Continuity of a Function with Two Variables (x,y): Homework Help and Equations

[Mathoholic!]

Homework Statement

To study the continuity of a function with two variables (x,y).

Homework Equations

f(x,y)=\frac{x^3}{x^2+y^2} if (x,y)\neq(0,0)
f(x,y)=0 if (x,y)=(0,0)

The Attempt at a Solution

I've tried going by the composition of functions but I can't seem to get anywhere...

 

[Mathoholic!]

Can I prove that the function is continuous by using the composition of functions like so(?):

g(t)=(t,t)
fog(t)=\frac{t}{2} with t\neq0
fog(t)=0 with t=0

Given that g(t) is continuous and fog(t) is continuous because when t→0, fog(t)=0.

 

[HallsofIvy]

That is not sufficient. You are, in effect, looking at the limit as you approach the origin along the line y= x. But in order to say that the function is continuous at the origin, you would have to show that you get the same limit as you approach the origin along any curve. While you can use that to show that a function is NOT continuous at the origin, by finding two different ways of aproaching the origin that give different limits, you cannot use it to show a function is continuous. You simply can't "try" every possible path.

For this function, I recommend you rewrite it in polar coordinates. That way, the distance to the given point, the origin, depends upon the single variable, r. If you can show that the limit, as r goes to 0, does not depend upon the variabe \theta, then the function is continuous at the origin.

 

[Mathoholic!]

For this function, I recommend you rewrite it in polar coordinates. That way, the distance to the given point, the origin, depends upon the single variable, r. If you can show that the limit, as r goes to 0, does not depend upon the variabe \theta, then the function is continuous at the origin.

When I rewrote the expression in polar coordinates it gave the following:

f(r,θ)=rcos³(θ)

When (x,y)→(0,0) , r→0, with r=(x²+y²)^1/2

Then f(r,θ)→0 when r→0.

Am I doing this right?

 

[SammyS]

When I rewrote the expression in polar coordinates it gave the following:

f(r,θ)=rcos³(θ)

When (x,y)→(0,0) , r→0, with r=(x²+y²)^1/2

Then f(r,θ)→0 when r→0.

Am I doing this right?

Yes that's right because cosine is bounded: |cosθ| ≤ 1 .