# Multivariate limit: (xy^2)/(x + y^2)

lim (xy^2)/(x + y^2)
(x,y)->(0,0)

i just came up with this limit to test my understanding. so approaching from y = mx, y = mx^2, and y = mx^3 all seemed to point out that the limit is 0. i tried approaching from the x and y axes and i got for approaching along the x axis: lim (as (x,y)->(0,0)) of 0 / x. would this be another instance of the limit being 0 or would it be an indeterminate form? because x is approaching 0 but it does not equal 0 so x would get smaller and smaller yet still be a finite number greater than 0 so the quotient would be 0. is this reasoning correct?

i have also tried polar coordinates and came up with: lim (as r->0) of (r^2 cos(x)sin(x))/(cos(x) + r sin^2 (x)). so when cos and sin are nonzero the limit becomes 0, no problem. when x = 0 the limit is also 0. but when x = pi/2 then i get lim (as r->0) of 0 / r which brings me back to the question i asked earlier about 0 / x. is this 0 or would it be indeterminate?

i know i can't say for sure that this limit exists but i'm starting to suspect it does if the 0 / x and 0 / r quotients indeed are also 0 when x and r approach 0 respectively.

## Answers and Replies

The limit as x→0 of 0/x is indeed zero, just as you reasoned. However, the multivariate limit you're investigating doesn't exist -- to see why, consider the path x = -y².

The limit as x→0 of 0/x is indeed zero, just as you reasoned. However, the multivariate limit you're investigating doesn't exist -- to see why, consider the path x = -y².

oh so it would become lim (as (x,y)->(0,0)) of -y^4 / 0 which is undefined for all values of y that approach 0. then i can say the limit does not exist because there is no neighborhood of (0,0) such that the function is defined and the difference between the limit and the function value is less than a given epsilon. would this be correct?

Close, but there are a few points to make:

1: You should say deleted neighborhood instead of neighborhood -- since a function not defined at (0, 0) can of course have a limit there.

2: Saying that there is some ε such that there is no deleted neighborhood for which the function is defined and |f(x) - L|<ε shows that the limit is not L. You still have to show that there is no such neighborhood regardless of what L is to prove that there is not limit. Granted, that's no harder than doing it for some specific L if your textbook requires that the function be defined in a deleted neighborhood of a point for the limit to be defined, however...

3: In general, one only requires that the point be an accumulation point of the domain for the limit to exist. In that case, the points actually ON the curve x = -y^2 wouldn't affect the existence of the limit at all, since they are not part of the domain. What would affect the existence of the limit is the points NEAR the curve, say, x = -y^2 + y^4.

Close, but there are a few points to make:

1: You should say deleted neighborhood instead of neighborhood -- since a function not defined at (0, 0) can of course have a limit there.

2: Saying that there is some ε such that there is no deleted neighborhood for which the function is defined and |f(x) - L|<ε shows that the limit is not L. You still have to show that there is no such neighborhood regardless of what L is to prove that there is not limit. Granted, that's no harder than doing it for some specific L if your textbook requires that the function be defined in a deleted neighborhood of a point for the limit to be defined, however...

3: In general, one only requires that the point be an accumulation point of the domain for the limit to exist. In that case, the points actually ON the curve x = -y^2 wouldn't affect the existence of the limit at all, since they are not part of the domain. What would affect the existence of the limit is the points NEAR the curve, say, x = -y^2 + y^4.

oh ok. i checked the limit when x = -y^2 + y^4 and it becomes -1 which is different from 0 so there indeed is no limit. i remember on another post i saw the following limit:
lim (as (x,y)->(0,0)) of (x+y) / (sin(x) + sin(y)). when y = -x the function becomes 0/0 and is undefined along that line. however the points on the line y=-x should not affect the existence of the limit because they are not a part of the function's domain, only points arbitrarily close to that line would correct?

Yes, that is correct.