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Calculus Limit Proof

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  1. Jun 27, 2016 #1
    1. The problem statement, all variables and given/known data
    I am posting this for another student who I noticed did not have the proof in the problem. Here is what she said. Lets try and help her out.

    I have been working on the problem below and I am stuck. I am stuck primarily because of the part where is says x=0. If x-0, it should cancel everything out. The derivative of 0 is 0 so will cancel everything out I think, so I am not sure if that is the reasoning and the proof behind it.

    cap-2-png.102533.png

    2. Relevant equations
    She had no relevant equations.

    3. The attempt at a solution
    She said:

    I thought I should start this way, but I am not 100% sure.
    upload_2016-6-26_20-51-50-png.102534.png
     
  2. jcsd
  3. Jun 27, 2016 #2

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    The meaning of "[itex]f(x)[/itex] is differentiable at [itex]x=0[/itex]" is that near [itex]x=0[/itex], we can write:

    [itex]\dfrac{f(x) - f(0)}{x} = f'(0) + \epsilon(x)[/itex]

    where [itex]\epsilon(x)[/itex] is a correction term that goes to zero as [itex]x \rightarrow 0[/itex]. So we can rewrite [itex]f(x)[/itex] as:

    [itex]f(x) = f(0) + x (f'(0) + \epsilon(x))[/itex]

    So try using that expression to rewrite [itex]f(ax)[/itex] and [itex]f(bx)[/itex].
     
  4. Jun 27, 2016 #3

    Mark44

    Staff: Mentor

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