Polar Coordinates and Limits
Consider
x4 + y 4
.
(x,y)→(0,0) x2 + y 2
lim
You try different paths (such as all lines through the origin, y = mx, a parabola or two, etc.)
and you notice you keep getting 0. This leads you to believe the limit exists and is 0. So
you decide to try polar coordinates: x = r cos(θ), y = r sin(θ), and take the limit as r → 0+ .
So for this one, you get
lim+
r→0
r4 cos4 (θ) + r4 sin4 (θ)
=
r2 cos2 (θ) + r2 sin2 (θ)
=
r4 (cos4 (θ) + sin4 (θ))
r→0
r2
lim+ r2 (cos4 (θ) + sin4 (θ)).
lim+
r→0
You want to say this limit is 0, and it is. But remember that we want to be 0 independent
of θ. Implicitly, what you’re using is that
0 ≤ |r2 (cos4 (θ) + sin4 (θ))| ≤ r2 | cos4 (θ) + sin4 (θ)| ≤ r2 · 2 = 2r2 .
Now as r → 0+ , 2r2 → 0, and so by squeeze theorem, r2 (cos4 (θ) + sin4 (θ)) → 0 as well. So
the limit is 0. The key is that the limit was of r2 times something bounded.
Here’s another one for you:
x2 y
.
(x,y)→(0,0) x2 + y 2
lim
Again, you try some different paths, like lines and such, and you keep getting 0. So you
want to prove this limit is 0. Well, you go to polar coordinates and you get
lim+
r→0
r3 cos2 (θ) sin(θ)
= lim r cos2 (θ) sin(θ).
r2 (sin2 (θ) + cos2 (θ)) r→0+
We want to say this goes to 0. Can we do this? Yes, because we have r times something
bounded. Indeed,
| cos2 (θ) sin(θ)| ≤ 1
since each of cos2 (θ) and sin(θ) is bounded by 1. Therefore
|r cos2 (θ) sin(θ)| ≤ |r| → 0
as r → 0+ . Therefore the whole limit is 0.
Now look at the limit we had today:
xy 3
lim
.
(x,y)→(0,0) x2 + y 6
1
Once we used polar coordinates and simplified, we had
lim+ r2
r→0
cos(θ) sin3 (θ)
.
cos2 (θ) + r4 sin6 (θ)
The r2 in front makes us think that the limit should go to 0, but in this case it doesn’t,
because there is no way to bound the remaining term
cos(θ) sin3 (θ)
cos2 (θ) + r4 sin6 (θ)
in the way we could with the cos4 (θ) + sin4 (θ) in the previous example. We need to bound
this by something independent of θ, and there is no way to do that. You could try and say
things like
cos(θ) sin3 (θ) 1
cos2 (θ) + r4 sin6 (θ) ≤ cos2 (θ) + r4 sin6 (θ) ,
but now the problem is as θ → π/2, this right hand side goes to infinity. And so we cannot
conclude that the limit is 0. Think of it as similar to the 0 · ∞ cases you had in one variable
limits.
So moral of the story: as long as you can bound the function independently of θ, you’re
ok. If not, then you need to try something else. Just so you can see one more ok example,
let’s say after the polar coordinate transformation, you got
lim+
r→0
r3 cos(θ)
.
1 + cos2 (θ)
Could we say this goes to 0? Well, we see the r3 on top. Can we bound the rest?
cos(θ) 1
≤
1 + cos2 (θ) 1 + cos2 (θ) ≤ 1,
as cos2 (θ) ≥ 0. Therefore our function satisfies
3 cos(θ) 3
r
1 + cos2 (θ) ≤ |r | → 0
as r → 0+ , so our limit is 0, as desired.
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