Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left.

\[

\frac{{x^2 - x + 12}}{{x + 3}}

\]

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- Aug 3rd 2008, 06:09 PMCrakahow to factorise this
Trying to jog my memory here, how would I go about factorising this, as I'm trying to find its limit as x approaches 3 from the left.

\[

\frac{{x^2 - x + 12}}{{x + 3}}

\] - Aug 3rd 2008, 06:20 PMskeeter
fyi, the numerator will not factor, but why factor at all? there is no discontinuity at x = 3 ... just plug in 3 for x and calculate the limit straight up.

- Aug 3rd 2008, 06:49 PMCraka
sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?

- Aug 3rd 2008, 07:02 PMmr fantastic
It's already been pointed out that the numerator does not factorise. So you can't cancel a common factor of x + 3.

The limit approaches -oo as x --> -3 from the left.

If you don't like this answer, I suggest you go back to the original question and check whether you made a mistake in copying it. If you haven't made a mistake in the equation posted, then I'm afraid you're stuck with the answer I've given. - Aug 3rd 2008, 07:27 PMCraka
Sorry to clarify, the question ask to find the value for the for the following expression.

$\displaystyle

\mathop {\lim }\limits_{x \to ( - 3)} \frac{{x^2 - x + 12}}{{x + 3}}

$

The answer given is -7. - Aug 3rd 2008, 07:47 PMmr fantastic
- Aug 3rd 2008, 08:05 PMCraka
Yes I have checked the question several times. That is the question being asked. From what you are saying it appears there is a mistake in the question being asked on the paper.

- Aug 3rd 2008, 09:50 PMmr fantastic
- Aug 5th 2008, 08:06 AMtutor
Apply the L'HOSPITAL RULE

differentiate with respect rule

u will get

=lim x-->-3 (2x-1)

=2(-3)-1

=-6-1

=-7 - Aug 5th 2008, 01:49 PMmr fantastic
NO!!

The limit does NOT have an indeterminant form 0/0. Using l'Hopital's Rule here is totally wrong.

This is the problem with l'Hopital's Rule - it's a lazy approach that is easily abused.

(And in a pre-algebra forum I doubt that l'Hopital's Rule has ever even been heard of).