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# Proving that a limit of two-variable function exists\does not exist Tweet

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1. Proving that a limit of two-variable function exists\does not exist
Hey,
The question is:
"Is there a value "a" such that the following function is continues:

If there isn't, prove it"

I think the limit does not exist, but I don't know how to prove it.

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2. Re: Proving that a limit of two-variable function exists\does not exist
is clearly continuous at every point other than (1,1), so it remains to check whether it can be continuous at (1,1). If tends to a limit as , then you can just set to be equal to that limit, and then it's continuous there by construction. If there is no limit (e.g. if it tends to infinity) then no such can exist and hence can never be continuous.

Does this help? Or were you stuck on actually taking the limit?
3. Re: Proving that a limit of two-variable function exists\does not exist
I was stuck in taking the limit, but I think I got it: If I take y=2-x the limit is 0 but if I take y=x^2 the limit is one. So, the limit does not exist.
Am I right?

Edit: for y=kx^2 (k is not 1) the limit is 1, and for k=1 the limit is zero. My bad.
Last edited by msokol; 25-06-2012 at 12:39.
4. Re: Proving that a limit of two-variable function exists\does not exist
With questions like this, the problem is almost always not that "it tends to infinity", but that if you take the limit in different directions you get different results. (e.g. you might find , but ).