# How do you find the end behavior of  [(x–1)(x+2)(x+5)] / [x(x+2)2]?

Nov 18, 2015

$\setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} \setminus \frac{\left(x - 1\right) \left(x + 2\right) \left(x + 5\right)}{2 x \left(x + 2\right)} = \setminus \pm \setminus \infty$

#### Explanation:

First of all, the expression can be simplified, since a factor $\left(x + 2\right)$ appears in both numerator and denominator. So, we have

$\setminus \frac{\left(x - 1\right) \cancel{\left(x + 2\right)} \left(x + 5\right)}{2 x \cancel{\left(x + 2\right)}} = \setminus \frac{\left(x - 1\right) \left(x + 5\right)}{2 x}$

From here, we can expand the numerator:

$\setminus \frac{\left(x - 1\right) \left(x + 5\right)}{2 x} = \setminus \frac{{x}^{2} + 4 x - 5}{2 x}$

To study the end behavior, simply factor our the highest power in both numerator and denominator:

$\setminus \frac{{x}^{2} + 4 x - 5}{2 x} = \setminus \frac{{x}^{2} \left(1 + \frac{4}{x} - \frac{5}{x} ^ 2\right)}{2 x} = \setminus \frac{x \left(1 + \frac{4}{x} - \frac{5}{x} ^ 2\right)}{2}$

Now, as $x \setminus \to \setminus \pm \setminus \infty$, you have that $\frac{4}{x}$ and $- \frac{5}{x} ^ 2$ tend to zero. So, the whole expression has the same behaviour as $\frac{x}{2}$ (which again is the same behaviour as $x$).

So, since $\setminus {\lim}_{x \setminus \to \setminus \pm \setminus \infty} x = \setminus \pm \setminus \infty$, this will also be the behaviour of your expression.