June 30th, 2017, 08:21 AM  #1 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  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. 
June 30th, 2017, 08:42 AM  #2 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
0. cos(1/y) is bounded and sinx > 0.

July 1st, 2017, 05:36 AM  #3 
Senior Member Joined: Nov 2015 From: Alabama Posts: 140 Thanks: 17 Math Focus: Geometry, Trigonometry, Calculus  
July 1st, 2017, 09:58 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,657 Thanks: 681 
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. Last edited by Country Boy; July 1st, 2017 at 10:01 AM. 
July 3rd, 2017, 02:02 AM  #5 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,134 Thanks: 88 
0$\displaystyle \leq$ sinxcos(1/y) $\displaystyle \leq$ sinx $\displaystyle \leq$ x all (x,y) > 0


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limit, multivariable 
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