# Question about limits and horizontal asymptotes

• mileena
In summary: 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→∞were both "horizontal asymptote = 0".It is a

## 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, ex 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!

Last edited:
mileena said:

## 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, ex 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?

Zondrina said:
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!

mileena said:
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.

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.

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→∞

## 1. What is the definition of a limit?

A limit is the value that a function approaches as the input (x) approaches a certain value. It represents the behavior of the function as x gets closer and closer to a particular value.

## 2. How do you find the limit of a function?

To find the limit of a function, you can either use the algebraic method (substituting the value of x into the function) or the graphical method (using a graph to estimate the limit). You can also use the limit laws, which allow you to manipulate the function algebraically to find the limit.

## 3. What is a horizontal asymptote?

A horizontal asymptote is a line that the graph of a function approaches as x approaches positive or negative infinity. It represents the long-term behavior of the function and can be found by looking at the end behavior of the function or by using the limit definition of an asymptote.

## 4. How do you determine the horizontal asymptote of a function?

The horizontal asymptote of a function can be determined by looking at the end behavior of the function as x approaches positive or negative infinity. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote will be at y=0. If the degrees are equal, the horizontal asymptote will be at the ratio of the leading coefficients. If the degree of the numerator is greater than the degree of the denominator, there will be no horizontal asymptote.

## 5. Can a function have more than one horizontal asymptote?

Yes, a function can have more than one horizontal asymptote. This can happen when the end behavior of the function approaches different values as x approaches positive or negative infinity. For example, a rational function may have two horizontal asymptotes if the degree of the numerator is one more than the degree of the denominator.