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Math Help - Compute the Limit: Question 1

  1. #1
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    Compute the Limit: Question 1

    Is my answer to the following question correct?

    \text{Calculate }\lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}

    = \frac{1^2+1-2}{1^2-1}

    = \frac{0}{0 }\text{Indeterminate}

    = \lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}

    = \lim_{x \rightarrow 1 } \frac{x-2}{-1}

    = \frac{1-2}{-1}

    = \frac{-1}{-1}

    = 1
    Last edited by sparky; September 20th 2011 at 05:38 PM.
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  2. #2
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    Re: Compute the Limit: Question 1

    Your second attempt is very strange, please review your algebra,

    \displaystyle \lim_{x\to 1}\frac{x^2+x-2}{x^2-1} = \lim_{x\to 1}\frac{(x+2)(x-1)}{(x-1)(x+1)} = \lim_{x\to 1}\frac{x+2}{x+1}= \dots
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  3. #3
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    Re: Compute the Limit: Question 1

    Is this correct now?

    \text{Calculate }\lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}

    = \frac{1^2+1-2}{1^2-1}

    = \frac{0}{0 }\text{Indeterminate}

    = \lim_{x \rightarrow 1 } \frac{x^2+x-2}{x^2-1}

    = \lim_{x \rightarrow 1 } \frac{(x+2) \cdot (x-1)}{(x+1) \cdot (x-1)}

    = \lim_{x \rightarrow 1 } \frac{x+2}{x+1}

    = \frac{1+2}{1+1}

    = \frac{3}{2}

    = 1\frac{1}{2}
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  4. #4
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    Re: Compute the Limit: Question 1

    That is much better.
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  5. #5
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    Re: Compute the Limit: Question 1

    Quote Originally Posted by pickslides View Post
    That is much better.
    Thanks so much pickslides
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