Single Variable Calculus, Chapter 7, 7.8, Section 7.8, Problem 16

Determine the $\displaystyle \lim_{x \to \infty} \frac{x + x^2}{1 - 2x^2}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

$\displaystyle \lim_{x \to \infty} \frac{x + x^2}{1 - 2x^2} = \frac{\infty + \infty^2}{1-2(\infty)^2} = - \frac{\infty}{\infty} \text{ Indeterminate}$

Thus, by Applying L'Hospital's Rule,
$\displaystyle \lim_{x \to \infty} \frac{x+x^2}{1-2x^2} = \lim_{x \to \infty} \frac{1+2x}{1-4x}$
We will still get an indeterminate form if we evaluate the limit we obtained. Again,
By applying L'Hospital's Rule.

$\begin{equation} \begin{aligned} \lim_{x \to \infty} \frac{1+2x}{1-4x} &= \lim_{x \to \infty} \frac{2}{-4}\\ \\ &= \frac{2}{-4}\\ \\ &= -\frac{1}{2} \end{aligned} \end{equation}$