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Math Help - "x^2 / (x-3)"

  1. #1
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    "x^2 / (x-3)"

    "x^2 / (x-3)"

    how can we simplfy this by making some cancellations so that we can get limit for 3?

    thanks...
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by zorspas View Post
    "x^2 / (x-3)"

    how can we simplfy this by making some cancellations so that we can get limit for 3?

    thanks...
    Hmmm if you want the limit when x tends to 3, you don't need to simplify...

    \lim_{x \to 3} x^2=9

    \lim_{x \to 3^+} \frac{1}{x-3}=+ \infty

    \lim_{x \to 3^-} \frac{1}{x-3}=- \infty
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  3. #3
    Senior Member nikhil's Avatar
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    Since right hand limit =+infinity and
    left hand limit =-infinity (already shown by Moo)
    therefor limit at x=3 does not exist.
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  4. #4
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    though there is no need for simplification, what can be done to simplify it?

    thanks...
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  5. #5
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    Quote Originally Posted by zorspas View Post
    though there is no need for simplification, what can be done to simplify it?

    thanks...
    \frac{x^2}{x-3} = \frac{[x^2 - 9] + 9}{x-3} = \frac{(x-3)(x+3) + 9}{x-3} = \frac{(x-3)(x+3)}{x-3} + \frac{9}{x-3} = x+3 + \frac{9}{x-3}.

    Or you can do polynomial long division and get the same result.
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