Solving Multivariable Limits: Tips & Tricks

[Wesleytf]

I'm taking multi-variable after having a while off from school, so forgive me if these are simple ones that I just don't "see"

Homework Statement

lim (x, y) --> 0, 0 $$\frac{x^2 y^2 e^y}{x^4+4y^2}$$

and

lim (x, y) --> (1, 1) $$\frac{x-y}{x^3-y}$$

Homework Equations

The Attempt at a Solution

The bottom one I feel doesn't exist because as x->0+, the ^3 is making it larger, where as when x->0-, the ^3 is making it smaller. I know this is poor logic; it's just a gut feeling about it. substituting y=mx or similar didn't get me anywhere. Plugging obviously doesn't work. I don't see anyway to simplify, but maybe there is a way. I also don't think polar coordinates will work for either.

I really think I only need a hint to the method of solution, so don't go solving the whole thing for me.

 

[Dick]

For the bottom one, suppose you approach along the path y=x^3? For the top one you don't have to worry about the factor e^y, why not? Can you handle the rest?

 

[Wesleytf]

For the bottom one, suppose you approach along the path y=x^3? For the top one you don't have to worry about the factor e^y, why not? Can you handle the rest?

For the top one, e^y -->1 as y->0. I wasn't sure if saying that and then doing the rest of the limit was a legal move. Should be easy now (I see one way by polar coordinates and then squeeze theorem, I think)

For the bottom one, I tried that, but doesn't that just 'break' the function--as in, we can't tell what it will be by that method? The way I thought the "two different limits for two different paths" property worked was that it would only work if you found two actual different paths. Taking y=x^3 makes one path not exist; is that enough to make the limit of the function not exist?

 

[Dick]

For the top one, e^y -->1 as y->0. I wasn't sure if saying that and then doing the rest of the limit was a legal move. Should be easy now (I see one way by polar coordinates and then squeeze theorem, I think)

For the bottom one, I tried that, but doesn't that just 'break' the function--as in, we can't tell what it will be by that method? The way I thought the "two different limits for two different paths" property worked was that it would only work if you found two actual different paths. Taking y=x^3 makes one path not exist; is that enough to make the limit of the function not exist?

If the function has a limit, it has to approach that limit along all paths. If the function doesn't even exist along a path, then the limit doesn't exist. If you think about what's happening 'close' to the path y=x^3 then you'll see the function becomes large without bound.

 

[Wesleytf]

If the function has a limit, it has to approach that limit along all paths. If the function doesn't even exist along a path, then the limit doesn't exist. If you think about what's happening 'close' to the path y=x^3 then you'll see the function becomes large without bound.

ha, I had already went back and put it on paper, and as soon as I did the behavior 'close' to y=x^3 became clear. thanks! hopefully my brain will uncrustify itself soon...

 

[Wesleytf]

follow up on the bottom one: Subbing y=x^3 will not work because it is not within the definition of a limit. However, comparing the substitutions y=x and y=1 does work to prove that it indeed DNE.