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Homework Help: Multivariable limit definition question

  1. Oct 15, 2012 #1


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    1. The problem statement, all variables and given/known data

    I'm reading through Taylor's advanced calculus and came across this question in section 7.2 :


    2. Relevant equations

    The definition of the partial of f with respect to some variable at some point (a,b), lets say the partial with respect to x of some f(x,y) would be :

    [itex]\frac{∂f}{∂x} (a, b) = \lim_{h→0} \frac{f(a+h, b) - f(a, b)}{h}[/itex]

    3. The attempt at a solution

    Okay, so the question wants me to compute the second partials given the conditions on f.

    So first I note that f is continuous everywhere for x,y ≠ 0. I also note that f(x,0) = f(0,y) = f(0,0) = 0 has been defined nicely for me as to make my function continuous at (0,0).

    I suppose first that I'll show that [itex]\frac{∂^2f}{∂y∂x} = -1[/itex]. So I first have to do a subsidiary calculation to find the first partial with respect to x.

    [itex]\frac{∂f}{∂x} (0, 0) = \lim_{h→0} \frac{f(h, 0) - f(0, 0)}{h} = \lim_{h→0} \frac{0 - 0}{h} = 0[/itex]

    Okay, that was not so bad since f was nicely defined for us. Now here comes my problem when I try to find the second partial. Note that when I use subscripts I refer to the partial of the function with respect to that variable. So [itex]f_x = \frac{∂f}{∂x}[/itex].

    [itex]\frac{∂^2f}{∂y∂x} (0, 0) = \lim_{k→0} \frac{f_x(0, k) - f_x(0, k)}{k} (***)[/itex]

    I'm labeling this equation (***) so I can return to it after I calculate fx(0,k). Now to calculate it :

    [itex]\frac{∂f}{∂x} (0, k) = \lim_{h→0} \frac{f(h, k) - f(0, k)}{h} = \lim_{h→0} \frac{h^2arctan(k/h) - h^2arctan(h/k)}{h} = 0[/itex]

    Now this produces a big problem when I plug it back into (***) as I don't get the answer of -1 which I'm apparently supposed to get :

    [itex]\frac{∂^2f}{∂y∂x} (0, 0) = \lim_{k→0} \frac{f_x(0, k) - f_x(0, k)}{k} = \lim_{k→0} \frac{0 - 0}{k} = 0[/itex]

    What am I doing wrong here? Have I missed something or have I done a calculation wrong?
  2. jcsd
  3. Oct 15, 2012 #2
    I think there is a typo here. The last limit should be

    [tex]\lim_{h\rightarrow 0} \frac{h^2arctan(k/h)-k^2arctan(h/k)}{h}[/tex]

    This limit is not zero, I think.

  4. Oct 15, 2012 #3


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    You sir, have made my day. That will teach me to speed through my calculations too quickly haha.

    So really I would get :

    [itex]\frac{∂f}{∂x} (0, k) = \lim_{h→0} \frac{f(h, k) - f(0, k)}{h} = \lim_{h→0} \frac{h^2arctan(k/h) - k^2arctan(h/k)}{h}[/itex]

    and switching to polar co-ordinates ill presume will give me my desired result?
  5. Oct 15, 2012 #4
    Yes, try to work in polar coordinates. I think that'll make things easy.
  6. Oct 15, 2012 #5


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    Hmm I've hit a hitch in my calculations again :

    For h=rcosθ and k = rsinθ we get :

    [itex]\frac{∂f}{∂x} (0, k) = \lim_{h→0} \frac{f(h, k) - f(0, k)}{h} = \lim_{h→0} \frac{h^2arctan(k/h) - k^2arctan(h/k)}{h} = \lim_{r→0} \frac{r^2cos^2θarctan(tanθ) - r^2sin^2arctan(cotθ)}{rcosθ}[/itex]

    Which tends to 0 as r tends to 0?
  7. Oct 15, 2012 #6
    Hmmm, maybe going over in polar coordinates was not a good idea after all. The problem is that [itex]h\rightarrow 0[/itex] doesn't exactly correspond to [itex]r\rightarrow 0[/itex].

    I actually don't think it is too hard to calculate the original limit explicitely:

    [tex]\lim_{h\rightarrow 0} \frac{h^2arctan(k/h)-k^2 arctan(h/k)}{h^2}=\lim_{h\rightarrow 0} h \cdot arctan(k/h) - k^2\lim_{h\rightarrow 0} \frac{arctan(h/k)}{h}[/tex]

    The first limit is 0 (why?). The second limit can be calculated by l'Hopitals rule.
  8. Oct 15, 2012 #7


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    The first limit is 0 because of the squeeze theorem. After applying L'Hospital's rule I get (2h)arctan(k/h) - k and as h→0, (2h)arctan(k/h) - k → -k just like I wanted.
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