Finding Horizontal Asymptote Process

[Glissando]

Homework Statement

Find the horizontal asymptote of f(x) = (x²-1)/(x²-4)

Homework Equations

Limits

The Attempt at a Solution

I'm pretty family with the process, I just get confused when there's an x² in the denominator and no x in the numerator ):

So I already figured out that 1 is the horizontal asymptote, therefore:

lim [f(x) -1]
x->+/- infinity

= lim 3/(x²-4)
x->+/-infinity

Here's where the problem starts...(just for positive infinity for now), does the above limit become:

lim 3/(infinity²-4)
x->+infinity

which I have no idea what it equals to ):

OR

lim 3/+inifinity(0-4) = 0^-
x->+infinity

I hope that makes sense ): Thanks for the help!

 

[Mark44]

Homework Statement

Find the horizontal asymptote of f(x) = (x²-1)/(x²-4)

Homework Equations

Limits

The Attempt at a Solution

I'm pretty family with the process, I just get confused when there's an x² in the denominator and no x in the numerator ):

So I already figured out that 1 is the horizontal asymptote, therefore:

lim [f(x) -1]
x->+/- infinity

= lim 3/(x²-4)
x->+/-infinity

I don't know how the above is connected to your original problem.

You can write your function in another way that makes it easier to find this limit.
f(x) = (x²-1)/(x²-4) = [x²(1 - 1/x²)]/[x²(1 - 4/x²)] = (1 - 1/x²)/(1 - 4/x²).

Now it should be clear that f(x) --> 1 as x --> infinity or as x --> -infinity.

Here's where the problem starts...(just for positive infinity for now), does the above limit become:

lim 3/(infinity²-4)
x->+infinity

which I have no idea what it equals to ):

OR

lim 3/+inifinity(0-4) = 0^-
x->+infinity

I hope that makes sense ): Thanks for the help!

 

[Glissando]

Now it should be clear that f(x) --> 1 as x --> infinity or as x --> -infinity.

Thanks for your quick reply (: What I did was that I minus 1 from the function to make the function approach 0, and from there to determine how the function approaches positive and negative infinity (from the bottom or the top). I hope that makes sense >. <!

 

[Mark44]

Thanks for your quick reply (: What I did was that I minus 1 from the function to make the function approach 0, and from there to determine how the function approaches positive and negative infinity (from the bottom or the top). I hope that makes sense >. <!

No, this doesn't make any sense at all. Subtracting 1 doesn't make the function approach 0. I have no idea where you got that 3 that you showed in your work. I hope you don't do this on your exam, because you will probably get no credit for an answer like this.

Also, the function doesn't approach positive or negative infinity -- the x values do this, and the function values approach 1.

 

[Glissando]

Hey Mark,

What I meant to say was that in doing the limit, there was no opportunity to determine how the curve will approach the asymptote since the value of the limit does not work out to zero. Therefore, since 1 is the horizontal asymptote, (the textbook says) to shift the curve down 1 unit so it will temporarily have an H.A. of y=0. Then use limits again, separating between x approaching positive and negative inifinity. It should turn out to be zero, but what I"m trying to figure out is which side of zero (positive or negative) and will tell how the curve approaches the asymptote individually on each side of the graph.

): hope that sort of helps.

 

[Mark44]

Now it's starting to make sense, but the book's approach seems like more work than is necessary. If you already know that a horizontal asymptote is y = 1, creating a new function y = f(x) - 1 will have the x-axis (the line y = 0) as a horizontal asymptote, as you said.

\lim_{x \to \infty}f(x) - 1 = \lim_{x \to \infty}\frac{x^2-1}{x^2 -4} - 1
=\lim_{x \to \infty}\frac{x^2-1}{x^2 -4} - \frac{x^2 -4}{x^2 -4}
=\lim_{x \to \infty}\frac{3}{x^2 -4}
Now I understand where the 3 came from.
Since the numerator is positive, and the denominator is positive for all x > 2, the value of this limit is 0, and the limit value is approached from above.

This means that the limit of the untranslated function is 1, and that the graph approaches the horizontal asymptote from above, as well.

To find the behavior as x approaches negative infinity, evaluate the same limit, but with x approaching negative infinity.

As already stated, this seems like more work than is necessary. By bringing out a factor of x² from the numerator and denominator, you get the horizontal asymptote directly and you can tell whether the graph approaches it from above or below.

\lim_{x \to \infty}\frac{x^2-1}{x^2 -4}= \lim_{x \to \infty}\frac{x^2(1 - 1/x^2)}{x^2(1 - 4/x^2)} = \lim_{x \to \infty}\frac{1 - 1/x^2}{1 - 4/x^2}= 1

For a given value of x, we are subtracting 1/x² from 1 in the numerator, and are subtracting 4/x² from 1 in the denominator. IOW, we are subtracting a larger amount from the denominator. This makes the denominator slightly smaller than the numerator, which makes the overall fraction slightly larger than 1. This is the same result we got earlier, but is more direct.

 

[Glissando]

Thank you so much Mark! That was crystal clear and made H.A.s a lot simpler! Thank you (:!:!)