How do you calculate the limit of ([sin(xy-2)]^2)/(xy-2) as (x,y) approaches (1,2) and as (x,y) approaches (3,1) ?

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How do you calculate the limit of ([sin(xy-2)]^2)/(xy-2) as (x,y) approaches (1,2) and as (x,y) approaches (3,1) ?

Mathematics
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Hold x or y constant and let the other go to the lim. Take lim as xy goes to (x,2)
does the (x,y) approaches (1,2) exist?
Following my instructions above is a tool to investigate if it exists or not

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it 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.
So take the partial with respect to x, then plug in your values, then take the partial to y and plug in.
Well, 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
The 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.
This is actually a clever question based on lim you have seen in Cal I. Remember (sin x)/x
oh, so I'm supposed to use the product of limits theorem? thx
Several methods, as i said above lim is [sin^2 (2x-2)]/(2x-2)
Other method: break up sin^2 and based on the theorem one part is equal to 1 multiplied by the lim of sin (xy-2). Once again hold, one constant as you find lim.
Actually 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.
After 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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