Epsilon Delta Proof of a Limit

• Sep 21st 2009, 05:58 PM
meks08999
Epsilon Delta Proof of a Limit
The problem is give a rigerous epsilon delta proof that lim (x,y) (0,0) (x^3 + y^3)/(x^2 + y^2)

Heres what I have. Given any ε > 0, we can find a corresponding δ > 0 such that if 0 < || (x,y) - (0,0)|| < δ , then (x^3 + y^3)/(x^2 + y^2) < ε

0 < √(x^2 +y^2) < δ

√(x^2) =|x|< δ
√(y^2) = |y| < δ

I'm stuck after this part. Is the above part even correct? If so, how do I finish i then?

Thanks
• Sep 21st 2009, 07:01 PM
NonCommAlg
Quote:

Originally Posted by meks08999
The problem is give a rigerous epsilon delta proof that lim (x,y) (0,0) (x^3 + y^3)/(x^2 + y^2)

Heres what I have. Given any ε > 0, we can find a corresponding δ > 0 such that if 0 < || (x,y) - (0,0)|| < δ , then (x^3 + y^3)/(x^2 + y^2) < ε

0 < √(x^2 +y^2) < δ

√(x^2) =|x|< δ
√(y^2) = |y| < δ

I'm stuck after this part. Is the above part even correct? If so, how do I finish i then?

Thanks

$\displaystyle \left|\frac{x^3+y^3}{x^2+y^2} \right| =\frac{|x^3+y^3|}{x^2+y^2} \leq \frac{x^2|x| + y^2|y|}{x^2+y^2} \leq |x| + |y| < 2 \delta.$ so we need to choose $\displaystyle \delta = \frac{\epsilon}{2}.$
• Sep 21st 2009, 07:37 PM
meks08999
Thanks for the help.