Epsilon/Delta Proof With 2 Variables

[Gooolati]

Homework Statement

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

is continuous at the origin

Homework Equations

\forall \epsilon > 0 \exists \delta > 0 s.t. if |(x,y)| < \delta then |f(x,y)| < \epsilon

(Since we are proving continuity at the origin)

The Attempt at a Solution

|(x,y)| < \delta \Leftrightarrow x^{2} + y^{2} &lt; \delta^{2}

then this means that |x^{2} - y^{2}| < \delta^{2}

so:

f(x,y) < \frac{x}{x^{2}+y^{2}}(\delta^{2})

and I feel like I'm close but then I'm stuck! All help appreciated thanks !

 

[Dick]

Homework Statement

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

is continuous at the origin

Homework Equations

\forall \epsilon > 0 \exists \delta > 0 s.t. if |(x,y)| < \delta then |f(x,y)| < \epsilon

(Since we are proving continuity at the origin)

The Attempt at a Solution

|(x,y)| < \delta \Leftrightarrow x^{2} + y^{2} &lt; \delta^{2}

then this means that |x^{2} - y^{2}| < \delta^{2}

so:

f(x,y) < \frac{x}{x^{2}+y^{2}}(\delta^{2})

and I feel like I'm close but then I'm stuck! All help appreciated thanks !

Have you tried thinking about it in polar coordinates? It's much easier.

 

[Gooolati]

hmm I have never used polar coordinates with an epsilon delta proof before

so x=rcosθ
and y=rsin

so f(x,y) is rcosθ(cos^{2}θ - sin^{2}θ)

and r<\delta and cosθ <= 1

so f(x,y) < \delta(cos^{2}θ - sin^{2}θ)

which = \deltacos(2θ) <= \delta(1) = \delta

set \delta = \epsilon

does this work?

 

[Dick]

hmm I have never used polar coordinates with an epsilon delta proof before

so x=rcosθ
and y=rsin

so f(x,y) is rcosθ(cos^{2}θ - sin^{2}θ)

and r<\delta and cosθ <= 1

so f(x,y) < \delta(cos^{2}θ - sin^{2}θ)

which = \deltacos(2θ) <= \delta(1) = \delta

set \delta = \epsilon

does this work?

Sure does. If r<ε then |f(r,θ)-f(0,0)|<ε.

 

[SammyS]

Sure does. If r<ε then |f(r,θ)-f(0,0)|<ε.

Isn't that |f(r,θ)-f(0,θ)| < ε ?

 

[Dick]

Isn't that |f(r,θ)-f(0,θ)| < ε ?

Oh, you know what I mean. But I don't see why you'd have to write it that way. f(0,θ)=0 and f(0,0)=0. Same thing. They are both the origin in xy coordinates.