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Math Help - Limits Question

  1. #1
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    Limits Question

    On a practice exam for our exam tomorrow our professor has a problem

    lim x -> 3- (x^2 2-x) / (2x^2 -5x -3)

    I plug in 3 and I wind up getting 3/0 since 0 can't be in the denominator it should be no limit??

    He has negative infinity as the correct answer?
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by KarlosK View Post
    On a practice exam for our exam tomorrow our professor has a problem

    lim x -> 3- (x^2 2-x) / (2x^2 -5x -3)

    I plug in 3 and I wind up getting 3/0 since 0 can't be in the denominator it should be no limit??

    He has negative infinity as the correct answer?
    \lim_{x \to 3^-} \frac{x^2 (2-x)}{2x^2 -5x -3}=\lim_{x \to 3^-} \frac{x^2 (2-x)}{(2x+1)(x-3)}

    Because x=3 does not make the numerator zero, it is going to be an infinity type of limit. To determine which type, think about the sign of x-3 as x approaches 3 from the left. So for x=2.999, note that x-3 is very close to zero, but still negative. Hence the overall fraction will approach negative infinity.
    I hope this helps!

    Good luck.
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  3. #3
    Member mathemagister's Avatar
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    Quote Originally Posted by KarlosK View Post
    On a practice exam for our exam tomorrow our professor has a problem

    lim x -> 3- (x^2 2-x) / (2x^2 -5x -3)

    I plug in 3 and I wind up getting 3/0 since 0 can't be in the denominator it should be no limit??

    He has negative infinity as the correct answer?
    To clarify, negative infinity is what the function is approaching as x\rightarrow 3^-, but since infinity does not exist, even the one-sided limit does not exist.

    Hope that clears it up

    Mathemagister
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