Limits help ...........10 pts for best answer?

a) |x^3| = |x| * x^2 ≤ |x| (x^2 + y^2), since y^2 ≥ 0.
Similarly, |y^3| ≤ |y| (x^2 + y^2).

b) Using part a,
|f(x,y)| = |x^3 + y^3| / (x^2 + y^2)
........≤ (|x^3| + |y^3|)/(x^2 + y^2), by triangle inequality
........≤ [|x| (x^2 + y^2) + |y| (x^2 + y^2)] / (x^2 + y^2), by part a
........= (|x| + |y|) (x^2 + y^2) / (x^2 + y^2)
........= |x| + |y|.

c) To show that lim((x,y)→ (0,0)) f(x,y) = 0:

Given ε > 0, we need to find δ > 0 such that
0 < √[(x - 0)^2 + (y - 0)^2] < δ ==> |f(x,y) - 0| < ε.
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To show this, note that
|f(x, y) - 0|
= |(x^3 + y^3) / (x^2 + y^2)|
≤ |x| + |y| by part b
= √x^2 + √y^2
≤ √(x^2 + y^2) + √(x^2 + y^2), since x^2, y^2 ≥ 0
= 2 √(x^2 + y^2).

So given ε > 0, let δ = ε/2. Then, 0 < √[(x - 0)^2 + (y - 0)^2] < δ
==> |f(x,y) - 0| ≤ 2 √(x^2 + y^2) < 2(ε/2) = ε, as required.
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d) Using polar coordinates,
lim((x,y)→ (0,0)) f(x, y)
= lim((x,y)→ (0,0)) (x^3 + y^3) / (x^2 + y^2)
= lim(r→0+) r^3 (cos^3(t) + sin^3(t)) / r^2
= lim(r→0+) r (cos^3(t) + sin^3(t))
= 0, by the Squeeze Theorem:

For this last part, use -1 ≤ cos t ≤ 1 and -1 ≤ sin t ≤ 1 for all t.
==> (-1)^3 + (-1)^3 ≤ cos^3(t) + sin^3(t) ≤ 1^3 + 1^3
==> -2 ≤ cos^3(t) + sin^3(t) ≤ 2 for all t.

So, -2r ≤ r(cos^3(t) + sin^3(t)) ≤ 2r for all t.
Since lim(r→0+) ±2r = 0, the result now follows from thee Squeeze Theorem.
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2) Try different paths to the origin.

Letting x = 0 yields lim(y→0) y^2/(0 + y^2) = 1.
Letting y = 0 yields lim(x→0) 0/(0 + x^2) = 0.

Since we get different limits on different paths to the origin,
the multivariable limit does not exist.

I hope this helps!