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Math Help - Proving a limit of a function

  1. #1
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    Proving a limit of a function

    Prove that (x^3-8)/(x^2+x-6) \rightarrow \infty as x\rightarrow\infty
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Slazenger3 View Post
    Prove that (x^3-8)/(x^2+x-6) \rightarrow \infty as x\rightarrow\infty
    For large enough  x,\; x^2+x-6 > x^2 \implies \frac{x^3-8}{x^2+x-6} < \frac{x^3-8}{x^2}=x-\frac8x .

    Now apply the limit comparison test.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by chiph588@ View Post
    For large enough  x,\; x^2+x-6 > x^2 \implies \frac{x^3-8}{x^2+x-6} < \frac{x^3-8}{x^2}=x-\frac8x .

    Now apply the limit comparison test.
    Except what you've shown is that our limit is less than infinity
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  4. #4
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Except what you've shown is that our limit is less than infinity
    Haha!

    Let's do this again:

     \frac{x^3-8}{x^2+x-6} = \frac{x^2+2x+4}{x+3}


    Let  x=t-3 :

     \frac{(t-3)^2+2(t-3)+4}{t} = \frac{t^2-4t+7}{t} = t-4+\frac7t

    So  \lim_{x\to\infty} \frac{x^3-8}{x^2+x-6} = \lim_{t\to\infty} \left(t-4+\frac7t\right) = \infty
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  5. #5
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    Quote Originally Posted by Slazenger3 View Post
    Prove that (x^3-8)/(x^2+x-6) \rightarrow \infty as x\rightarrow\infty
    A common proof (valid for any coefficients) would go along these lines: dividing numerator and denominator by x^2, we have \frac{x^3-8}{x^2+x-6}= \frac{x-\frac{8}{x^2}}{1+\frac{1}{x}-\frac{6}{x^2}}. Under this form, the limits of the terms are immediate to see; in particular, the numerator goes to +\infty and the denominator converges to 1, hence the ratio goes to +\infty.
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