Find horizontal asymptotes of a radical function

[LANS]

1. The problem statement, all variables and given/known data
Find the horizontal asymptotes for the following equation:



2. Relevant equations

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



3. 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]

1. The problem statement, all variables and given/known data
Find the horizontal asymptotes for the following equation:



2. Relevant equations

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



3. 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.