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

**Combinatus** · February 10th 2011, 02:20 PM

The problem is stated as follows:

---

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

---

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 $f$ 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?

**chisigma** · February 10th 2011, 05:22 PM

For any possible $\theta$ and $\phi$ is...

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

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

Kind regards

$\chi$ $\sigma$

**Combinatus** · February 11th 2011, 01:00 PM

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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