thanks for the explanation of limit, but i want someone to help check out this problem for me

lim x->3 (x^2 - x - 12)/ (x - 3)
i got 5 as the answer

okay, so i cant use substitution.

i factor the (x^2 - x - 12) and got (x - 4) and (x + 3)

[(x - 4)(x + 3)]/ (x - 3)

i cant cancel anything out though
so i use multiply by the conjugate

[(x^2 - x - 12)/(x - 3)] [(x + 3) / (x + 3)] -->
(x^3 + 2x^2 - 15x -36) / (x^2 - 9)

then
numerator:
1/x^2 (x^3 + 2x^2 - 15x -36) -->
x + 2 - 15/x - 36/x^2

denominator:
1/x^2 (x^2 - 9) -->
1 - 9/x^2

got:
[x + 2 - 15/x - 36/x^2] over [1 - 9/x^2] -->
[x + 2 - 0 - 0] over [1 - 0] -->
3+ 2 = 5

is this right?

Are you sure you copied the problem correcty? It makes more sense if you are asked the limit of
(x^2 + x - 12)/ (x - 3),
For (x^2 - x - 12)/ (x - 3), as x>3 , the function becomes 6/0, which is infinity.

You should have gone no further than here:

[(x - 4)(x + 3)]/ (x - 3)

At this point you just examine the numerator and denominator. If the limit of the numerator is N and the limit of denominator is D, and if D is not zero, then the limit is:

N/D

If D is zero, like in this case, then there are two cases. Either N is non-zero, in which case the limit does not exist. Or N is also zero. In that case you need to investigate furher.

In this case N is nonzero, so the limit does not exist.

drwls - yes im sure i copied it right. i know it would have been much easier if it was (x^2 + x - 12). i don't know if this is a typo made by my instructor

Count Iblis - oh yea i forgot all about that rule. wow thanks a lot, wish i had realized it earlier to save tons of time

You're welcome! I'm glad I could help clarify things for you. It's always important to carefully double-check the problem statement, as even a small typo can lead to different outcomes and solutions. If you encounter similar problems in the future, remember to apply the rules and properties of limits systematically to find the correct solution. Don't hesitate to ask if you have any more questions!