I'm talking mulitvariable calculus right now, but I cant check my work (i think) because i cant figure out how to do limits with (x,y)->(a,b). Is it possible?

How is lim(x,y)->(a,b)(f(x, y)) different from limx->a(limy->b(f(x,y)) ?

Quote: Original post by Extrarius
How is lim(x,y)->(a,b)(f(x, y)) different from limx->a(limy->b(f(x,y)) ?

Oh, interesting. Is that always the same thing?

Thanks.

I don't even know if it is the same because I've never done multivariable limits. I was just asking if there is a difference =-)

It is the same. If the limit exists also you can iterate them in arbitrary order although this isn't always the case: look at (x-y)/(x+y) as they both approach zero. Iterating x then y gives -1 and y then x give +1. In that case the limit doesn't exist.

thanks!

lim(f(x,y)) as (x,y) -> (0,0) does not always exist just because lim(f(x,0)) as (x,0) -> (0,0) = 0 and lim(f(0,y)) as (0,y) -> (0,0) = 0

You need to check cases OTHER than just going along the x-axis (when y = 0) and the y-axis (when x =0)

If you have James Stuart's Multivariable Calculus (either the third and fourth editions) look at Chapter 11 - section 2, example 2.

f(x,y) = (x*y)/(x^2+y^2)

lim(f(x,0)) as (x,0) -> (0,0) = 0 and lim(f(0,y)) as (0,y) -> (0,0) = 0, but...

... What if we take the limit by going along the line y=x?

so, f(x,x) = (x*x)/(x^2+x^2) = (x^2)/(x^2+x^2)

lim(f(x,x)) as (x,y) -> (0,0) = 1/2

Therefore, since different limits are obtained at the same point, the given limit of f(x,y) -> (0,0) does NOT exist.

In actuality, the are an infinite number of paths one could check and if just ONE of them does not equal the others then the limit does not exist.

So, you could say let y = m*x (this would cover all lines) on the xy-plane (x or y not equal to zero)

So then f(x,y) = f(x,m*x)

Let, f(x,x*m) = (x*(x*m))/(x^2+(x*m)^2) = (x^2)/(x^2+(x*m)^2)

And, lim((x*m*x)/(x^2+(x*m)^2) as (x,x*m) -> (0,0) = (m/(m^2 + 1))

If the 'm' (slope) of the line is any number not equal to zero then there will be a infinite amount of results that are defined.

So, the point is just because lim(f(x,0)) as (x,0) -> (0,0) = 0 and lim(f(0,y)) as (0,y) -> (0,0) = 0

Does not ALWAYS mean the limit of the function exists or is zero.

So, if you enter lim(lim((x*y)/(x^2+y^2),x,0),y,0) in your TI-89 it outputs 0.

But, that answer is NOT correct. In fact, the limit DOES NOT exist.

In conclusion, do NOT trust your calculator by ONLY checking the limit along x=0 and y = 0 such as what this command "lim(lim((x*y)/(x^2+y^2),x,0),y,0)" does, you rather need to check other paths and make sure they also give zero.

Also, all one needs to find is ONE counter-example to show or prove that something is false in mathematics.

Do NOT worry about this answer / post... I edited the original one, just can't find how to delete this one.

This line "Therefore, since different limits are obtained at the same point, the given limit of f(x,y) -> (9,0,) doe NOT exist."

Should read "Therefore, since different limits are obtained at the same point, the given limit of f(x,y) -> (0,0,) does NOT exist. "