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Evaluating a limit using L'Hopital

  1. Nov 12, 2011 #1
    1. The problem statement, all variables and given/known data
    For[tex]Q(x) = x^k + \sum_{n=0}^{k-1} a_n x^n[/tex]

    Find [tex]\lim_{x→∞}\left({Q(x)^{\frac{1}{k}} - x}\right)[/tex]


    2. Relevant equations
    L'Hopital


    3. The attempt at a solution
    [tex]Q(x) - x^k = \sum_{n=0}^{k-1} a_n x^n[/tex]
    [tex]Q(x) - x^k = (Q(x)^{\frac{1}{k}})^k - x^k = \left(Q(x)^{\frac{1}{k}} - x\right)\left([Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1} \right)[/tex]

    [tex]Q(x)^{\frac{1}{k}} - x = \frac{Q(x) - x^k}{[Q(x)^{\frac{1}{k}}]^{k-1} + x[Q(x)^{\frac{1}{k}}]^{k-2} + ... + x^{k-2}Q(x)^{\frac{1}{k}} + x^{k-1}}[/tex]

    Then [itex]\lim_{x→∞}{\left(Q(x)^{\frac{1}{k}} - x\right)} = \lim_{x→∞}\frac{f(x)}{g(x)}[/itex], and if I can show [itex]\lim_{x→∞}g(x) = ∞[/itex], I can use L'Hopital.

    Am I headed in the right direction? I can see this either working out nicely after a lot of work or being an incredible waste of time. Am I overlooking a simpler way to evaluate this limit?
     
  2. jcsd
  3. Nov 13, 2011 #2

    micromass

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    Try to factor things as follows:

    [tex]x^k-Q(x)=x^{k-1}\left(a_{k-1}+\frac{a_{k-2}}{x}+...+\frac{a_0}{x^{k-1}}\right)[/tex]

    and

    [tex](Q(x)^\frac{1}{k})^{k-1}=x^{k-1}(\sqrt[k]{1+\frac{a_{k-1}}{x}+\frac{a_{k-2}}{x^2}+...+\frac{a_0}{x^{k}}})^{k-1}[/tex]

    Try to factor all terms like that and use that things like [itex]\frac{1}{x^p}\rightarrow 0[/itex].
     
  4. Nov 13, 2011 #3
    Okay, so that will help me show that the limit of the denominator is ∞, but then taking the derivative is going to be a huge pain...
     
  5. Nov 13, 2011 #4

    micromass

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    No, you don't need to differentiate anything!!

    Do something like this

    \begin{align}
    \lim_{x\rightarrow +\infty}{\frac{x^3+x}{(\sqrt{x^2+x+1})^3}}
    &= \lim_{x\rightarrow +\infty} {\frac{ x^3(1+\frac{1}{x^2}) }{ x^3(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}})^3} }\\
    &= \lim_{x\rightarrow +\infty} {\frac{ 1+\frac{1}{x^2} }{(\sqrt{1+\frac{1}{x}+\frac{1}{x^2}})^3} }\\
    &= 1
    \end{align}
     
  6. Nov 13, 2011 #5
    Oh, okay. The problem hint suggested I use L'Hopital, so I guess I was thinking "in the box" when it came to evaluating the limit.

    Thanks for the suggestions!
     
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