# How do you find lim (2x^2-4x+1)/(3x^2+5x-6) as x->oo?

Mar 2, 2017

${\lim}_{x \to \infty} \frac{2 {x}^{2} - 4 x + 1}{3 {x}^{2} + 5 x - 6} = \frac{2}{3}$

#### Explanation:

We can manipulate the limit as follows

${\lim}_{x \to \infty} \frac{2 {x}^{2} - 4 x + 1}{3 {x}^{2} + 5 x - 6} = {\lim}_{x \to \infty} \frac{2 {x}^{2} - 4 x + 1}{3 {x}^{2} + 5 x - 6} \cdot \frac{\frac{1}{x} ^ 2}{\frac{1}{x} ^ 2}$

$\text{ } = {\lim}_{x \to \infty} \frac{2 - \frac{4}{x} + \frac{1}{x} ^ 2}{3 + \frac{5}{x} - \frac{6}{x} ^ 2}$

$\text{ } = \frac{2}{3}$

As both $\frac{1}{x} \rightarrow 0$ and $\frac{1}{x} ^ 2 \rightarrow 0$ as $x \rightarrow \infty$