pital's rule, but I made some extra assumptions and such which dont seem to be reasonable. Any nudges in the right direction would be great. My work is pretty useless...and ultimately it seems like I wouldnt be proving the right thing anyway. I feel like I am overthinking (as usual) what would be an otherwise simple and elegant proof. If anyone really wants to see my attempt, I will post it, but it is almost certainly incorrect. Thanks in advance.


    edit: I know intuitively that it must exist, since f is continuous and f' exists everywhere but 0 (for sure). And since the two-sided limit exists at 0, either f' is continuous at 0 or has a removable discontinuity there, but it must be defined there because f is continuous at 0 and does not have a corner there. There cannot be a corner at f \left( 0 \right) because the two-sided limit exists, so we know that the derivative must exist there (as f is continuous without a corner)... but I do not see how L'H˘pital would come into play in a formal proof?
    Last edited by xifentoozlerix; May 13th 2008 at 06:25 PM.
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  2. #2
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    Quote Originally Posted by xifentoozlerix View Post
    Suppose that f is continuous everywhere, f' exists for x\neq0, and \mathop {\lim }\limits_{x \to 0} f'\left( x \right) exists. Prove that f'\left( 0 \right) exists.
    The first step that I like to do a lot of times is to formally state the problem. I find that it helps:
    "Let f:\mathbb{R}\mapsto \mathbb{R} be continous and let f be differenciable on (-\infty, 0)\cup (0,\infty) (possibly at x=0 also, but we do not know this). Suppose also that \lim_{x\to 0}f'(x) exists. Prove that f is differenciable at zero".

    We want to show " f is differenciable at zero", which means we want to show that the limit \lim_{x\to 0}\frac{f(x) - f(0)}{x-0} exists.

    Note the numerator is \lim_{x\to 0}f(x) - f(0) = 0. This is where we are using the countinuity condition. The denominator is zero as well. Thus, we can use L'Hopital here because the numerator is differenciable. Thus, \lim_{x\to 0}\frac{f'(x)}{1}. And this exists by the stated problem.
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  3. #3
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    Thank you so much....after sleeping on it I had a similar idea, but you make it seem so simple now (it always seems that way though...simple when you know how).
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