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Math Help - Evaluating limits.

  1. #1
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    Evaluating limits.

    Evaluate the limit, if it exists. (If it does not exist, enter NONE).




    I know according to rule 5 of the law of limits, this cant work by substition because the denominator would equal 0. I know i need to start this off by factoring but im not sure how to start this process.

    Thanks.
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  2. #2
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    I think the next step would look somethin like this..

    (x-2)(x+3)
    --------------
    x-2

    ...but im not exactly positive.
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  3. #3
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    That's correct. Continue on.
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  4. #4
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    I got it, i made a mistake swapping a + sign for a -.... Thank you!
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  5. #5
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    Remember that when x= a makes polynomial P(x) zero- that is if P(a)= 0, then x-a must be a factor of P(x). Here, you saw correctly that the denominator is 0 when x= 2. That should make you look immediately at the numerator!

    If the numerator were NOT 0 at x= 2, that would tell you that the limit does not exist and you can stop.

    If the numerator is 0 at x= 2, which is the case here, 2^2+ 2- 6= 0, you know that you can factor out x-2. Once you know that, it should be clear that x^2+ x- 6= (x- 2)(x+ 3).

    Now you can think \frac{x^2+ x- 6}{x- 2}= \frac{(x-2)(x+3)}{x-2}= x+ 3 as long as x\ne 2.

    Notice that "as long as x\ne 2". Of course, \frac{x^2+x-6}{x-2} is "undefined" at x= 2 while x+3 is equal to 5 at x= 2 so we cannot just say that \frac{x^2+ x- 6}{x- 2}= x+3. Fortunately there is an (often overlooked) theorem about limits that says "If f(x)= g(x) for all x except x=a, then \lim_{x\to a} f(x)= \lim_{x\to a} g(x)".

    Knowing that \frac{x^2+x-6}{x-2}= x+ 3 for all x except x= 2 is enough to tell us they have the same limit at x= 2 and, of course, \lim_{x\to 2} x+ 3 is easy.
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  6. #6
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    great tip, i wasnt aware of that trick to determine if there is a limit! Awsome, i love this place.
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