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How to prove this limit problem?

  1. Nov 13, 2014 #1
    1. The problem statement, all variables and given/known data
    Let f : (a, b) → R be a differentiable function and x0 ∈ (a, b). For any h > 0 small, there
    exists θ ∈ (0, 1), depending on h, such that
    f(x0 + h) = f(x0) + hf'(x0 + θh).

    If f is twice differentiable at x0 with f''(x0) != 0, prove that
    (a) lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2=(1/2)*f''(x0)
    (b) lim h→0 θ =1/2
    2. Relevant equations
    lim h→0 (f(c+h)-f(c))/h = f'(c)
    L'Hopital's rule

    3. The attempt at a solution
    For (a),
    lim h→0 (f(x0 + h) − f(x0) − f'(x0)h)/h2
    =lim h→0 (hf'(x0 + θh)-h*f'(x0))/(h2)
    =lim h→0 (f'(x0+θh)-f'(x0))/(2h)
    I have no idea what I should do in the next step because the denominator is not θh.Or did I make any mistake in my steps?
     
  2. jcsd
  3. Nov 13, 2014 #2
    If you're allowed (and it looks like you are) I would use L'Hopital for (a) and then use that result from along with work similar to that which you've done to get (b).

    There is a slight error in you work; the last line should be $$\lim_{h\rightarrow 0}\frac{f'(x_0+\theta h)-f'(x_0)}{h}.$$ I'm not sure how you got the 2 in the denominator.

    Also note that, since ##\theta## is nonzero, $$\frac{f'(x_0+\theta h)-f'(x_0)}{h}=\theta\cdot\frac{f'(x_0+\theta h)-f'(x_0)}{\theta h}.$$

    Edit: L'Hopital is not needed for (a); you could get the job done using a fairly straightforward ##\epsilon##-##\delta## argument. But it's much easier to just use l'Hopital if it's available. You might need to justify why you're allowed to use it (i.e. saying more than just 0/0) depending on how much of a stickler the grader is for that kind of thing.

    Edit 2: The work that you've done is useful, but I would use it for part (b) rather than part (a). In my opinion, part (a) is more easily done without using the fact that ##f(x_0+h)=f(x_0)+f'(x_0+\theta h)h##.
     
    Last edited: Nov 13, 2014
  4. Nov 13, 2014 #3

    pasmith

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    Homework Helper

    The problem with l'hopital is that the numerator is [itex]f'(x_0 + \theta(h)h) - f'(x_0)[/itex] and we don't know that [itex]\theta[/itex] is differentiable in some punctured open neighbourood of [itex]0[/itex].
     
  5. Nov 13, 2014 #4
    Sorry for the confusion. I intend for l'Hopital to be applied immediately to $$\lim_{h\rightarrow0}\frac{f(x_0+h)-f(x_0)-f'(x_0)h}{h^2}$$ without any algebraic rearrangement or use of the fact that ##f(x_0+h)-f(x_0)=hf'(x_0+\theta h)##. The only difficulty here is showing that ##g(h)=f(x_0+h)## is differentiable in a neighborhood of ##0##. It's true, but it's not trivial. I would put it somewhere in the realm of not-hard-to-show-but-worth-mentioning.
     
  6. Nov 13, 2014 #5
    For part (a), you should try l'hopital's rule directly for the left side of the equation. For part (b), you can use your first attempt's last line and do the trick: multiply by [itex]\theta / \theta [/itex] like gopher_p mentioned and use the fact that [itex]f''(x_0) \neq 0[/itex] to show (b) (you need this fact because [itex]\theta[/itex] is a function of [itex]h[/itex]).

    It is a trivial application of the chain rule with a certain restriction on [itex]h[/itex] (exercise).
     
  7. Nov 14, 2014 #6
    Actually, I'm not even sure if (b) can be proven with the info given because proving that the limit of [itex]\theta[/itex] as [itex]h[/itex] goes to 0 might be a problem. You can at least say: if the limit of [itex]\theta[/itex] exists, it is 1/2.
     
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