Is f(x,y) continuous at (0,0)?

[Jonmundsson]

Homework Statement

We define the function [itex]f: \mathbb{R}^2 \to \mathbb{R} [/itex] as

[itex]
\begin{equation}
f(x,y) = \frac{xy^2 ln(x^2 + y^2)}{x^2 + y^2}
\end{equation}
[/itex] if [itex](x,y) \neq (0,0)[/itex]. Also note that [itex]f(0,0) = 0[/itex].

Show that [itex]f[/itex] is continuous at [itex](0,0)[/itex]

Homework Equations

The Attempt at a Solution

Polar coords don't work and I don't see a good way to utilize the squeeze theorem which leaves me with delta epsilon. I'm terrible with delta epsilon proofs so I'm wondering if someone can get me started and I'll take it from there.

Thanks.

 

[SammyS]

Homework Statement

We define the function [itex]f: \mathbb{R}^2 \to \mathbb{R} [/itex] as

[itex]
\begin{equation}
f(x,y) = \frac{xy^2 ln(x^2 + y^2)}{x^2 + y^2}
\end{equation}
[/itex] if [itex](x,y) \neq (0,0)[/itex]. Also note that [itex]f(0,0) = 0[/itex].

Show that [itex]f[/itex] is continuous at [itex](0,0)[/itex]

Homework Equations

The Attempt at a Solution

Polar coords don't work and I don't see a good way to utilize the squeeze theorem which leaves me with delta epsilon. I'm terrible with delta epsilon proofs so I'm wondering if someone can get me started and I'll take it from there.

Thanks.

What is the problem with polar coord's?

 

[Jonmundsson]

I get [itex]ln(r^2)[/itex] which is undefined as [itex]r \to 0[/itex]

 

[Dick]

Homework Statement

We define the function [itex]f: \mathbb{R}^2 \to \mathbb{R} [/itex] as

[itex]
\begin{equation}
f(x,y) = \frac{xy^2 ln(x^2 + y^2)}{x^2 + y^2}
\end{equation}
[/itex] if [itex](x,y) \neq (0,0)[/itex]. Also note that [itex]f(0,0) = 0[/itex].

Show that [itex]f[/itex] is continuous at [itex](0,0)[/itex]

Homework Equations

The Attempt at a Solution

Polar coords don't work and I don't see a good way to utilize the squeeze theorem which leaves me with delta epsilon. I'm terrible with delta epsilon proofs so I'm wondering if someone can get me started and I'll take it from there.

Thanks.

I think polar coordinates will work. Why do you think they don't?

 

[Jonmundsson]

I figured it out. I feel pretty dumb now.

It's just like [itex] \displaystyle \lim _{x \to 0} x sin(1/x) = 0[/itex] but [itex]\displaystyle \lim _{x \to 0} sin(1/x)[/itex] is undefined.

Thanks for the help.

 

[SammyS]

I figured it out. I feel pretty dumb now.

It's just like [itex] \displaystyle \lim _{x \to 0} x sin(1/x) = 0[/itex] but [itex]\displaystyle \lim _{x \to 0} sin(1/x)[/itex] is undefined.

Thanks for the help.

Yes.

I assume you got

r ln(r²) sin²(θ) cos(θ) .​

Then took the limit of that as r → 0 .

 

[Dick]

I figured it out. I feel pretty dumb now.

It's just like [itex] \displaystyle \lim _{x \to 0} x sin(1/x) = 0[/itex] but [itex]\displaystyle \lim _{x \to 0} sin(1/x)[/itex] is undefined.

Thanks for the help.

I think it's more like a l'Hopital's rule proof. But glad you got it.