One way to analyze this problem is to change from a rectangular system (x,y) to a polar system (r , θ) via
x = r cos(θ) , y = r sin(θ). The expression to be analyzed becomes:
[r3 cos3(θ) + r3 sin3(θ) ]/r2 = r[cos3(θ) + sin3(θ) ]
The expression in the square brackets is bounded between -1 and +1. As (0,0) is approached, r approaches zero.
So regardless of what value of θ (direction of approach) is considered, the limit is zero.
Richard P.
tutor
Rather than using an epsilon and delta argument, I would suggest invoking the squeeze theorem. One has
-2 r < r[cos3(θ) + sin3(θ) ] < 2 r
Since the limit of r as the origin is approached is zero, by the squeeze theorem, the limit of
r[cos3(θ) + sin3(θ)
is also zero.
Report
10/08/16
Pia R.
one last question. okay so in that case, is it possible to use squeeze thm without the use of polar coordinates?
-x^3/x^2 ≤ (x^3+y^3)/(x^2+y^2) ≤ x^3/x^2
-x ≤ (x^3+y^3)/(x^2+y^2) ≤ x
0=lim (x,y) -> (0,0) -x ≤ (x^3+y^3)/(x^2+y^2) ≤ lim (x,y) -> (0,0) x = 0
0 ≤ (x^3+y^3)/(x^2+y^2) ≤ 0
By squeeze Theorem
lim (x,y) -> (0,0) (x^3+y^3)/(x^2+y^2) = 0
Report
10/08/16
Pia R.
10/07/16