Horizontal asymptotes - approaches from above or below?

  • Thread starter fp252
  • Start date
  • Tags
    Horizontal
In summary, the conversation discusses the difficulty in determining whether the curve approaches a horizontal asymptote from above or below. The individual shares their calculations for determining the direction of approach, but is unsure if they have made a mistake. They also ask for clarification on the concept of limits and the use of infinity in the calculations.
  • #1
fp252
10
0
I seem to be having a lot of difficulty finding whether for a horizontal asymptote, whether the curve approaches the asymptote from above or below.


For example, for the problem [tex]y = \frac{6x + 1}{1 - 2x}[/tex], I know that:

For the vertical asymptote, x = 1/2, and that [tex]\lim_{x \to \frac{1}{2}^{+}} = - \infty[/tex], while [tex]\lim_{x \to \frac{1}{2}^{-}} = \infty[/tex].


Meanwhile, for the horizontal asymptote, there is one at y = -3. However, I do not know understand whether it approaches from above or below.

This is how I tried to find it:
As x --> -3 from the left, eg. x = -10, [tex]\frac{6(-3)+1}{1-2(-10)}[/tex] = [tex]\frac{-17}{21}[/tex] = -0.81
therefore approaches from above, because -0.81 is greater than -3.

As x --> -3 from the right, eg. x = 10, [tex]\frac{6(-3)+1}{1-2(10)}[/tex] = [tex]\frac{-17}{-19}[/tex] = 0.95
therefore also approaches from above, because -0.81 is greater than -3.


However, my graphing calculator says otherwise. The graph shows that, left of the vertical asymptote, the curve approaches from above; to the right, it approaches from the bottom. What am I doing wrong?
 
Physics news on Phys.org
  • #2
fp252 said:
Meanwhile, for the horizontal asymptote, there is one at y = -3. However, I do not know understand whether it approaches from above or below.

This is how I tried to find it:
As x --> -3 from the left, eg. x = -10, [tex]\frac{6(-3)+1}{1-2(-10)}[/tex] = [tex]\frac{-17}{21}[/tex] = -0.81
therefore approaches from above, because -0.81 is greater than -3.

As x --> -3 from the right, eg. x = 10, [tex]\frac{6(-3)+1}{1-2(10)}[/tex] = [tex]\frac{-17}{-19}[/tex] = 0.95
therefore also approaches from above, because -0.81 is greater than -3.

Why are you substituting x=-3 into the numerator and then substituting [itex]x=\pm10[/itex] into the denominator?
 
  • #3
Hmm, that's what I saw in my class notes. I assumed that the numerator's x-value remained to be -3, while the denominator was supposed to have a "test" value.

Ok, so I'm guessing I'm supposed to be subbing [itex]x=\pm10[/itex] into both numerator and denominator?
 
  • #4
fp252 said:
Ok, so I'm guessing I'm supposed to be subbing [itex]x=\pm10[/itex] into both numerator and denominator?

Yes!:smile:

After all, it's y that approaches -3 not x.
 
  • #5
Thanks for clarifying, gabbagabbahey!

I don't know how I've been pondering over this simple mistake, haha... :redface:
 
  • #6
Also, another question: is what I shown above involving the calculations of [tex]\lim_{x \to - \infty}[/tex] and [tex]\lim_{x \to +\infty}[/tex]?
 
Last edited:
  • #7
[itex] \displaystyle \lim_{x\to\infty} \frac{6x+1}{1-2x} [/itex]

[itex] \displaystyle \lim_{x\to\infty} \frac{6+\frac{1}{x}}{\frac{1}{x}-2} [/itex]

Does that help?
 

1. What is a horizontal asymptote?

A horizontal asymptote is a horizontal line that a function approaches as the input values get larger or smaller. It is a boundary that the function will never cross, but it can get closer and closer to it.

2. How do you determine if a function has a horizontal asymptote?

To determine if a function has a horizontal asymptote, you need to look at the behavior of the function as the input values get infinitely large or small. If the function approaches a specific value or horizontal line, then that is the horizontal asymptote.

3. Can a function have more than one horizontal asymptote?

Yes, a function can have more than one horizontal asymptote. For example, a rational function can have up to two horizontal asymptotes, one for the behavior as the input values get infinitely large and another for the behavior as the input values get infinitely small.

4. How do you determine if a function approaches a horizontal asymptote from above or below?

To determine if a function approaches a horizontal asymptote from above or below, you need to look at the behavior of the function as the input values approach the horizontal asymptote. If the function values are increasing as the input values get closer to the asymptote, then the function approaches from above. If the function values are decreasing, then the function approaches from below.

5. Can a function intersect its horizontal asymptote?

No, a function cannot intersect its horizontal asymptote. The horizontal asymptote serves as a boundary that the function will never cross, but it can get arbitrarily close to it.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
487
  • Calculus and Beyond Homework Help
Replies
9
Views
930
  • Calculus and Beyond Homework Help
Replies
4
Views
913
  • Calculus and Beyond Homework Help
Replies
12
Views
701
Replies
9
Views
960
  • Calculus and Beyond Homework Help
Replies
2
Views
265
  • Calculus and Beyond Homework Help
Replies
5
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
480
  • Calculus and Beyond Homework Help
Replies
2
Views
669
  • Precalculus Mathematics Homework Help
Replies
18
Views
1K
Back
Top