Calculus: Early Transcendentals, Chapter 2, Review Exercises, Section Review Exercises, Problem 16

Determine the limit $\displaystyle \lim_{x \to -\infty} \frac{1 - 2x^2 - x^4}{5 + x - 3x^4}$

By dividing both numerator and denominator by $x^4$ and by using the property of limits, we have

$
\begin{equation}
\begin{aligned}
\lim_{x \to -\infty} \frac{1 - 2x^2 - x^4}{5 + x - 3x^4} &= \lim_{x \to -\infty} \left( \frac{\frac{1}{x^4} - \frac{2}{x^2} -1}{\frac{5}{x^4} + \frac{1}{x^3} - 3} \right)\\
\\
&= \frac{\displaystyle \lim_{x \to -\infty}\frac{1}{x^4} - 2 \cdot \lim_{x \to -\infty} \frac{1}{x^2} - \lim_{x \to -\infty} 1
}{\displaystyle 5 \cdot \lim_{x \to -\infty} \frac{1}{x^4} + \lim_{x \to -\infty} \frac{1}{x^3} - \lim_{x \to -\infty} 3}\\
\\
&= \frac{0 - 2 \cdot 0 - 1}{5 \cdot 0 + 0 - 3}\\
\\
&= \frac{-1}{-3} = \frac{1}{3}
\end{aligned}
\end{equation}
$


Therefore,
$\displaystyle \lim_{x \to -\infty} \left( \frac{1 - 2x^2 - x^4}{5 + x - 3x^4} \right) = \frac{1}{3}$