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June 30th, 2017, 08:21 AM   #1
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multivariable limit

Hello

I have a limit that I am stuck on. It is

the limit (x,y) -> (0,0) for sin(x)cos(1/y)

Sandwich theorem is screaming at me, but I am not sure on what to do after setting up an inequality to prove the limit is 0. Perhaps I should try using the traditional definition of a limit to prove this.

Jacob

Edit:

Could I say

-1<_sin(x)cos(1/y)<_1

then multiply all three terms by (x^2)(y^2)

The left and right hand side both have a limit equal to 0. So then sin(x)cos(1/y) -->0 as (x,y)--> (0,0)

Maybe this is the right way to go about it.

Last edited by SenatorArmstrong; June 30th, 2017 at 08:42 AM.
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June 30th, 2017, 08:42 AM   #2
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0. cos(1/y) is bounded and sinx -> 0.
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July 1st, 2017, 05:36 AM   #3
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Quote:
Originally Posted by zylo View Post
0. cos(1/y) is bounded and sinx -> 0.
Yeah I know the solution is zero, but I was looking for more evidence of that since division by zero is undefined. Just plugging in the numbers would not work, I think the squeeze theorem in necessary.

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July 1st, 2017, 09:58 AM   #4
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What does "division by 0" have to do with this problem? Yes, there is a 1/y and y is "going to 0" but that has nothing to do with y actually being equal to 0.

The definition of "" is " if and only if, given any there exist a such that if then ". People often forget the "0< |x- a|" part but it should be there.

The limit as "y goes to a" depends on what the function is like close to a but has nothing to do with what happens to the function at a.

For example, is very obvious. But if f(x) is defined to be 3x- 2 for all x except 5 and defined to be any other number for x= 5, it is not so obvious, but still true, that the limit is still 13. Because of that "0< |x- a|", the value at x= 5 is entirely irrelevant to the limit.

By the way, that is why we can say that . which is equal to x+ 2 for all x except x= 2 since that would make the fraction . Again the value (or lack of value) at x= 2 is not relevant to the limit.
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Last edited by Country Boy; July 1st, 2017 at 10:01 AM.
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July 3rd, 2017, 02:02 AM   #5
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0$\displaystyle \leq$ |sinxcos(1/y)| $\displaystyle \leq$ |sinx| $\displaystyle \leq$ |x| all (x,y) -> 0
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