# Thread: Horizontal asymptote of this function

1. ## 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?

2. 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!

3. 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.

4. Hello, youngb11!

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

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

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

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 $\,x\!:$

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

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

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

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

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

5. Thanks a lot! And yes that was very elegant Very easy to understand thanks again!

6. 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 )