# math

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

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