Find the limit if it exists. lim as (x,y) -->(0,0) of (x^3)(y^2)/(x^2+y^2)?

Answer 1

Let f(x, y) = x^3 y^2 / (x^2 + y^2).

For (x, y) ≠ (0, 0), let r = √(x^2 + y^2).
Then 0 ≤ |x|, |y| ≤ r.
So | x^3 y^2 | ≤ r^5
and | f(x, y) | = | x^3 y^2 | / r^2 ≤ r^5 / r^2 = r^3.

So for any ε > 0, let δ = ε^(1/3). Then 0 < ||(x, y)|| < δ => | f(x, y) | < δ^3 = ε. So the limit is 0.

Answer 2

The easiest way is to convert to polar coordinates
= r^5 cos^3 t *sin^2t/r^2 = r^3 cos ^3t *sin ^2t
As r==>0 this expression ==>0 independent of angle t so the limit is 0

Answer 3

as (x,y) -->(0,0), that means x-->y

(x^3)(y^2)/(x^2+y^2) becomes
(x^3)(x^2)/(x^2 + x^2)
x^5/2x^2
x^3/2
0

So, (x,y) -->(0,0) the limit is 0