$\displaystyle f(x)=\sqrt{x^2+4x} - \sqrt{x^2+x}$
How would I find the horizontal asymptote of this function?
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!
Hello, youngb11!
Thanks for the compliment, Chaz!
This is definitely not elegant, but it is more algebraically complete.
$\displaystyle f(x)=\sqrt{x^2+4x} - \sqrt{x^2+x}$
$\displaystyle \text{How would I find the horizontal asymptote of this function?}$
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}}} $
. . . . . . . . . . $\displaystyle \displaystyle =\;\frac{3}{\sqrt{1+0} + \sqrt{1+0}} \;=\;\frac{3}{2}$
$\displaystyle \text{The horizontal asymptote is: }\:y \,=\,\frac{3}{2}$