Find horizontal asymptotes of a radical function

[LANS]

Homework Statement

Find the horizontal asymptotes for the following equation:

Homework Equations

$$f(x) = \sqrt{x^2+4x}-\sqrt{x^2+x}$$

The Attempt at a Solution

First I factored f(x):
$$f(x) = \sqrt{x}\sqrt{x+4}-\sqrt{x+1}$$
Then I conjugated it:
$$f(x) = \frac{x(x+4-x+1)}{\sqrt{x}\sqrt{x+4}-\sqrt{x+1}}$$
That's as far as I've been able to get. Any help would be appreciated.

edit: I "cheated" by plugging in big numbers and found the asymptote is y= -1.5

 

[lanedance]

Homework Statement

Find the horizontal asymptotes for the following equation:

Homework Equations

$$f(x) = \sqrt{x^2+4x}-\sqrt{x^2+x}$$

The Attempt at a Solution

First I factored f(x):
$$f(x) = \sqrt{x}\sqrt{x+4}-\sqrt{x+1}$$
Then I conjugated it:
$$f(x) = \frac{x[x+4-x+1]}{\sqrt{x}\sqrt{x+4}-\sqrt{x+1}}$$
That's as far as I've been able to get. Any help would be appreciated.

edit: I "cheated" by plugging in big numbers and found the asymptote is y= -1.5

do you mean (plus sign on denominator & brackets)
$$f(x) = \frac{x[x+4-x+1]}{\sqrt{x}(\sqrt{x+4}+\sqrt{x+1})}$$

i would start with
$$f(x) = \frac{3x}{\sqrt{x^2+4x}+\sqrt{x^2+x}}$$

now try taking x outside the denominator and cancelling with numerator (or equivalently multiply both by 1/x)

then take the limit as x goes to +- infinity

 

[LANS]

I fixed the brackets, and I'll try that tomorrow (I'm going to bed now). Thanks.