# Thread: how to factorise this

1. ## how 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.

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

2. 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.

3. sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?

4. Originally Posted by Craka
sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?
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.

5. Sorry to clarify, the question ask to find the value for the for the following expression.

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

6. Originally Posted by Craka
Sorry to clarify, the question ask to find the value for the for the following expression.

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

Did you check the question like I asked?

To get that answer the numerator must be $\displaystyle x^2 - x {\color{red}-} 12$ NOT $\displaystyle x^2 - x {\color{red}+} 12$ as you've posted several times now.

$\displaystyle x^2 - x - 12 = (x - 4)(x + 3)$.

7. 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.

8. Originally Posted by Craka
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.
Looks like it.

9. Apply the L'HOSPITAL RULE
differentiate with respect rule
u will get
=lim x-->-3 (2x-1)
=2(-3)-1
=-6-1
=-7

10. Originally Posted by Craka
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.

$\frac{{x^2 - x + 12}}{{x + 3}}$
Originally Posted by Craka
sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?
Originally Posted by tutor
Apply the L'HOSPITAL RULE
differentiate with respect rule
u will get
=lim x-->-3 (2x-1)
=2(-3)-1
=-6-1
=-7
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).