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Math Help - [SOLVED] logic for determining limits with absolute value.

  1. #1
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    [SOLVED] logic for determining limits with absolute value.

    Given
    f(x)=\frac{x^2-1}{|x-1|}

    \lim_{x\to1+}{f(x)}

    \lim_{x\to1-}{f(x)}

    Is the numerator reducible? I am just not seeing the method I should start with, a push in the correct direction would be appreciated.
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  2. #2
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    Quote Originally Posted by hechz View Post
    Given
    f(x)=\frac{x^2-1}{|x-1|}

    \lim_{x\to1+}{f(x)}

    \lim_{x\to1-}{f(x)}

    Is the numerator reducible? I am just not seeing the method I should start with, a push in the correct direction would be appreciated.
    \frac{x^2-1}{|x-1|} = \frac{(x+1)(x-1)}{|x-1|}

    recall the definition of absolute value.

    for x < 1 ...

    \frac{(x+1)(x-1)}{|x-1|} = \frac{(x+1)(x-1)}{-(x-1)} = -(x+1)

    for x > 1 ...

    \frac{(x+1)(x-1)}{|x-1|} = \frac{(x+1)(x-1)}{x-1} = x+1
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  3. #3
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    Thanks... I have no idea why I didn't see such a basic identity.


    /goes to re-read HS algebra book
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