Continuous and differentiable points of f(x,y,z) (Existence of multivariable limit?)by Combinatus Tags: continuous, differentiable, existence, limit, multivariable, points 

#1
Feb1011, 05:13 PM

P: 37

1. The problem statement, all variables and given/known data
Find the continuous points P and the differentiable points Q of the function [tex]f[/tex] in [tex]{R}^3[/tex], defined as [tex]f(0,0,0) = 0[/tex] and [tex]f(x,y,z) = \frac{xy(1\cos{z})z^3}{x^2+y^2+z^2}, (x,y,z) \ne (0,0,0)[/tex]. 2. Relevant equations 3. The attempt at a solution If you want to look at the limit I'm having trouble with, just skip a few paragraphs. I'm mostly including the rest in case anyone is in the mood to point out flaws in my reasoning. Differentiating [tex]f[/tex] with respect to x, y and z, respectively (when [tex](x,y,z) \ne (0,0,0)[/tex] will make it apparent that all three partials will contain a denominator of [tex](x^2+y^2+z^2)^2[/tex] and a continuous numerator. Thus, these partials are continuous everywhere except in [tex](0,0,0)[/tex], and it follows that [tex]f[/tex] is differentiable, and consequently, also continuous in all points [tex](x,y,z) \ne (0,0,0)[/tex]. Investigating if [tex]f[/tex] is differentiable at [tex](0,0,0)[/tex], we investigate the limit [tex]\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}}}.[/tex] Evaluating along the line [tex]x = y = z[/tex], that is, [tex]h_1 = h_2 = h_3[/tex], it is found after a bit of work and one application of l'Hôpital's rule that the limit from the right does not equal the limit from the left, and hence, [tex]f[/tex] is not differentiable in [tex](0,0,0)[/tex]. To prove continuity of [tex]f[/tex], we want to show that [tex]\lim_{(x,y,z) \to (0,0,0)}f(x,y,z) = 0[/tex]. Since I haven't found any good counterexamples to this, I've tried to prove it with the epsilondelta definition instead, with little luck. We see that [tex]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,[/tex] getting me nowhere. Trying with spherical coordinates instead, we get [tex]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.[/tex] I'm not sure how to proceed. Suggestions? 



#2
Feb1011, 10:17 PM

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P: 7,395

Notice that [tex]f(x,y,0)=0[/tex]. Also look at [tex]\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y,\,z)\right)\ .[/tex] WolframAlpha evaluates this as ‒z. For small values of z, [tex]1\cos(z)\ \to\ \frac{z^2}{2}[/tex] 



#3
Feb1111, 05:02 AM

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P: 1,584

Look at the path: x=y=z=t and then eamine the limit as t tends to zero.




#4
Feb1111, 01:24 PM

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Continuous and differentiable points of f(x,y,z) (Existence of multivariable limit?)
Look at [tex]\lim_{z\to0} f(x,\,y,\,z)[/tex].
This limit is zero. 



#5
Feb1111, 03:39 PM

P: 37

Using the [tex]\left\sin^2 {\phi} \cos{\theta} \sin{\theta} (1\cos{(\rho \cos{\phi})})  \rho \cos^3 {\phi}\right[/tex] part from my use of polar coordinates, I guess it should be pretty clear that, since (x,y,z) → (0,0,0) implies ρ → 0 for any angles θ and Φ, we get that this expression goes to 0, thus showing the limit. 



#6
Feb1111, 05:39 PM

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P: 1,584

So what you have found is that the limit is dependent on the path you take. What does that suggest to you?




#7
Feb1111, 06:16 PM

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#8
Feb1111, 06:26 PM

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So he did, my bad. I should have read what he had done in more detail



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