losfomot Posted September 21, 2005 Posted September 21, 2005 Here is the question: f(x)= (x^2-36)/(x-6) for 0<x<6 and 2x for 6<x<12 How should f(6) be defined in order for f to be continuous? I originally said 'no limit' but apparantly that is wrong. The answer is actually 12... which is fine for the second equation because 2x=2*6=12... but I don't understand how the first equation could equal 12? If I replace x with 6 in the first equation I get 0/0 could someone help me out with this?
matt grime Posted September 21, 2005 Posted September 21, 2005 you can factorize the numerator. the point was you were supposed to work out the limit of the first definition as x tends to 6,. not simply bung in 6 and notice it was a divde by zero thing. as the first part isn't defined at x=6 this doesn't make any sense to do.
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