I am a Calculus I student just starting out.

I understand (I think) the concept of limit, which we started with.

Now, though, we're doing limits involving infinity, and I'm starting to get confused.

I can handle finding vertical and horizontal asymptotes of regular rational functions, but when they start throwing natural logs and trig functions in there, I'm getting lost.

For example:

In this problem I am supposed to find vertical asymptotes, if any; the behavior of the graph on each side of each vertical asymptote; and the horizontal asymptotes, if any.

The problem: f(x) = 4 tan

^{ -1 }x -1

Now, my thinking is that, since the graph of the inverse tangent is continuous on all x, then the graph of inverse tangent multiplied by four and then subtracting one is continuous on all x. That seems to be correct, as far as the answer in my book is concerned; I was thinking there would be no vertical asymptotes, and the book agrees.

However, when it comes to horizontal asymptotes, I get confused. On this problem, I was thinking that, as x approaches infinity, the inverse tangent of x would approach 1, so the limit was 3. As x approaches negative infinity, the inverse tangent of x would approach -1, so the limit was -5. Therefore, I thought that the horizontal asymptotes would be where f(x) = -5 and 3. The book, though, gives the answer as +/- 2(pi) - 1.

Can anybody help explain any of this? Sorry if I'm not being too clear about what I don't understand. I'm so confused at this point that I'm not even really sure what it is that I'm confused about. Thank you!

Anonymous

February 14 2007, 15:05:01 UTC 10 years ago

Anonymous

November 15 2009, 19:34:52 UTC 7 years ago

a tan-1 (X)+d, in which a affects the horizontal asymptotes of the graph, and d is the vertical shift.

because d is -1, the graph of arctan (inverse tan) is shifted down the y-axis 1.

the horizontal asymptotes of the equation is found using the following: a(pi/2), since the horizontal asymptotes of the arctan graph are at +/- (pi/2).

4(pi/2) gives you 2pi, which means the horizontal asymptotes are at +/- 2 pi.

the -1 part is because you shifted the graph of arctan down one on the y axis, therfore, the asymptoes are at +/- 2(pi) -1. (2pi shifted down the y-axis one and -2pi shifted down the y-axis one)

you can think of it as (pi/1) - (1/1), multiply by pi to get common demoninators (2pi^2/pi)-(pi/pi). simplify (2pi^2-pi)/pi. factor pi out of numerator (pi(2pi-1))/pi. simplify, pi cancels out. 2pi-1