What limit do you get, approaching the origin along that line?Kaura said:using the line x=y gives x^2/sin(2x)
andrewkirk said: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.
What are the simplest lines you might try?Kaura said:The limit for x^2/sin(2x) is 0
I cannot think of another line at the moment though
Kaura said:figuring out how to show that it does not
Job done?Kaura said:using the line -x=y gives -x^2/sin(0)
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 said:Job done?
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 said: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.
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 said: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.
The limit of xy/sin(x+y) as x and y approach 0 is 0. This can be found by using the squeeze theorem or by converting the function to polar coordinates and taking the limit as r approaches 0.
No, L'Hopital's rule cannot be used to solve the limit of xy/sin(x+y) at (0,0) because the function is not in an indeterminate form.
The limit of xy/sin(x+y) at (0,0) is a commonly used example in calculus to demonstrate the application of the squeeze theorem and polar coordinates in finding limits of functions. It also shows the importance of understanding the behavior of a function at critical points.
The graph of xy/sin(x+y) at (0,0) has a hole at (0,0) and approaches 0 as x and y approach 0. This is consistent with its limit, which is also 0.
Yes, the limit of xy/sin(x+y) at (0,0) can be extended to higher dimensions. In fact, it can be extended to any number of variables as long as the function remains continuous and the limit approaches 0 as all the variables approach 0.