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

How would I find the horizontal asymptote of this function?

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- Mar 9th 2011, 07:15 PMyoungb11Horizontal asymptote of this function
$\displaystyle f(x)=\sqrt{x^2+4x} - \sqrt{x^2+x}$

How would I find the horizontal asymptote of this function? - Mar 9th 2011, 07:26 PMTheChaz
This is ugly, but might be the best way....

(Considering "the bottom" to be "1")

Multiply top and bottom by the conjugate ... $\displaystyle \sqrt{x^2 + 4x} + \sqrt{x^2 + x}$

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

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 PMyoungb11
- Mar 9th 2011, 08:05 PMSoroban
Hello, youngb11!

Thanks for the compliment, Chaz!

This is definitely not elegant, but it is more algebraically complete.

Quote:

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

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

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

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

. . $\displaystyle \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 \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 $\displaystyle \,x\!:$

. . $\displaystyle \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}}} $

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

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

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

- Mar 9th 2011, 08:06 PMyoungb11
Thanks a lot! And yes that was very elegant:p Very easy to understand thanks again!

- Mar 9th 2011, 08:10 PMTheChaz
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 ;) )