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Math Help - Limits

  1. #1
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    Limits

    Use the epsilon definition of the limit that \lim_{x\rightarrow \infty }f(x)=L if for every positive real number \epsilon, there exists a real number M such that \left | f(x)-L\right |< \epsilon whenever x>M, to prove \lim_{x\rightarrow \infty }\frac{x+3}{x^2-3}=0
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  2. #2
    MHF Contributor Calculus26's Avatar
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    I'm assuming your function is (x+3)/(x^2-9)

    See attachment
    Attached Thumbnails Attached Thumbnails Limits-limit2.jpg  
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  3. #3
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    thanks for the help. So is it not possible to do the initial question?
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  4. #4
    MHF Contributor Calculus26's Avatar
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    No it is very possible though a little more difficult-- my impression is that at this level it would seem the one I suggested would be more relevant.

    However for the original

    | (x + 3)/(x-3)(x+3)| < |2x/(x+3)(x-3| since for x > 3 x + x > x+3


    |2x/(x+3)(x-3)|< | 2x/(x)(x-3)| smaller denominator bigger fraction x < x+3

    | 2x/(x)(x-3)| = 2/(x-3) x > 3

    2/(x-3) < e

    1/(x-3) < e/2


    again if x- 3 > e/2

    1/(x-3) < e/2

    M = 3 + e/2
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