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 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. Thanks from romsek
July 1st, 2017, 05:36 AM   #3
Senior Member

Joined: Nov 2015
From: Alabama

Posts: 140
Thanks: 17

Math Focus: Geometry, Trigonometry, Calculus
Quote:
 Originally Posted by zylo 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.

Thank you!

 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 "$lim_{x\to a} f(x)= L$" is "$lim_{x\to a} f(x)= L$ if and only if, given any $\epsilon> 0$ there exist a $\delta> 0$ such that if $0< |x- a|< \delta$ then $|f(x)- L|< \epsilon$". 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, $\lim_{x\to 5} 3x- 2= 13$ 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 $\lim_{x\to 2}\frac{x^2- 4}{x- 2}= \limit_{x\to 2 } x+ 2= 4$. $\frac{x^2- 4}{x- 2}= \frac{(x- 2)(x+ 2)}{x- 2}$ which is equal to x+ 2 for all x except x= 2 since that would make the fraction $\frac{0}{0}$. Again the value (or lack of value) at x= 2 is not relevant to the limit. Thanks from SenatorArmstrong 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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