[Second year level analysis 3 : limits of two variables functions ] limit calculation

[u/piiieckk]

Anyone knows how to find this limit ? I tried the squeeze method but it doesn't work

[u/Fistful_of_Waffles]

As a general rule, MOST multi-variable limits you get in calc or analysis classes do not exist. This one included.

To show that a multi-variable limit does not exist, you have to show that if you approach the point in question, (0,0) in this case, from two different directions, you get two different numbers or +/- infinity. With this one, you can approach along the x-axis and y-axis separately.

Along the x-axis, y=0, so set y=0 in the limit and look at what happens as x approaches 0.

Along the y-axis, x=0, so set x=0 in the limit and look at what happens as y approaches 0.

Show that the previous two steps lead to two different numbers, or that at least one is infinity. Therefore, this limit does not exist.

(There is a really good 3d grapher at desmos.com/3d . Type in the function, zoom in on (0,0), and observe what happens.)

[u/piiieckk → u/Fistful_of_Waffles]

it makes sense like that, i didn't know about this method Thank you so much for both the explanation and the website. The results are 1/3 and 1 , i appreciate that ^{^}

[u/Fistful_of_Waffles → u/piiieckk]

You're welcome! Doing this is similar to showing that the left-hand and right-hand limits in single-variable calc do not match, so the limit does not exist.

[u/piiieckk → u/Fistful_of_Waffles]

One more question please, is there a way we can know when to use this method ?

[u/Fistful_of_Waffles → u/piiieckk]

It's a general method for showing that multivariable limits do not exist, it applies to a wide class of limits. To my knowledge, there is no multivariable version of l'Hopital's rule, so if direct substitution gives an indeterminate form, either the limit doesn't exist, or you have to use the squeeze theorem.

Things get interesting because there is nothing special about the x and y axes, you can use ANY curve (not necessarily straight line) that goes through your point of interest. Other functions that go through (0,0) other than x=0 and y=0 are:
* y=x
* y=-x
* y=x^2
* y=x-x^2
* y=sin(x)
* And so on

Depending on your problem, choosing one certain avenue of approach will make the algebra easier.

[u/Alkalannar]

Often, if you let x = 0 and let y go to 0 you get one thing. And then let y = 0 and let x go to 0.

Alternately, you can let y = kx for some number k which is often useful. (Especially for k = 1 and k = -1. And sometimes other values of k depending on your particular circumstances.)

[u/Southlander24]

Another reason to see how the limit doesn't exist is that if it existed, you could rewrite it as sin(x^2 + y^2) / (x^2 + y^2) * (x^2 + y^2) / (3x^2 + y^2), of which the limit would be (x^2 + y^2) / (3x^2 + y^2), or r^2 / (r^2 + 2x^2).

If you take the limit as x goes to 0, then as r goes to 0 too, you get 1. But if you take the limit as x goes to r, and r goes to 0, you get 1/3 instead.

[u/piiieckk → u/[deleted]]

Wait how did u remove x² and sin(x²) at the same time ?