- #1
PeteyCoco
- 38
- 1
I started learning how to do these things today and boy, they take some interesting logic. Anyway, here's my attempt at one:
prove that the limit as (x,y) → (0,0) of [(x^2)(siny)^2]/(x^2 + 2y^2) exists
Here's what I did:
0<√(x^2 + y^2) < δ, |[(x^2)(siny)^2]/(x^2 + 2y^2) - 0| < ε.
x^2 ≤ x^2 + y^2 since y^2 ≥ 0
so it follows that
(x^2)(siny)^2]/(x^2 + 2y^2) ≤ (siny)^2 < ε
siny < y for all y > 0
(siny)^2 < y^2
(siny)^2 < x^2 + y^2 = δ^2
(x^2)(siny)^2]/(x^2 + 2y^2) ≤ (siny)^2 < δ^2 < ε ∴ limit exists at (0,0)
Does this work?
prove that the limit as (x,y) → (0,0) of [(x^2)(siny)^2]/(x^2 + 2y^2) exists
Here's what I did:
0<√(x^2 + y^2) < δ, |[(x^2)(siny)^2]/(x^2 + 2y^2) - 0| < ε.
x^2 ≤ x^2 + y^2 since y^2 ≥ 0
so it follows that
(x^2)(siny)^2]/(x^2 + 2y^2) ≤ (siny)^2 < ε
siny < y for all y > 0
(siny)^2 < y^2
(siny)^2 < x^2 + y^2 = δ^2
(x^2)(siny)^2]/(x^2 + 2y^2) ≤ (siny)^2 < δ^2 < ε ∴ limit exists at (0,0)
Does this work?