# What is the limit of (x^2 + 1) / (2x^2 -3x -2) as x goes to infinity?

I found $\frac{1}{2}$
You can collect ${x}^{2}$ to get:
${\lim}_{x \to \infty} \frac{{x}^{2} \left(1 + \frac{1}{x} ^ 2\right)}{{x}^{2} \left(2 - \frac{3}{x} - \frac{2}{x} ^ 2\right)} = {\lim}_{x \to \infty} \frac{\cancel{{x}^{2}} \left(1 + \frac{1}{x} ^ 2\right)}{\cancel{{x}^{2}} \left(2 - \frac{3}{x} - \frac{2}{x} ^ 2\right)} = {\lim}_{x \to \infty} \frac{1 + \frac{1}{x} ^ 2}{2 - \frac{3}{x} - \frac{2}{x} ^ 2} =$ as $x \to \infty$ then $\frac{1}{x} ^ n \to 0$ so:
$= \frac{1 + 0}{2 - 0 - 0} = \frac{1}{2}$