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Describing asymptotic behavior in terms of limits involving infinity

Page history last edited by PBworks 15 years, 11 months ago

Describing Asymptotic Behavior In Terms of Limits Involving Infinity


Horizontal Asymptotes


Such that

lim f(x) = L


lim f(x) = L


Translation: When x approaches +/-, there is a y value, L, which can be determined. This value is never a point on the graph, but rather L, the value from the limit, is where the horizontal asymptote is located.



SOURCE: http://www.mathnstuff.com/math/spoken/here/2class/300/fx/library/lines.htm

In this example, there are both vertical and horizontal asymptotes. Let's focus on the horizontal one. Here, we see that as x approaches infinity on both sides of the graph (+ and -), there is a horizontal asymptote at y=3. In the equation, , x is said to approach infinity. To see how this is applied to the graph, plug in infinity for x into the equation. The result is an extremely small, negligible fraction, and 3. From this, the obvious horizontal asymptote, as x approaches infinity, must be 3, as it is the only numerical value left from the equation.



Vertical Asymptotes


Such that

lim f(x) = +-

x C+


lim f(x) = +-

x C-


Translation: As the x value approaches the asymptote, a vertical line at x=C, from either the left or right hand side of the equation, the y value approaches +/-. In the equation of the limit, this will be seen as x approaching a value that makes the denominator equal to zero, so the limit will approach infinity.



SOURCE: http://www.mathnstuff.com/math/spoken/here/2class/300/fx/library/lines.htm

In this example, there is a vertical asymptote at x=2. As the x values approach 2 from both sides of the graph (+ and -), the y values veer off to + and - . To analyze this in terms of the equation, set the denominator of the equation, x-2 equal to 0; 0 = x-2. From this, you can find the x value of the vertical asymptote, as x=2.


SHORTCUTS for discovering Vertical and Horizontal asymptotes by looking at the equation


Horizontal: Only works as x approaches , and gives you a number when the highest exponents in numerator and denominator are equal. In essence, you divide the leading coefficient (of the highest exponential term) in the numerator by the leading coefficient (of the highest exponential term) in the denominator to find the numerical y-value of the horizontal asymptote. In short, if the degrees of both numerator and denominator are equal, then L (the horizontal asymptote) = ratio of leading coefficients.

Example Problem: What is the value of the horizontal asymptote from of (12 - 7x )/ (3-5)

a. 12/5

b. 4

c. 7/3

d. 1/3

Answer: b, because 12/3=4, and both share the same exponential value, so their coefficients can be divided, as each is leading in the numerator and denominator.


Vertical: As described above, by factoring each part of the limit equation, both numerator and denominator, and setting the denominator equal to 0, you can discover the x value of the vertical asymptote. However, there are times when this method is not foolproof, as the next example problem will prove.

Example Problem: lim

x +/- C


Find the vertical asymptote and draw a graph.


Answer: x=2 is a vertical asymptote for f(x)

Explanation: Given that , we see that the function f is discontinuous at x=-2 and x=2. Since , x=-2 is a "removable" discontinuity. Since and , x=2 is the vertical asymptote of f.



SOURCE: http://archives.math.utk.edu/visual.calculus/1/vertical.1/index.html


Slant (Oblique) Asymptotes


If the degree of the numerator is one more than the degree of the denominator, you have neither a horizontal or vertical asymptote, but a slant asymptote. To find the value of the asymptote, you must divide the numerator by the denominator using polynomial division, and you can ignore the remainder. The equation you get as a result will be the asymptote on the graph, which looks odd, but that's ok, it's supposed to be that way.


Example Problem: Draw a graph of and define the slant asymptote of .


Answer: The slant asymptote is at y=-3x-3, and the graph is drawn such:


SOURCE: http://www.purplemath.com/modules/asymtote3.htm



More Practice Problems

__Multiple Choice__


If f(x) and g(x) are such that


lim f(x) as x a =




lim g(x) as x a = 0




a. lim [ f(x) . g(x) ] as x a is always equal to 0

b. lim [ f(x) . g(x) ] as x a is never equal to 0

c. lim [ f(x) . g(x) ] as x a may be or

d. lim [ f(x) . g(x) ] as x a may be equal to a finite value.


Answer :


c and d. Try the following functions:


f(x) = 1 / x and g(x) = 2x as x approaches 0.


f(x) = 1 / x 2 and g(x) = x as x approaches 0.


SOURCE: http://www.analyzemath.com/calculus_questions/limits.html



Find the horizontal asymptote of the graph of


a. 0

b. +/-1

c. +/-4

d. nonexistent


Answer: b, there is a horizontal asymptote at y=+/-1.


SOURCE: http://www.libraryofmath.com/limits-to-infinity-and-horizontal-asymptotes.html


Free Response


Determine the following limits, and graph the vertical asymptotes of the equation of f(x).


Answer: . There are asymptotes at x=4 and x=-3.

The limits are: and



SOURCE: http://archives.math.utk.edu/visual.calculus/1/vertical.1/index.html



For a lot of good starting material on limits, http://www.mecca.org/~halfacre/MATH/limits.htm

For a good animated step-by-step tutorial on horizontal asymptotes, http://archives.math.utk.edu/visual.calculus/1/horizontal.5/3.html

For an animated tutorial on vertical asymptotes, http://archives.math.utk.edu/visual.calculus/1/vertical.4/3.html

For an applet to calculate limits for a variety of equations, http://math.dartmouth.edu/~klbooksite/appfolder/203unit/Limits.html



Bryce N.

i like cupcakes

Comments (13)

Anonymous said

at 9:14 pm on Jan 27, 2008

When you're editing your page, there's a pull-down menu that has different font sizes (ie. x-small, small, medium, etc.) It's sort of near the middle of the screen. -Zahra

Anonymous said

at 8:25 pm on Jan 30, 2008


Anonymous said

at 1:42 am on Jan 31, 2008

still cramming eh?

Anonymous said

at 1:42 am on Jan 31, 2008

it's marissa btw

Anonymous said

at 2:06 am on Jan 31, 2008

oh, hey. yeah, livin' the good life at 2 am. i love myself.

Anonymous said

at 2:07 am on Jan 31, 2008

i actually had a lot done, it just takes me a while to get the picture files cut and uploaded.

Anonymous said

at 2:07 am on Jan 31, 2008

wait, if you were already done were you just checking these wikis at 1.42 am for fun? oh, marissa...

Anonymous said

at 9:48 am on Feb 8, 2008

Your explanation and graph for horizontal asymptotes were both very informative and helped me understand the concept.

Anonymous said

at 10:07 am on Feb 8, 2008

Bryce, your diagrams and verbal description of asymptotic behavior clarified my understanding, and the practice problems were clear and useful. The aesthetics of the page as well as the organization work really well, and it was pleasing to read and look at.
I wish that you had not included the picture of the cupcake because now I am craving one.
Also, it's probably part of the Ron Paul ReLOVEution.

Anonymous said

at 1:46 pm on Feb 8, 2008

Hey Bryce,
Your page is really good! I especially like your visual calculus links at the bottom--they have really good explanations. I have just a small formatting suggestion: For this problem, 12x^2 - 7x / 3x^2-5, you might want to put parentheses around the demoninator and numerator so that it is clearer. Otherwise--awesome! ~Rebecca

Anonymous said

at 1:49 pm on Feb 8, 2008

Such a great page. I almost have nothing to say; I love the explanations that you give after the multiple choice problems to show us how to get to the correct answer. The graphs at the beginning helped refresh my memory of limits, and the hints/shortcuts were a very helpful review. I liked your inclusion of slanted asymptotes, as we have not discussed those as much in class. The cupcake was lovely as well.

Anonymous said

at 7:53 am on Feb 21, 2008

I, too, love the cupcake picture. I also agree with Rebecca -- your formatting issues with the equations create some problems, particularly in the example Rebecca pointed out. I like the links and the problems. Mr. S

Alexis Brown said

at 12:25 pm on Feb 2, 2009

This is Alexis. Your explanation of the various types of asymptotes is very clear. I find your page to be extremely helpful, especially with all your examples and the graphs. Your links were really helpful too and there was a great problems too. It is a hard subject to explain, but you do it well. Great job.

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