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anonymous
 5 years ago
How do you calculate the limit of ([sin(xy2)]^2)/(xy2) as (x,y) approaches (1,2) and as (x,y) approaches (3,1) ?
anonymous
 5 years ago
How do you calculate the limit of ([sin(xy2)]^2)/(xy2) as (x,y) approaches (1,2) and as (x,y) approaches (3,1) ?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hold x or y constant and let the other go to the lim. Take lim as xy goes to (x,2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does the (x,y) approaches (1,2) exist?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Following my instructions above is a tool to investigate if it exists or not

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it looks like it doesn't, but it's a situation where you have a 0/0 case, so it wouldn't hurt to do a l'hospital's expession for x and y using partial derivs. If, the two limits in each variable are different then the limit does not exist. If they are equal then it does.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So take the partial with respect to x, then plug in your values, then take the partial to y and plug in.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, when you did that, did you get that it existed? I got that it did not; @ daomoyan: I don;t think you can useL'Hopitals in this case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The limit does exist. Even though there is a hole; you can still find the limit because you are only approaching that point (but you never actually get there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is actually a clever question based on lim you have seen in Cal I. Remember (sin x)/x

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh, so I'm supposed to use the product of limits theorem? thx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Several methods, as i said above lim is [sin^2 (2x2)]/(2x2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Other method: break up sin^2 and based on the theorem one part is equal to 1 multiplied by the lim of sin (xy2). Once again hold, one constant as you find lim.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually because there is doubt. You check in both ways. One time hold x. One time hold y. And if answer is different, it confirms that you are actually right, the lim does not exist.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0After further review, you are correct, the lim does not exist because when hold x constant you get one value, when you hold y constant you get another. The results are different, confirms that you are right, the lim does not exist.
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