Question

Find all horizontal asymptotes for the function

Answer 1

\[f(x)=\frac{ 3x^2-2 }{ x \sqrt{x^2+7} }\]

Answer 2

Horizontal? Hmm.
https://www.desmos.com/calculator/hr74bkfdkc

Here is a quick picture of what the function looks like.
You can see that as x get's larger and larger in each direction, we get closer and closer to an asymptote in the horizontal direction.

So to find horizontal asymptotes, we'll have to take the limit of the function in each direction, left and right.

\[\large \lim_{x \rightarrow \infty}\frac{3x^2-2}{x\sqrt{x^2+7}}\]And the other one,\[\large \lim_{x \rightarrow -\infty}\frac{3x^2-2}{x\sqrt{x^2+7}}\]

Answer 3

Do you understand how to solve this limit?
We can take a couple different approaches.
I'm trying to decide which way will make more sense to you. Hmm

Answer 4

Couldn't you just look at the biggest exponents and ignore everything else?

Answer 5

I slightly know how to solve the limit . this was on the test and i made completely huge mistakes.
i dont know where, or have no clue

Answer 6

Ah ok, let's find out where :)
I'll show you how I would solve it at least.

Answer 7

Thank you!

Answer 8

@galacticwavesXX Yes that is the best way to approach these types of problems. Maybe we should do that instead of doing some confusing math :D heh

Answer 9

When we approach infinity, all of the terms will become insignificant except the leading terms (If their powers match). We can simply look at the coefficients on the leading terms in the numerator and denominator and that will tell us what this limit is approaching.

The square root is making it a little confusing though, let's do something about that real quick.
\[\large \lim_{x \rightarrow \infty}\frac{3x^2-2}{x\sqrt{x^2+7}}\]Let's factor an x^2 out of each term under the root.\[\large \lim_{x \rightarrow \infty}\frac{3x^2-2}{x\sqrt{x^2(1+\frac{7}{x^2})}}=\lim_{x \rightarrow \infty}\frac{3x^2-2}{x\cdot x\sqrt{(1+\frac{7}{x^2})}}=\lim_{x \rightarrow \infty}\frac{3x^2-2}{x^2\sqrt{(1+\frac{7}{x^2})}}\]

From here we can see that the leading term in the top and bottom matches. They're both x^2.

The coefficient on the top x^2 is \(\large 3\).
The coefficient on the bottom x^2 is \(\large \sqrt 1\).

So our limit is approaching \(\large \dfrac{3}{1}=3\).

Answer 10

Hmm maybe factoring that x^2 out wasn't necessary.. lemme know if that was confusing :\

Answer 11

Um, what about the top?
would it be divided by x^2?

Answer 12

It was necessary so you can ignore everything else in the denominator. If you hadn't the limit would have been 3x

Answer 13

All you are taking once the bottom is factored out properly is \[\frac{ 3x^2 }{ x^2 }=3\]

the \[x^2 \]cancel out that is why you are left with 3

Answer 14

Oh i got it!
so glacticwaves do you agree with zepdrixs answer?

Answer 15

yup 3 is correct!!

Answer 16

Thank you both of you!
I

Answer 17

Welcome anytime!!

Answer 18

I would love to give you madel to both..

Answer 19

sure go ahead