Discontinuity
Author: Joseph Lo

Consider the function \[f(x,y) = \begin{cases} \dfrac{2xy}{x^2+y^2} & x \neq 0 \\ 0 & x = 0 \end{cases}\].

The limit as \((x,y) \to (0,0)\) along any trace through \((0,0)\) exists, but the value of the limit depends on the direction of the approach. Therefore, \[\lim\limits_{(x,y)\to(0,0)} f(x,y)\] does not exist. and therefore \(f(x,y)\) is discontinuous at \((0,0)\).

θ

$f(x,y) = \frac{2xy}{x^2+y^2}

f(x, y) = ((2 * x) * y / (x^(2) + y^(2)))
a = Surface(s, t, f(s, t), s, 1.0E-10, 2, t, 1.0E-10, 2)
b = Surface((-s), t, f((-s), t), s, 1.0E-10, 2, t, 1.0E-10, 2)
c = Surface((-s), (-t), f((-s), (-t)), s, 1.0E-10, 2, t, 1.0E-10, 2)
d = Surface(s, (-t), f(s, (-t)), s, 1.0E-10, 2, t, 1.0E-10, 2)
angle = 0 (slider from 0 to 2, step 0.05)
theta: (angle * π)
Plane = Surface((s * cos(theta)), (s * sin(theta)), t, s, -3, 3, t, -2, 2)
Limit: (0, 0, sin((2 * theta)))
B = Point(xAxis)
O = (0, 0)
text3: "






Limit of f as (x,y) → (0,0) 
along the trace = " + (sin((2 * theta))) + "
 "
text2: "direction of the trace:
θ = " + angle + "π"
u = Vector((cos(theta), sin(theta)), (0, 0))
v = Vector(((-cos(theta)), (-sin(theta))), (0, 0))
A = Point(u)
α = Angle(B, O, A)
C: ((0.35 * cos(theta / 2)), (0.35 * sin(theta / 2)))
tracecheck = true
g = CurveCartesian((t * cos(theta)), (t * sin(theta)), f((t * cos(theta)), (t * sin(theta))), t, 0, 3)
e = CurveCartesian((t * cos(theta)), (t * sin(theta)), f((t * cos(theta)), (t * sin(theta))), t, -3, 0)