# Math Help - function of 2 variables: limits

1. ## function of 2 variables: limits

Hello,

Is there some way to determine if the limit of a two variable function exists and plugging in is not an option?

I have tried doing limits along different lines, but if they all come up the same answer, that still does not tell me the limit exists, right? I have just proven it does not not exist, as I cannot solve for infinitely many lines.

2. Originally Posted by sprinks13
Hello,

Is there some way to determine if the limit of a two variable function exists and plugging in is not an option?

I have tried doing limits along different lines, but if they all come up the same answer, that still does not tell me the limit exists, right? I have just proven it does not not exist, as I cannot solve for infinitely many lines.
there are a lot of ways to show this. sometimes you can simplify so that you can plug in, sometimes not. give us the particular example so we can help you

note: you are correct, you cannot prove that the limit exists by showing that it is the same along finitely many lines. however, to show that it doesn't exist, you have to show that you get a different answer along some path. if you keep getting the same thing, it suggests that the limit exists and that it is that number, but you can't tell for sure, you have to prove it

3. lim(x,y)-->(0,0) [(sin(x)-y)/x^3]

4. Originally Posted by sprinks13
lim(x,y)-->(0,0) [(sin(x)-y)/x^3]
boy, this was a tough one. i have to think about it some more. i'm going to bed. but as of this point, think it does not exist. i get different answers when i approach along the line $x = \cos^{-1}y$ and when i approach along $x = \sec^{-1}y$. but maybe i made a mistake (try it yourself). i'm really tired at the moment. i tried to estimate the answer using a 2-D graph (i don't know how accurate that'll be) and it seemed from the graph that the asnwer would be 0, so i don't know what to trust. i'd trust the answer i got from the lines though, if i had to choose. good luck.

5. The limit doesn't exist. Following the path y = sin(x), the limit is 0. Following the path y = x, the limit is -1/6. For the limit to exist, it's value must be independent of the path; Jhevon's late night intuition was right

6. Originally Posted by TD!
The limit doesn't exist. Following the path y = sin(x), the limit is 0. Following the path y = x, the limit is -1/6. For the limit to exist, it's value must be independent of the path; Jhevon's late night intuition was right
nice! the lines you go along are a lot nicer choices than mine