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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

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)