Question

I need the horizontal asymptote for

f(x) = (x)/(x^2 -4)

Can someone please show me. I already figured out that

The y-intercept is 0.
The vertical asymptotes are x = -2, x = 2

So now all i need is horizontal asymptote for

f(x) = (x)/(x^2 -4)

Answer 1

As x>>inf, the f(x) approaches zero, both directions.

Answer 2

What does this mean in words please? Thanks!

Answer 3

What is the value of the

horizontal asymptote for

f(x) = (x)/(x^2 -4)

This is my question?

Answer 4

Mr. Bob understood your question and has already supplied the answer. You may want to reread his answer.

For your benefit, I supply you herewith the link to another thread that may possibly give you some more hint.
http://www.jiskha.com/display.cgi?id=1249414112

Answer 5

So does this mean that the horizontal asymptote for

f(x) = (x)/(x^2 -4) Is not 0?



A horizontal asymptote is the y-value approaches as x approaches } ‡. Does this include 0? y= 0 ???

Answer 6

I do not know if you have read the link I cited earlier. However, the link includes another link to a sketch of the function which gives you a much better idea of how the function behaves as x approches ±∞.

In my school days, we were supposed to make these sketches free-hand by inspection of the function, no calculators, and no calculations. In fact, one of the objectives of the purpose of the posted question is probably to familiarize the student with how functions behave under different conditions, namely identify asymptotes, x and y-intercepts, concavity/convexity, behaviour of the function for x=±∞, etc.

Answer 7

Forgot to supply the link to the sketch, distracted by babbling, sorry.
http://i263.photobucket.com/albums/ii157/mathmate/Jessica-1.png

Answer 8

Thanks! So then the final answer is that

(x)/(x^2-4)

(x)/(x-2) (x+2)
Standard form then this = y = 0?

Answer 9

I am not sure in what context to which you refer as standard form.

The form (x)/((x-2)(x+2)) helps you to identify the vertical asymptotes, as they show where the denominator becomes zero, while the original expression of x/(x²-1) makes the horizontal asymptote evident, assuming you are familiar with the l'H0‹0pital rule.

The expression (x)/(x-2) (x+2) is algebraically inaccurate because the last factor should be part of the denominator. The expression implies that it is in the numerator. Parentheses are required to clarify and rectify it:
(x)/((x-2)(x+2))

Yes, the horizontal asymptote is at y=0, as indicated by Mr.Bob initially.