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Math Help - Definition of Limit proof: x -> infinity of sqrt(x) - x = -infinity

  1. #1
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    Definition of Limit proof: x -> infinity of sqrt(x) - x = -infinity

    First post! yay!

    Okay so yesterday I had a problem set due and it had two definition of limit at infinity proofs on it. I ended up leaving them... well not blank but all I did was write the definition.

    the question was:
    prove the following statements using the appropriate definition of limit:
    (i) Limit as x-> infinity sqrt(x) - x = -infinity

    I got as far as:

    Given N<0, there exists M>0 such that
    x>M implies sqrt(x) - x < N

    which I believe is the appropriate definition.
    I always draw a blank on these proofs. The first thing I did when I made my account here was print the sticky on definition of limit proofs. I am going through it now.

    Anyways any help on this would be amazing, Thanks!
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  2. #2
    MHF Contributor chisigma's Avatar
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    Take into account that is...

    \displaystyle \sqrt{x}-x = - \frac{x\ \sqrt{x} - x}{\sqrt{x}} = - x\ \frac{\sqrt{x}-1}{\sqrt{x}}

    Kind regards

    \chi \sigma
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  3. #3
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    If N\ge 3 then N^2>2N or N^2-N>N.

    If J\in \mathbb{Z}^-~\&~J<-2 then if x\ge J^2 can you conclude that \sqrt{x}-x^2<J~?
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  4. #4
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    Hey thanks for the help guys, sorry I haven't been able to follow up on my question, I had a bunch of assignments due on Monday. So I was able to prove this limit as follows:
    I started playing with the inequality sqrt(x) - x < N. I multiplied by -1 to get x-sqrt(x)>-N --> sqrt(x)(sqrt(x) - 1) > -N. then since N<0 by assumption, --> -N>0.

    Now if sqrt(x) - 1 > sqrt(-N) and sqrt(x) > sqrt(x) - 1 > sqrt(-N) then we get sqrt(x)(sqrt(x)-1) > sqrt(-N)sqrt(-N) = -N.

    so I worked with sqrt(x) - 1 > sqrt(-N) ---> x > (1 + sqrt(-N))^2. So I concluded by choosing M = (1+sqrt(-N))^2 and showed that it works.

    Can someone just confirm with me that this proof works? and also how do i post the math symbols like you guys did above me?
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