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Math Help - Limits

  1. #1
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    Limits

    Just need a couple of problems double checked for me.

    lim (x^2+4x-5)/x^2-1
    x->1

    I am saying that this limit does not exist because of the 0 numberator.

    In this one, I have forgotten the rules for square roots.

    lim [(sq rt 2x-1) -3]/x-5
    x->5-
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  2. #2
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    Quote Originally Posted by becky
    Just need a couple of problems double checked for me.

    lim (x^2+4x-5)/x^2-1
    x->1
    You have,
    \lim_{x\to 1}\frac{x^2+4x-5}{x^2-1}
    Express as,
    \frac{(x-1)(x+5)}{(x-1)(x+1)}
    Which cancels,
    \lim_{x\to 1}\frac{x+5}{x+1}=\frac{6}{2}=3
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  3. #3
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    Quote Originally Posted by becky

    lim [(sq rt 2x-1) -3]/x-5
    x->5-
    You have,
    \lim_{x\to 5}\frac{\sqrt{2x-1}-3}{x-5}
    Rationalize numerator,
    \frac{\sqrt{2x-1}-3}{x-5}\cdot \frac{\sqrt{2x-1}+3}{\sqrt{2x-1}+3}
    Thus,
    \frac{2x-10}{(x-5)(\sqrt{2x-1}+3)}
    Thus,
    \frac{2(x-5)}{(x-5)(\sqrt{2x-1}+3)}
    Thus,
    \lim_{x\to 5}\frac{2}{\sqrt{2x-1}+3}=\frac{2}{3+3}=1/3
    ----------------
    Basic rule, is that if you have a fraction that does not lead to division by zero upon substitution then you can substitute.
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