# Thread: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

1. ## 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!
$\displaystyle \text{Evaluate }\underset{x\to \pm \infty }{\mathop{\lim }}\,\frac{{{\left( 1+x \right)}^{5}}-{{\left( 1-x \right)}^{5}}}{{{\left( 1+x \right)}^{5}}+{{\left( 1-x \right)}^{5}}}.$

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

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

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

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

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

$\displaystyle \text{ Graph shows }\underset{x\to -\infty }{\mathop{\lim }}\,\frac{{{\left( 1+x \right)}^{5}}-{{\left( 1-x \right)}^{5}}}{{{\left( 1+x \right)}^{5}}+{{\left( 1-x \right)}^{5}}}=-\infty$

2. ## 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.

3. ## 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 --- $\displaystyle \left( \frac{1}{0})\right$ is NOT indeterminate. So I can??? do ---

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

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

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

This gives the right answers. But is it 100% right?

4. ## Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

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

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.

5. ## Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

Hello, mathminor827!

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

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

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

Divide numerator and denominator by $\displaystyle x^4$

. . $\displaystyle \lim_{x\to\pm\infty}\frac{\frac{10}{x^3} + \frac{20}{x} + 2x}{\frac{2}{x^4} + \frac{20}{x^2} + 10} \;=\;\lim_{x\to\pm\infty}\frac{0+0+2x}{0+0+10} \;=\;\lim_{x\to\pm\infty}\frac{x}{5} \;=\;\pm\infty$

6. ## Re: Where's My Mistake? Limiting Behavior of Quintic Rational Polynomial

Originally Posted by thesmurfmaster
$\displaystyle \frac{2}{0}=+\infty$

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 $\displaystyle \frac{2}{0^+} = +\infty$ because I was taking limit as $\displaystyle x \to + \infty$.

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