Finding Continuous & Differentiable Points of f in {R}^3

[Combinatus]

Homework Statement

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).

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 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?

 

[SammyS]

Homework Statement

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}

 

[hunt_mat]

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

 

[SammyS]

Look at \lim_{z\to0} f(x,\,y,\,z).

This limit is zero.

 

[Combinatus]

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.)

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.

 

[hunt_mat]

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

 

[SammyS]

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

Actually, he found that in this case, it's not dependent on the path.

 

[hunt_mat]

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