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Homework Help: Show (x^3+2x)/(2x+1) < x^2 as x -> infinite

  1. Sep 29, 2008 #1
    Show (x^3+2x)/(2x+1) < x^2 as x --> infinite

    1. The problem statement, all variables and given/known data
    Show (x^3+2x)/(2x+1) < x^2 as x --> infinite


    2. Relevant equations
    This is not a formal proof. I just need to reduce the left side of the inequality to the point it is fairly obvious that it is less than x^2 as x-->infinite.


    3. The attempt at a solution

    so far I turned it into
    (x^3-1)/(2x+1) + 1 but it is still not broken down enough to be obvious. I'm not sure what to do next.
     
  2. jcsd
  3. Sep 30, 2008 #2

    Mentallic

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    Re: Show (x^3+2x)/(2x+1) < x^2 as x --> infinite

    I'm not sure how "obvious" you want it to be, but this should do if it isn't going to be a formal proof -

    [tex]\frac{x^{3}+2x}{2x+1}-x^{2}<0[/tex]

    [tex]\frac{x^{3}+2x-2x^{3}-x^{2}}{2x+1}<0[/tex]

    [tex]\frac{-x^{3}-x^{2}+2x}{2x+1}<0[/tex]

    [tex]\frac{-x(x+2)(x-1)}{2x+1}<0[/tex]

    The polynomial of x with degree 3 in the numerator dominates the linear polynomial in the denominator, so from here it can be seen that as [tex]x\rightarrow\infty[/tex], the value of the fraction [tex]\rightarrow\infty[/tex] and since the highest degree of x is negative, the value will approach [tex]-\infty[/tex] which is < 0
     
  4. Sep 30, 2008 #3

    statdad

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    Re: Show (x^3+2x)/(2x+1) < x^2 as x --> infinite

    Since [tex] x \to \infty [/tex] you can assume that [tex] x > 1 [/tex] so that [tex] 2x < x^3 [/tex]

    then, for such [tex] x [/tex]
    [tex]
    \frac{x^3 + 2x}{2x+1} < \frac{x^3 + x^3}{2x} = \frac{2x^3}{2x} = x^2
    [/tex]
     
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