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

Dec 4, 2016

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

#### Explanation:

You can separate the numerator in two parts, one of which can be divided by (x+1). For instance add and subtract $1$ and you get:

$\frac{{x}^{3} + 2}{x + 1} = \frac{{x}^{3} + 1 + 2 - 1}{x + 1} = \frac{{x}^{3} + 1}{x + 1} + \frac{1}{x + 1}$

As $\overline{x} = - 1$ is a root of ${x}^{3} + 1$, then $\left(x + 1\right)$ must be a factor, and in fact:

$\left({x}^{3} + 1\right) = \left({x}^{2} - x + 1\right) \left(x + 1\right)$

so that:

$\frac{{x}^{3} + 2}{x + 1} = \frac{\left({x}^{2} - x + 1\right) \left(x + 1\right)}{x + 1} + \frac{1}{x + 1} = \left({x}^{2} - x + 1\right) + \frac{1}{x + 1}$

Now go for the limit: the first addendum tends to infinity, the second addendum tends to zero.

${\lim}_{x \to \infty} \frac{{x}^{3} + 2}{x + 1} = {\lim}_{x \to \infty} \left[\left({x}^{2} - x + 1\right) + \frac{1}{x + 1}\right] = + \infty$