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Math Help - Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

  1. #1
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    Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

    Hi all ---

    I'm trying to compute the limits of the following function. After using a computer to check my work, I know my solution is wrong. But the problem is --- I can't find my mistake! Can someone please point it out? Thanks!
      \text{Evaluate }\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{\left( 1+x \right)}^{5}}-{{\left( 1-x \right)}^{5}}}{{{\left( 1+x \right)}^{5}}+{{\left( 1-x \right)}^{5}}}.

    \underline{\text{My work:}}\text{ I divide top and bottom by the highest power of the rational function --- }{{x}^{5}}.

    \text{Then }\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{\frac{{{\left( 1+x \right)}^{5}}-{{\left( 1-x \right)}^{5}}}{{{x}^{5}}}}{\frac{{{\left( 1+x \right)}^{5}}+{{\left( 1-x \right)}^{5}}}{{{x}^{5}}}}=\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{\frac{{{\left( 1+x \right)}^{5}}}{{{x}^{5}}}-\frac{{{\left( 1-x \right)}^{5}}}{{{x}^{5}}}}{\frac{{{\left( 1+x \right)}^{5}}}{{{x}^{5}}}+\frac{{{\left( 1-x \right)}^{5}}}{{{x}^{5}}}}

    =\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{\left( \frac{1+x}{x} \right)}^{5}}-{{\left( \frac{1-x}{x} \right)}^{5}}}{{{\left( \frac{1+x}{x} \right)}^{5}}+{{\left( \frac{1-x}{x} \right)}^{5}}} =\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{\left( 1+\frac{1}{x} \right)}^{5}}-{{\left( \frac{1}{x}-1 \right)}^{5}}}{{{\left( \frac{1}{x}+1 \right)}^{5}}+{{\left( \frac{1}{x}-1 \right)}^{5}}}

     =\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{\left( 1+0 \right)}^{5}}-{{\left( 0-1 \right)}^{5}}}{{{\left( 0+1 \right)}^{5}}+{{\left( 0-1 \right)}^{5}}}=\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{1-(-1)}{1-1} =\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{2}{0}=+\infty

      \text{This is right as }x\to \infty .\text{ But WRONG as }x\to -\infty .

    \text{ Graph shows }\underset{x\to -\infty }{\mathop{\lim }}\,\frac{{{\left( 1+x \right)}^{5}}-{{\left( 1-x \right)}^{5}}}{{{\left( 1+x \right)}^{5}}+{{\left( 1-x \right)}^{5}}}=-\infty
    Last edited by mathminor827; June 7th 2012 at 04:51 PM.
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  2. #2
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    Cool Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

    The denomination is a polynomial in x^4. So dividing by x^5 causes a division by zero that did not exist in the original equation.

    You can solve this problem by applying L'hopital's rule 4 times.
    Thanks from mathminor827
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  3. #3
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    Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

    Hi Kiwi_Dave --- Thank you very much. I understand your first sentence.
    But I don't think you can can apply L'hopital's rule. Here's why ---

    If I'm right -- sorry if I'm not --- \left( \frac{1}{0})\right is NOT indeterminate. So I can??? do ---

    =\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{\left( \frac{1+x}{x} \right)}^{5}}-{{\left( \frac{1-x}{x} \right)}^{5}}}{{{\left( \frac{1+x}{x} \right)}^{5}}+{{\left( \frac{1-x}{x} \right)}^{5}}} =\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{\left( 1+\frac{1}{x} \right)}^{5}}-{{\left( \frac{1}{x}-1 \right)}^{5}}}{{{\left( \frac{1}{x}+1 \right)}^{5}}+{{\left( \frac{1}{x}-1 \right)}^{5}}}

       & \text{so }\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{\left( 1+\frac{1}{x} \right)}^{5}}-{{\left( \frac{1}{x}-1 \right)}^{5}}}{\text{(terms with power  4)}}=\frac{1-(-1)}{{{0}^{+}}}=\frac{2}{{{0}^{+}}}=\infty

     \text{so }\underset{x\to -\infty }{\mathop{\lim }}\,\frac{{{\left( 1+\frac{1}{x} \right)}^{5}}-{{\left( \frac{1}{x}-1 \right)}^{5}}}{\text{(terms with power  4)}}=\frac{1-(-1)}{{{0}^{-}}}=\frac{2}{{{0}^{-}}}=-\infty

    This gives the right answers. But is it 100% right?
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  4. #4
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    Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

    \frac{2}{0}=+\infty

    How did you conclude this? You can see in your original equation that the denominator will only be positive if x goes to positive infinity, if it goes to negative infinity it will be negative; and since the numerator is positive in either case the limit will be negative if x goes to negative infinity.
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  5. #5
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    Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

    Hello, mathminor827!

    \text{Evaluate: }\;\lim_{x\to \pm \infty }\frac{(1+x)^5- (1-x)^5}{(1+x)^5+( 1-x)^5}

    \begin{array}{ccccc} \text{Numerator:} & (1 + x)^5 - (1-x)^5 &=& 10x + 20x^3 + 2x^5 \\ \text{Denominator:} & (1+x) + (1-x)^5 &=& 2 + 20x^2 + 10x^4 \end{array}

    We have: . \lim_{x\to\pm\infty}\frac{10x + 20x^3 + 2x^5}{2 + 20x^2 + 10x^4}


    Divide numerator and denominator by x^4

    . . \lim_{x\to\pm\infty}\frac{\frac{10}{x^3} + \frac{20}{x} + 2x}{\frac{2}{x^4} + \frac{20}{x^2} + 10} \;=\;\lim_{x\to\pm\infty}\frac{0+0+2x}{0+0+10} \;=\;\lim_{x\to\pm\infty}\frac{x}{5} \;=\;\pm\infty
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    Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

    Quote Originally Posted by thesmurfmaster View Post
    \frac{2}{0}=+\infty

    How did you conclude this? You can see in your original equation that the denominator will only be positive if x goes to positive infinity, if it goes to negative infinity it will be negative; and since the numerator is positive in either case the limit will be negative if x goes to negative infinity.
    Hi thesmurfmaster --- Thank you for your answer. I got  \frac{2}{0^+} = +\infty because I was taking limit as  x \to + \infty .

    Hi Soroban --- Thank you for your answer. I understand your solution in which you divided by  x^4 . But does my original one work? I divided by  x^2 .
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