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Math Help - Derivative - Limits

  1. #1
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    Question Derivative - Limits

    Hello, I have a problem with which I seem to be getting nowhere. I have to find the value of the limit, and the answer says that it is 3/2

    lim 5x^2-7x+2/x^2-1

    lim x-> 1

    Thanks for the help.
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  2. #2
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    Quote Originally Posted by cokeclassic View Post
    Hello, I have a problem with which I seem to be getting nowhere. I have to find the value of the limit, and the answer says that it is 3/2

    lim 5x^2-7x+2/x^2-1

    lim x-> 1

    Thanks for the help.
    Notice that

    5x^2 - 7x + 2 = (5x - 2)(x - 1) and x^2 - 1 = (x + 1)(x - 1).


    So \frac{5x^2 - 7x + 2}{x^2 - 1} = \frac{(5x - 2)(x - 1)}{(x + 1)(x - 1)}

     = \frac{5x - 2}{x + 1}.


    Therefore

    \lim_{x \to 1} \frac{5x^2 - 7x + 2}{x^2 - 1} = \lim_{x \to 1} \frac{5x - 2}{x + 1}

     = \frac{5(1) - 2}{1 + 1}

     = \frac{3}{2}.
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  3. #3
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    Quote Originally Posted by cokeclassic View Post
    Hello, I have a problem with which I seem to be getting nowhere. I have to find the value of the limit, and the answer says that it is 3/2

    lim 5x^2-7x+2/x^2-1

    lim x-> 1

    Thanks for the help.
    Start by noting that \lim_{x \rightarrow 1} \frac{5x^2 - 7x + 2}{x^2 - 1} = \lim_{x \rightarrow 1} \frac{(5x - 2)(x - 1)}{(x - 1)(x+1)}
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  4. #4
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    Quote Originally Posted by mr fantastic View Post
    Start by noting that \lim_{x \rightarrow 1} \frac{5x^2 - 7x + 2}{x^2 - 1} = \lim_{x \rightarrow 1} \frac{(5x - 2)(x - 1)}{(x - 1)(x+1)}
    Beat ya :P
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  5. #5
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    Notice that the reason you could not just set x= 1 and evaluate is because both numerator and denominator are equal to 0 when x= 1. You should then recognize that if a polynomial is 0 at x= a, (x- a) is a factor. Recognizing that x-1 must be a factor of 5x^2- 7x+ 2 you can just divide it by x-1 to find the other factor.
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