# How do you find the limit of  (x^2+2x-1)/(3+3x^2) as x approaches oo?

Apr 27, 2016

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

#### Explanation:

When finding the limit as $x \to \infty$ of the ratio of polynomials in $x$, dividing by ${x}^{n}$, where $n$ is the greater of the degrees of the polynomials, allows for directly evaluating the limit.

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

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

$= \frac{1 + \frac{2}{\infty} - \frac{1}{\infty}}{3 + \frac{3}{\infty}}$

$= \frac{1 + 0 + 0}{3 + 0}$

$= \frac{1}{3}$

Note that with the degrees of the polynomials being equal, we have the result as the ratio of the coefficients of the first terms in the polynomial. This makes sense as when $x$ grows large, the contribution of the terms with lesser powers becomes negligible compared to the first term.