need to know how to do problem #2 from http://www.math.psu.edu/files/140FinalsamA.pdf
could you also show what it would equal for a one sided limit from either side?
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factor out of a negative from the absolute value so it's (x|x-3|)/x-3 since absolute value of -1 is just 1... now you want to cancel the x-3 so you have to get rid of the absolute value bars... when you have absolute value bars you have two cases, either
-x(x-3) or x(x-3) on top. In both cases, the x-3 will cancel, leaving -x in case 1 and +x in the other so the limit dne
ok thanks. so it doesn't exist as a one sided (i.e. approaching 3 from the + side or approaching 3 from the - side) either, correct?
I take that back, it equals -x when it approaches from the + side and x when it approaches from the - side