# Horizontal asymptote of this function

• Mar 9th 2011, 07:15 PM
youngb11
Horizontal asymptote of this function
$f(x)=\sqrt{x^2+4x} - \sqrt{x^2+x}$

How would I find the horizontal asymptote of this function?
• Mar 9th 2011, 07:26 PM
TheChaz
This is ugly, but might be the best way....
(Considering "the bottom" to be "1")
Multiply top and bottom by the conjugate ... $\sqrt{x^2 + 4x} + \sqrt{x^2 + x}$

Then you'll have 3x on top. As $x \to \inf$, the bottom will become closer to just $\sqrt{x^2} + \sqrt{x^2} = 2\sqrt{x^2} = 2x$

So we're looking at 3x over 2x --> 3/2.

Yeah... I'm betting soroban can do it more elegantly!
• Mar 9th 2011, 07:32 PM
youngb11
Quote:

Originally Posted by TheChaz
This is ugly, but might be the best way....
(Considering "the bottom" to be "1")
Multiply top and bottom by the conjugate ... $\sqrt{x^2 + 4x} + \sqrt{x^2 + x}$

Then you'll have 3x on top. As $x \to \inf$, the bottom will become closer to just $\sqrt{x^2} + \sqrt{x^2} = 2\sqrt{x^2} = 2x$

So we're looking at 3x over 2x --> 3/2.

Yeah... I'm betting soroban can do it more elegantly!

Thanks a lot for the explanation! I think I'll be able to figure it out.
• Mar 9th 2011, 08:05 PM
Soroban
Hello, youngb11!

Thanks for the compliment, Chaz!
This is definitely not elegant, but it is more algebraically complete.

Quote:

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

$\text{How would I find the horizontal asymptote of this function?}$

We want: . $\displaystyle \lim_{x\to\infty}f(x)$ .if it exists.

As TheChaz suggested, multiply top and bottom by the conjugate:

. . $\displaystyle \frac{\sqrt{x^2+4x} - \sqrt{x^2+x}}{1} \cdot \frac{\sqrt{x^2+4x} + \sqrt{x^2+x}}{\sqrt{x^2+4x} + \sqrt{x^2+x}}$

. . $\displaystyle =\;\frac{(x^2+4x) - (x^2+x)}{\sqrt{x^2+4x} + \sqrt{x^2+x}} \;=\;\frac{3x}{\sqrt{x^2+4x} + \sqrt{x^2+x}}$

Divide top and bottom by $\,x\!:$

. . $\displaystyle =\;\frac{\dfrac{3x}{x}}{\dfrac{\sqrt{x^2+4x}}{x} + \dfrac{\sqrt{x^2+x}}{x}} \;=\;\dfrac{3}{\sqrt{\dfrac{x^2+4x}{x^2}} + \sqrt{\dfrac{x^2+x}{x^2}}}$

. . $\;=\;\dfrac{3}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{4x}{x ^2}} + \sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}} \;=\;\dfrac{3}{\sqrt{1+\dfrac{4}{x}} + \sqrt{1+\dfrac{1}{x}}}$

$\displaystyle \text{Therefore: }\;\lim_{x\to\infty} \dfrac{3}{\sqrt{1+\dfrac{4}{x}} + \sqrt{1+\dfrac{1}{x}}}$

. . . . . . . . . . $\displaystyle =\;\frac{3}{\sqrt{1+0} + \sqrt{1+0}} \;=\;\frac{3}{2}$

$\text{The horizontal asymptote is: }\:y \,=\,\frac{3}{2}$

• Mar 9th 2011, 08:06 PM
youngb11
Thanks a lot! And yes that was very elegant:p Very easy to understand thanks again!
• Mar 9th 2011, 08:10 PM
TheChaz
I'm just wondering how you saw the thread so quickly! (Of course, I'm also surprised/terrified by your answers from time to time ;) )