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Thread: Continuous and differentiable points of f(x,y,z) (Existence of multivariable limit?)

  1. #1
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    Continuous and differentiable points of f(x,y,z) (Existence of multivariable limit?)

    The problem is stated as follows:

    ---

    Find the continuous points P and the differentiable points Q of the function $\displaystyle f$ in $\displaystyle {R}^3$, defined as

    $\displaystyle f(0,0,0) = 0$

    and
    $\displaystyle f(x,y,z) = \frac{xy(1-\cos{z})-z^3}{x^2+y^2+z^2}, (x,y,z) \ne (0,0,0)$.

    ---


    If you want to look at the limit I'm having trouble with, just skip a few paragraphs to the sentence that begins with a red word. I'm mostly including the rest in case anyone is in the mood to point out flaws in my reasoning.


    Differentiating $\displaystyle f$ with respect to x, y and z, respectively (when $\displaystyle (x,y,z) \ne (0,0,0)$ will make it apparent that all three partials will contain a denominator of $\displaystyle (x^2+y^2+z^2)^2$ and a continuous numerator. Thus, these partials are continuous everywhere except in $\displaystyle (0,0,0)$, and it follows that $\displaystyle f$ is differentiable, and consequently, also continuous in all points $\displaystyle (x,y,z) \ne (0,0,0)$.

    Investigating if $\displaystyle f$ is differentiable at $\displaystyle (0,0,0)$, we investigate the limit

    $\displaystyle \lim_{(h_1,h_2,h_3) \to (0,0,0)}{\frac{f(h_1,h_2,h_3) - f(0,0,0) - h_1 f_1(0,0,0) - h_2 f_2(0,0,0) - h_3 f_3(0,0,0)}{\sqrt{{h_1}^2 + {h_2}^2 + {h_3}^2}}} = \lim_{(h_1,h_2,h_3) \to (0,0,0)}{\frac{h_1 h_2 (1-\cos{h_3}) - {h_3}^3}{({h_1}^2 + {h_2}^2 + {h_3}^2)^{3/2}}}.$

    Evaluating along the line $\displaystyle x = y = z$, that is, $\displaystyle h_1 = h_2 = h_3$, it is found after a bit of work and one application of l'H˘pital's rule that the limit from the right does not equal the limit from the left, and hence, $\displaystyle f$ is not differentiable in $\displaystyle (0,0,0)$.

    To prove continuity of $\displaystyle f$, we want to show that $\displaystyle \lim_{(x,y,z) \to (0,0,0)}f(x,y,z) = 0$. Since I haven't found any good counter-examples to this, I've tried to prove it with the epsilon-delta definition instead, with little luck.

    We see that

    $\displaystyle |f(x,y,z) - 0| = \left|\frac{xy(1-\cos{z})-z^3}{x^2 + y^2 + z^2}\right| \le \left|\frac{xy(1-\cos{z})-z^3}{z^2}\right|,$

    getting me nowhere.

    Trying with spherical coordinates instead, we get

    $\displaystyle |f(x,y,z)-0| = \left|\frac{{\rho}^2 {\sin^2 \phi} \cos{\theta} \sin{\theta} (1-\cos{(\rho \cos{\phi})}) - {\rho}^3 \cos^3 {\phi}}{{\rho}^2 \sin^2 {\phi} \cos^2 {\theta} + {\rho}^2 \sin^2 {\phi} \sin^2 {\theta} + {\rho}^2 \cos^2 {\phi}}\right| = \left|\sin^2 {\phi} \cos{\theta} \sin{\theta} (1-\cos{(\rho \cos{\phi})}) - \rho \cos^3 {\phi}\right|.$

    I'm not sure how to proceed. Suggestions?
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  2. #2
    MHF Contributor chisigma's Avatar
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    For any possible $\displaystyle \theta$ and $\displaystyle \phi$ is...

    $\displaystyle \displaystyle \lim_{\rho \rightarrow 0} |\sin^{2} \phi\ \cos \theta\ \sin \theta\ \{1-\cos (\rho\ \phi)\} - \rho\ \cos^{3} \phi|=0$

    ... so that $\displaystyle f(*,*,*)$ is continous in $\displaystyle [0,0,0]$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
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    Brilliant explanation! Thank you!

    (Although I guess this makes me wonder a bit about what delta to choose for a formal proof.)
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