Question

Find all horizontal asymptotes for (4x)/( √x^2+9)

Answer 1

Strange looking function you have there. Do y ou actually mean\[f(x)=\frac{ 4x }{ (\sqrt{x})^2+9 }?\]

If you can either assure me that I have read your question correctly, or otherwise provide the correct function, I'd be happy to continue helping you.

Answer 2

No the radical covers all of x^2+9 @mathmale

Answer 3

\(\color{black}{\displaystyle f(x)=\frac{4x}{\sqrt{x^2+9}} }\) is probably the function considered.

Answer 4

Yes @SolomonZelman

Answer 5

\(\color{black}{\displaystyle \lim_{x\to \infty}\frac{4x}{\sqrt{x^2+9}}= \lim_{x\to \infty} \sqrt{\frac{16x^2}{x^2+9}}=4}\)

\(\color{black}{\displaystyle \lim_{x\to -\infty}\frac{4x}{\sqrt{x^2+9}}= -\lim_{x\to \infty} \sqrt{\frac{16x^2}{x^2+9}}=-4}\)

Answer 6

that is the basic work behind your two horizontal asymptotes.
If no questions are asked I will assume you got it.

Answer 7

I sort of forgot how to use limits to infinity. I see why you made 4x to 16x^2 but can you explain how you got 4

Answer 8

L'Hospital's rule wice

Answer 9

twice

Answer 10

Remember that rule? (I can certainly explain it, but Dr. Google is filled with a variety of tutorials with definitions and examples for it, so ....)

Answer 11

You can multiply both top and bottom by 1/x if you want to avoid L'Hospital's rule

Answer 12

basically,

\(\color{black}{\displaystyle \color{black}{\displaystyle \lim_{x\to \infty}\frac{4x}{\sqrt{x^2+9}}= \lim_{x\to \infty} \sqrt{\frac{16x^2}{x^2+9}}= \sqrt{\lim_{x\to \infty}\frac{16x^2}{x^2+9}}=\sqrt{\lim_{x\to \infty}\frac{32x}{2x}}} }\)

\(\color{black}{\displaystyle =\sqrt{\lim_{x\to \infty}\frac{32}{2}} =\sqrt{\lim_{x\to \infty}16}=4 }\)

Answer 13

alternatively, \[\frac{4x}{\sqrt{x^2}}=\frac{4x}{|x|}=4, x>0, -4, x<0\]

Answer 14

I like that!

Answer 15

Not saying that as an instructor to a student kind of manner, but other way around.

Answer 16

Hospital's rule ...

\(\rm http://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx \)

Answer 17

Thank you.