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$^{2} < \delta^{2}$

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$^{2} < \delta^{2}$

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 = $\delta$cos(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 = $\delta$cos(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.