Question about limits and horizontal asymptotes

[mileena]

Homework Statement

Find the limit and any horizontal asymptotes:

lim 4/(e^-x)
x→∞

Homework Equations

Principle 1: A limit defines a horizontal asymptote whenever x→∞ or x→-∞.

Principle 2: If a limit goes to ∞ or -∞, there won't be a horizontal asymptote.

The Attempt at a Solution

lim 4/(e^-x) = ∞
x→∞

since as x gets bigger, e^x also gets bigger.

Horizontal Asymptote = 0

Given the above answer, if correct, then I have a problem with the two above principles I learned. Should I email my professor about these two principles? I am trying to learn these first before I do the actual homework.

Thanks!

 

[STEMucator]

Homework Statement

Find the limit and any horizontal asymptotes:

lim 4/e^-x
x→∞

Homework Equations

Principle 1: A limit defines a horizontal asymptote whenever x→∞ or x→-∞.

Principle 2: If a limit goes to ∞ or -∞, there won't be a horizontal asymptote.

The Attempt at a Solution

lim 4/e^-x = ∞
x→∞

since as x gets bigger, e^x also gets bigger.

Horizontal Asymptote = 0

Given the above answer, if correct, then I have a problem with the two above principles I learned. Should I email my professor about these two principles? I am trying to learn these first before I do the actual homework.

Thanks!

So this is your function if I'm not mistaken : $\frac{4}{e^{-x}} = 4e^x$.

As $x → ∞$, $4e^x → ∞$. So there are no horizontal asymptotes happening in this direction.

Why do you think $y=0$ is a horizontal asymptote for your function? Can you elaborate?

 

[mileena]

So this is your function if I'm not mistaken : $\frac{4}{e^{-x}} = 4e^x$.

As $x → ∞$, $4e^x → ∞$. So there are no horizontal asymptotes happening in this direction.

Why do you think $y=0$ is a horizontal asymptote for your function? Can you elaborate?

Hi Zondrina!

Yes, you interpreted the formula correctly.

I didn't realize horizontal asymptotes were subject to a direction. I thought they applied to the whole function, so I saw that as x→-∞, the function approached 0. So I thought that was a horizontal asymptote (y = 0) for the function.

EDIT: I will also go back and edit the function I posted in my first post to remove the ambiguity! I have to learn how to use the LaTeX function on this forum, but it was complicated when I first tried to even write a simple fraction!

 

[STEMucator]

Hi Zondrina!

Yes, you interpreted the formula correctly.

I didn't realize horizontal asymptotes were subject to a direction. I thought they applied to the whole function, so I saw that as x→-∞, the function approached 0. So I thought that was a horizontal asymptote (y = 0) for the function.

EDIT: I will also go back and edit the function I posted in my first post to remove the ambiguity! I have to learn how to use the LaTeX function on this forum, but it was complicated when I first tried to even write a simple fraction!

When you went to quote my second post, it will allow you to see how I wrote the latex. You can use that to get a feel for it. Here's a good tutorial to teach you some basics as well ( complements of micromass ) :

https://www.physicsforums.com/showpost.php?p=3977517&postcount=3

Also you are correct when you say this :

I didn't realize horizontal asymptotes were subject to a direction. I thought they applied to the whole function, so I saw that as x→-∞, the function approached 0. So I thought that was a horizontal asymptote (y = 0) for the function.

I was just checking if you explicitly knew to take $x→-∞$ instead. You have to check both directions with these kinds of problems.

 

[mileena]

Thank you!

So I hope this is right:

as x→∞, the limit is also ∞, and there is no horizontal asymptote. But as x→-∞, both the limit and horizontal asymptote is 0.

For the overall function, the horizontal asymptote is y = 0.

And I also will go back and check the LaTeX. I would love to learn to write fractions and radicals, etc., correctly so they are easier to read for everyone involved. I have just been scared to try something new.

 

[mileena]

Also, technically, Principle 1 above:

"Principle 1: A limit defines a horizontal asymptote whenever x→∞ or x→-∞."

is violated here as x→∞, since the limit of the function here is ∞ but there is no horizontal asymptote.

Maybe I should email my professor and ask her to clarify this. It worked for the examples she gave in class when she said the principle:

lim(3x³ - 1000x²)
x→∞

lim(-x⁵ - x² + x - 10)
x→∞