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

Mar 17, 2017

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

#### Explanation:

When evaluating the limit at $\pm \infty$ of a rational function you can ignore all the terms except the ones of highest order at the numerator and denominator:

${\lim}_{y \to \infty} \frac{{y}^{4} - {y}^{3} + 1}{2 {y}^{4} - 4 {y}^{2} + 5} = {\lim}_{y \to \infty} {y}^{4} / \left(2 {y}^{4}\right) = \frac{1}{2}$

It's easy to see why this is the case: start noting that around $\pm \infty$ we can assume that $y \ne 0$ so also ${y}^{4} \ne 0$.

We can then divide both the numerator and denominator by ${y}^{4}$:

${\lim}_{y \to \infty} \frac{{y}^{4} - {y}^{3} + 1}{2 {y}^{4} - 4 {y}^{2} + 5} = {\lim}_{y \to \infty} \frac{\frac{{y}^{4} - {y}^{3} + 1}{y} ^ 4}{\frac{2 {y}^{4} - 4 {y}^{2} + 5}{y} ^ 4} = {\lim}_{y \to \infty} \frac{1 - \frac{1}{y} + \frac{1}{y} ^ 4}{2 - \frac{4}{y} ^ 2 + \frac{5}{y} ^ 4} = \frac{1 - 0 + 0}{2 - 0 + 0} = \frac{1}{2}$