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

In summary, the author found that the limit for the function f is not dependent on the path taken to get there.
  • #1
Combinatus
42
1

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?
 
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  • #2


Combinatus said:

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].
...

I'm not sure how to proceed. Suggestions?
A couple of things to look at:

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


Look at the path: x=y=z=t and then eamine the limit as t tends to zero.
 
  • #4


Look at [tex]\lim_{z\to0} f(x,\,y,\,z)[/tex].

This limit is zero.
 
Last edited:
  • #5


hunt_mat said:
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.)


SammyS said:
Look at [tex]\lim_{z\to0} f(x,\,y,\,z)[/tex].

This limit is zero.

Good find!

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


So what you have found is that the limit is dependent on the path you take. What does that suggest to you?
 
  • #7


hunt_mat said:
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.
 
  • #8


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

1. What is the definition of a continuous point in {R}^3?

A point in {R}^3 is considered continuous if the limit of the function at that point exists and is equal to the value of the function at that point. In other words, if you can draw a smooth curve through the point without lifting your pencil, it is considered continuous.

2. How can I determine if a point is continuous in {R}^3?

To determine if a point is continuous in {R}^3, you can use the limit definition of continuity. This involves taking the limit of the function as it approaches the point from both the left and right sides. If the limits are equal, then the point is continuous. Alternatively, you can plot the function and visually inspect if there are any breaks or jumps at the point.

3. What is the difference between a continuous and differentiable point in {R}^3?

A continuous point in {R}^3 is one where the limit of the function exists and is equal to the value of the function at that point. A differentiable point, on the other hand, not only has a well-defined limit, but also has a well-defined derivative. In other words, the function is smooth and has a tangent line at that point.

4. How can I determine if a point is differentiable in {R}^3?

To determine if a point is differentiable in {R}^3, you can use the definition of the derivative. This involves taking the limit of the function as it approaches the point from both the left and right sides, and then comparing it to the slope of the tangent line at that point. If the two values are equal, then the point is differentiable. Alternatively, you can take the partial derivatives of the function and see if they exist at that point.

5. Can a point be continuous but not differentiable in {R}^3?

Yes, a point can be continuous but not differentiable in {R}^3. This can occur if the limit of the function exists at the point, but the function is not smooth and does not have a well-defined tangent line at that point. An example of this is the function f(x) = |x|, which is continuous at x = 0 but not differentiable at that point.

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