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

 P: 40 1. The problem statement, all variables and given/known data Find the continuous points P and the differentiable points Q of the function $$f$$ in $${R}^3$$, defined as $$f(0,0,0) = 0$$ and $$f(x,y,z) = \frac{xy(1-\cos{z})-z^3}{x^2+y^2+z^2}, (x,y,z) \ne (0,0,0)$$. 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 $$f$$ with respect to x, y and z, respectively (when $$(x,y,z) \ne (0,0,0)$$ will make it apparent that all three partials will contain a denominator of $$(x^2+y^2+z^2)^2$$ and a continuous numerator. Thus, these partials are continuous everywhere except in $$(0,0,0)$$, and it follows that $$f$$ is differentiable, and consequently, also continuous in all points $$(x,y,z) \ne (0,0,0)$$. Investigating if $$f$$ is differentiable at $$(0,0,0)$$, we investigate the limit $$\lim_{(h_1,h_2,h_3) \to (0,0,0)}{\frac{f(h_1,h_2,h_3) - f(0,0,0) - h_1 f_1(0,0,0) - h_2 f_2(0,0,0) - h_3 f_3(0,0,0)}{\sqrt{{h_1}^2 + {h_2}^2 + {h_3}^2}}} = \lim_{(h_1,h_2,h_3) \to (0,0,0)}{\frac{h_1 h_2 (1-\cos{h_3}) - {h_3}^3}{({h_1}^2 + {h_2}^2 + {h_3}^2)^{3/2}}}.$$ Evaluating along the line $$x = y = z$$, that is, $$h_1 = h_2 = h_3$$, 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, $$f$$ is not differentiable in $$(0,0,0)$$. To prove continuity of $$f$$, we want to show that $$\lim_{(x,y,z) \to (0,0,0)}f(x,y,z) = 0$$. 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 $$|f(x,y,z) - 0| = \left|\frac{xy(1-\cos{z})-z^3}{x^2 + y^2 + z^2}\right| \le \left|\frac{xy(1-\cos{z})-z^3}{z^2}\right|,$$ getting me nowhere. Trying with spherical coordinates instead, we get $$|f(x,y,z)-0| = \left|\frac{{\rho}^2 {\sin^2 \phi} \cos{\theta} \sin{\theta} (1-\cos{(\rho \cos{\phi})}) - {\rho}^3 \cos^3 {\phi}}{{\rho}^2 \sin^2 {\phi} \cos^2 {\theta} + {\rho}^2 \sin^2 {\phi} \sin^2 {\theta} + {\rho}^2 \cos^2 {\phi}}\right| = \left|\sin^2 {\phi} \cos{\theta} \sin{\theta} (1-\cos{(\rho \cos{\phi})}) - \rho \cos^3 {\phi}\right|.$$ I'm not sure how to proceed. Suggestions?
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HW Helper
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P: 7,810
 Quote by Combinatus 1. The problem statement, all variables and given/known data Find the continuous points P and the differentiable points Q of the function $$f$$ in $${R}^3$$, defined as $$f(0,0,0) = 0$$ and $$f(x,y,z) = \frac{xy(1-\cos{z})-z^3}{x^2+y^2+z^2}, (x,y,z) \ne (0,0,0)$$. ... I'm not sure how to proceed. Suggestions?
A couple of things to look at:

Notice that  $$f(x,y,0)=0$$.

Also look at  $$\lim_{y\to 0}\left(\lim_{x\to 0}f(x,\,y,\,z)\right)\ .$$  WolframAlpha evaluates this as ‒z.

For small values of |z|,  $$1-\cos(z)\ \to\ \frac{z^2}{2}$$
 HW Helper P: 1,583 Look at the path: x=y=z=t and then eamine the limit as t tends to zero.
 Emeritus Sci Advisor HW Helper PF Gold P: 7,810 Continuous and differentiable points of f(x,y,z) (Existence of multivariable limit?) Look at $$\lim_{z\to0} f(x,\,y,\,z)$$. This limit is zero.
P: 40
 Quote by hunt_mat Look at the path: x=y=z=t and then eamine the limit as t tends to zero.
Wouldn't it be possible for the limit to be different along some other path? (Although in this particular case, there isn't.)

 Quote by SammyS Look at $$\lim_{z\to0} f(x,\,y,\,z)$$. This limit is zero.
Good find!

Using the $$\left|\sin^2 {\phi} \cos{\theta} \sin{\theta} (1-\cos{(\rho \cos{\phi})}) - \rho \cos^3 {\phi}\right|$$ 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.
 HW Helper P: 1,583 So what you have found is that the limit is dependent on the path you take. What does that suggest to you?
Emeritus