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?

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.