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 $