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## Homework Statement

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].

## Homework Equations

## 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 counter-examples to this, I've tried to prove it with the epsilon-delta 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?