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Homework Help: Limit as x tends to infinity, without Laurent

  1. Jan 31, 2015 #1
    1. The problem statement, all variables and given/known data
    I want to find the following limit, ## \lim_{x \rightarrow \infty } x( \sqrt{ x^{2} +9} -x) ##, without using the Laurent series
    2. Relevant equations

    3. The attempt at a solution
    I used the Laurent Series to expand the square root, giving ## x((x+\frac{9}{2x})-x)##, then giving the limit as ##\frac{9}{2}## . How would one go about this question without using the above method?
  2. jcsd
  3. Jan 31, 2015 #2

    Simon Bridge

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    ... to me that reads:$$\lim_{x\to\infty}x\sqrt{x^2-x+9}$$ ... I don't think that converges: perhaps you meant to write something different.

    Mostly someone would deal with a limit like that by remembering how the functions work for the limit in question.
    i.e. as x gets very big, large powers of x will come to dominate a sum.

    So: $$\lim_{x\to\infty} \frac{x}{\sqrt{x^4-2x+9000}}=\lim_{x\to\infty}\frac{1}{x} = 0$$ ... which you can check on a calculator.

    I find it suggestive that the 9 under the root is a square number - so if you want to be more rigorous-ish, try completing the square, look for a substitution maybe?
    Last edited: Jan 31, 2015
  4. Jan 31, 2015 #3


    Staff: Mentor

    The OP has something different. I don't believe it was edited, but perhaps it was.

    @Skeptic, multiply by ##\sqrt{x^2 + 9} + x## over itself, and you'll be able to take the limit.
  5. Jan 31, 2015 #4

    Simon Bridge

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    Oh I see - my display is not showing the line over the square-root in post #1, and it didn't drag to quote in my reply either - but I do see it quoted in post #3.
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