Question

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

Answer 1

`\lim_{x\to\pm\infty} \frac{(x-1)(x+2)(x+5)}{2x(x+2)} = \pm\infty`

Explanation:

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

`\frac{(x-1)cancel((x+2))(x+5)}{2xcancel((x+2))} = \frac{(x-1)(x+5)}{2x}`

From here, we can expand the numerator:

`\frac{(x-1)(x+5)}{2x} = \frac{x^2+4x-5}{2x}`

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

`\frac{x^2+4x-5}{2x} = \frac{x^2(1+4/x-5/x^2)}{2x} = \frac{x(1+4/x-5/x^2)}{2}`

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

So, since `\lim_{x\to\pm\infty} x = \pm\infty`, this will also be the behaviour of your expression.