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Math Help - Limts True or False Help

  1. #1
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    Limts True or False Help

    Hey all here is an easy calc. question but i can't seem to understand how to answer it

    Question:

    True or False? If the statement is true, explain why; if it is false, then explain how to determine the value of c and then, evaluate the limit.

    There is no number c such that \lim_{x \rightarrow\ -1} \frac{x^2 + cx - c + 2}{x^2 - 1} <br />
exists.


    The Part that I need help with:
    I know that answer is false but how do i evaluate c?

    Last edited by jospams; January 28th 2010 at 04:05 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by jospams View Post
    Hey all here is an easy calc. question but i can't seem to understand how to answer it

    Question:

    True or False? If the statement is true, explain why; if it is false, then explain how to determine the value of c and then, evaluate the limit.

    There is no number c such that \lim_{x \rightarrow\ -1} \frac{x^2 + cx - c + 2}{x^2 - 1}<br />


    The Part that I need help with:
    I know that answer is false but how do i evaluate c?

    There is no statement here to be false.
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    There is no statement here to be false.
    are you sure(someone in the class said it was false and i have to factor something to get the answer)? why is it true?
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  4. #4
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    The limit can only exist if there is a factor of (x+1) in the numerator to cancel with the (x+1) term in the denominator. If such a factor exists in the numerator, then the discontinuity at x=-1 will be removable (meaning the limit does exist) instead of infinite (meaning the limit does not exist).

    Therefore we must find what value of c causes x^2+cx-c+2 to be divisible by x+1. You can do this by using long division to find the remainder, and setting the remainder equal to zero.
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  5. #5
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    If you want to avoid long division, another method is to just set up the equality:

    x^2+cx-c+2 = (x+1)(x+k) where k is some constant...

    Therefore:

    x^2+cx-c+2 = x^2+(k+1)x+k

    So,

    c=k+1 and -c+2=k

    Solving this gives us c=3/2 (which is the same answer I got with long division method).
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  6. #6
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    Quote Originally Posted by Drexel28 View Post
    There is no statement here to be false.
    Yes, there was. The statement was "There is no number c such that ...".

    (Unless it was edited to change the statement before I got here.)
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Yes, there was. The statement was "There is no number c such that ...".

    (Unless it was edited to change the statement before I got here.)
    Exists wasn't there before. It just said

    "There is no c such that \lim\cdots"
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  8. #8
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    The question has been answered. Thread closed.
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