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Math Help - how to factorise this

  1. #1
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    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.

    \[
    \frac{{x^2 - x + 12}}{{x + 3}}
    \]
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  2. #2
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    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.
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    sorry should be as x approaches -3 (not 3). How would I go about finding the limit then in this case?
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    Quote Originally Posted by Craka View Post
    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.
    Last edited by mr fantastic; August 3rd 2008 at 07:48 PM. Reason: Deleted some irrelevancies.
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  5. #5
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    Sorry to clarify, the question ask to find the value for the for the following expression.

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

    The answer given is -7.
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  6. #6
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    Quote Originally Posted by Craka View Post
    Sorry to clarify, the question ask to find the value for the for the following expression.

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

    The answer given is -7.
    Did you check the question like I asked?

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

    x^2 - x - 12 = (x - 4)(x + 3).
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  7. #7
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    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.
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  8. #8
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    Quote Originally Posted by Craka View Post
    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.
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  9. #9
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    Apply the L'HOSPITAL RULE
    differentiate with respect rule
    u will get
    =lim x-->-3 (2x-1)
    =2(-3)-1
    =-6-1
    =-7
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  10. #10
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    Quote Originally Posted by Craka View Post
    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}}

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