How Do You Evaluate the Limit of xy/sin(x+y) as (x,y) Approaches (0,0)?

[Kaura]

Homework Statement

limit (x -> 0 y -> 0) of xy/sin(x+y)

Homework Equations

None that come to mind but maybe Lopital's Rule

The Attempt at a Solution

I know that the limit does not exist but I am having trouble figuring out how to show that it does not

using the line x=y gives x^2/sin(2x)
using the line -x=y gives -x^2/sin(0)

I am not sure what to do from here or if I am just completely missing what to do

 

[andrewkirk]

using the line x=y gives x^2/sin(2x)

What limit do you get, approaching the origin along that line?

Now, can you think of a different line along which to approach the origin that gives you a different limit. If you can, you will have proven that the general limit does not exist - because if it did, the limits for all different lines of approach would be the same.

L'Hopital's rule will be useful in finding the limit for both lines.

 

[Kaura]

What limit do you get, approaching the origin along that line?

Now, can you think of a different line along which to approach the origin that gives you a different limit. If you can, you will have proven that the general limit does not exist - because if it did, the limits for all different lines of approach would be the same.

L'Hopital's rule will be useful in finding the limit for both lines.

The limit for x^2/sin(2x) is 0
I cannot think of another line at the moment though

 

[andrewkirk]

The limit for x^2/sin(2x) is 0
I cannot think of another line at the moment though

What are the simplest lines you might try?

 

[Kaura]

For -x=y it gives -x^2/sin(0) which is undefined I think is there another line that is better though?

 

[andrewkirk]

Those are both sloping lines. Try a non-sloping line.

 

[haruspex]

figuring out how to show that it does not

using the line -x=y gives -x^2/sin(0)

Job done?

 

[andrewkirk]

Job done?

Not if one follows the convention that, where the domain of a function is not explicitly stated, it is assumed to be the set of points on which the given formula has a well-defined value. In that case the whole line -x=y is excluded from the domain.

 

[haruspex]

Not if one follows the convention that, where the domain of a function is not explicitly stated, it is assumed to be the set of points on which the given formula has a well-defined value. In that case the whole line -x=y is excluded from the domain.

Ok, but it still provides the basis for the proof of nonconvergence. Sloping straight lines won't do it. We need a curve which is asymptotically y=-x at (0,0).

 

[andrewkirk]

Sloping straight lines won't do it but, if my calcs are correct, a non-sloping line will, given the observation the OP has already made about the limit along the line y=x.

 

[haruspex]

Sloping straight lines won't do it but, if my calcs are correct, a non-sloping line will, given the observation the OP has already made about the limit along the line y=x.

Given the symmetry between x and y, any line can be represented by y=ax. The function is ax²/sin(x(1+a)). If a is not -1 then this clearly tends to 0 as x tends to 0.

 

[andrewkirk]

@haruspex: My mistake. You're right. Somehow I'd got it into my head that the numerator was x+y rather than xy.