Understanding Function Behavior: A Deep Dive

Dec 7, 2025 by Admin 45 views

Hey math enthusiasts! Ever wonder what a function is really doing as its input gets super, super big, or maybe super, super small? We're talking about limits at infinity, and it's a crucial concept for understanding the long-term behavior of functions. Today, guys, we're going to break down the function $f(x)= rac{2 x}{1-x^2}$ and figure out exactly where its graph is headed as $x$ goes off to infinity. This isn't just some abstract math puzzle; understanding these limits helps us with things like graphing, finding asymptotes, and even in more advanced calculus topics. So, buckle up, and let's dive into the fascinating world of function behavior!

Exploring the Function $f(x)= rac{2 x}{1-x^2}$

Alright, let's get down to business with our specific function: $f(x)= rac{2 x}{1-x^2}$ . We want to know what happens to the value of $f(x)$ as $x$ becomes an infinitely large positive number or an infinitely large negative number. This is what we call the limit as $x$ approaches infinity. Think about plugging in gigantic numbers for $x$ – like a million, or a billion, or even a googolplex! What does the output look like?

To tackle this, we can use a neat trick. When we're dealing with rational functions (that's just a fancy word for a fraction where the numerator and denominator are polynomials), we can divide both the numerator and the denominator by the highest power of $x$ that appears in the denominator. In our case, the highest power of $x$ in the denominator ( $1-x^2$ ) is $x^2$ . So, let's divide every term in the numerator and the denominator by $x^2$ :

f(x) = rac{ rac{2x}{x^2}}{ rac{1}{x^2} - rac{x^2}{x^2}}

Now, let's simplify this expression:

f(x) = rac{ rac{2}{x}}{ rac{1}{x^2} - 1}

Look at that! We've transformed our function into something a bit more revealing. Now, let's consider what happens as $x$ approaches infinity. As $x$ gets incredibly large, terms like $rac{2}{x}$ and $rac{1}{x^2}$ get incredibly small. In fact, they approach zero! Think about it: if you divide a small number (like 2) by a humongous number (like a billion), the result is practically zero.

So, as $x o ext{infinity}$ , we have:

rac{2}{x} o 0

and

rac{1}{x^2} o 0

Substituting these values back into our transformed function:

ext{As } x o ext{infinity}, f(x) o rac{0}{0 - 1} = rac{0}{-1} = 0

What does this tell us? It means that as $x$ gets larger and larger (in either the positive or negative direction), the value of $f(x)$ gets closer and closer to 0. This is a fundamental insight into the function's end behavior. We call this a horizontal asymptote. In this case, the line $y=0$ (which is just the x-axis) is a horizontal asymptote for the graph of $f(x)$ . This means the graph will hug the x-axis as it extends far to the left or far to the right.

It's also super important to remember that this analysis holds true whether $x$ is approaching positive infinity ( $x o ext{+\infty}$ ) or negative infinity ( $x o - ext{\infty}$ ). The terms $rac{2}{x}$ and $rac{1}{x^2}$ still go to zero in both scenarios. So, the graph of $f(x)$ approaches the x-axis from both sides as $x$ moves away from the origin indefinitely.

Analyzing the Options: What Does the Graph Do?

Now that we've done the heavy lifting and figured out the limit of our function as $x$ approaches infinity, let's revisit those options you were given. We're looking for the statement that accurately describes the behavior of $f(x)= rac{2 x}{1-x^2}$ as $x$ heads towards infinity.

A. The graph approaches -2 as $x$ approaches infinity. Based on our calculations, this is incorrect. We found that the function approaches 0, not -2.
B. The graph approaches 0 as $x$ approaches infinity. Bingo! This matches our findings perfectly. As $x$ gets infinitely large, $f(x)$ gets infinitely close to 0.
C. The graph approaches 1 as $x$ approaches infinity. This is also incorrect. The value 1 is not where our function is heading in the long run.
D. The Discussion category is mathematics. While it's true that this discussion falls under the category of mathematics, this statement doesn't describe the behavior of the function as $x$ approaches infinity. It's a meta-commentary about the topic, not an analysis of the function's graph.

So, the correct statement is B. This is why understanding limits at infinity is so powerful – it gives us a clear picture of where the function is going on those vast, unending stretches of the x-axis.

Beyond Infinity: Understanding Asymptotes

Our journey into the behavior of $f(x)= rac{2 x}{1-x^2}$ has already revealed a key feature: a horizontal asymptote at $y=0$ . But let's dig a little deeper, shall we? Asymptotes are like invisible guides for our graph, telling us where the function is heading or what values it's getting close to. For rational functions, we often look for horizontal, vertical, and sometimes even slant (or oblique) asymptotes.

We've already nailed the horizontal asymptote by examining the limit as $x o ext{±\infty}$ . Remember, we divided by the highest power of $x$ in the denominator ( $x^2$ ) to simplify:

f(x) = rac{ rac{2}{x}}{ rac{1}{x^2} - 1}

And as $x o ext{±\infty}$ , this simplified to $rac{0}{0-1} = 0$ . So, $y=0$ is our horizontal asymptote. This is a common scenario when the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator. In our case, the degree of $2x$ (which is 1) is less than the degree of $1-x^2$ (which is 2).

What about vertical asymptotes? These occur where the denominator of a rational function is zero, but the numerator is not zero at that same point. For $f(x)= rac{2 x}{1-x^2}$ , we set the denominator to zero:

1 - x^2 = 0

Solving for $x$ , we get:

x^2 = 1

x = ext{±}1

Now, we need to check if the numerator, $2x$ , is non-zero at these points. At $x=1$ , the numerator is $2(1)=2$ , which is not zero. At $x=-1$ , the numerator is $2(-1)=-2$ , which is also not zero. Therefore, we have vertical asymptotes at $x=1$ and $x=-1$ . This means the graph of the function will shoot off towards positive or negative infinity as $x$ gets extremely close to 1 or -1 from either side. The function is undefined at these points, and the graph can't cross these vertical lines.

Since the degree of the numerator is not equal to or greater than the degree of the denominator, we don't need to worry about slant or oblique asymptotes. Those typically occur when the degree of the numerator is exactly one greater than the degree of the denominator.

Visualizing the Graph

Let's try to sketch what this graph might look like, putting all our findings together. We know:

Horizontal Asymptote: $y=0$ (the x-axis).
Vertical Asymptotes: $x=1$ and $x=-1$ .
Behavior as $x o ext{±\infty}$ : The graph approaches $y=0$ .

Let's also test a few points to get a feel for the graph's shape:

When $x=0$ : $f(0) = rac{2(0)}{1-0^2} = rac{0}{1} = 0$ . So, the graph passes through the origin $(0,0)$ . This is interesting because the origin lies on our horizontal asymptote! This means the graph crosses the x-axis at $x=0$ .
When $x=2$ : $f(2) = rac{2(2)}{1-2^2} = rac{4}{1-4} = rac{4}{-3} = - rac{4}{3}$ . So, the point $(2, -4/3)$ is on the graph.
When $x=-2$ : $f(-2) = rac{2(-2)}{1-(-2)^2} = rac{-4}{1-4} = rac{-4}{-3} = rac{4}{3}$ . So, the point $(-2, 4/3)$ is on the graph.

Consider the intervals defined by the vertical asymptotes: $(- ext{infinity}, -1)$ , $(-1, 1)$ , and $(1, ext{infinity})$ .

Interval $(1, ext{infinity})$ : As $x$ is slightly larger than 1 (e.g., $x=1.1$ ), the denominator $(1-x^2)$ will be negative and small, making $f(x)$ a large negative number. As $x$ gets very large, $f(x)$ approaches 0 from below (it's negative for large positive $x$ , as we saw with $x=2$ ). So, in this region, the graph comes down from the x-axis, goes towards negative infinity as $x o 1^+$ , and then curves back up towards the x-axis from below as $x o ext{infinity}$ .
Interval $(-1, 1)$ : We know it passes through $(0,0)$ . For $x$ slightly less than 1 (e.g., $x=0.9$ ), the denominator is positive and small, making $f(x)$ a large positive number. For $x$ slightly greater than -1 (e.g., $x=-0.9$ ), the denominator is positive and small, making $f(x)$ a large positive number. Wait, let's recheck $x=-0.9$ . $f(-0.9) = rac{2(-0.9)}{1 - (-0.9)^2} = rac{-1.8}{1 - 0.81} = rac{-1.8}{0.19}$ , which is a large negative number. My apologies, guys! Let's re-evaluate this middle section carefully. The numerator $2x$ is positive for $x>0$ and negative for $x<0$ . The denominator $1-x^2$ is positive for $-1 < x < 1$ and negative for $x>1$ or $x<-1$ . So, in $(-1, 0)$ , $2x$ is negative and $1-x^2$ is positive, meaning $f(x)$ is negative. In $(0, 1)$ , $2x$ is positive and $1-x^2$ is positive, meaning $f(x)$ is positive. This confirms it passes through the origin, is negative for $x$ just below 0, and positive for $x$ just above 0. As $x o 1^-$ , $f(x) o rac{2}{0^+} o + ext{infinity}$ . As $x o -1^+$ , $f(x) o rac{-2}{0^+} o - ext{infinity}$ . So, the graph comes from negative infinity near $x=-1$ , goes up through $(0,0)$ , and then shoots up to positive infinity as $x$ approaches 1.
Interval $(- ext{infinity}, -1)$ : For $x$ slightly less than -1 (e.g., $x=-1.1$ ), the denominator $(1-x^2)$ is negative and small, making $f(x)$ a large positive number. For large negative $x$ , $f(x)$ approaches 0 from above (it's positive for large negative $x$ , as we saw with $x=-2$ ). So, the graph comes from the x-axis (from above), shoots up to positive infinity as $x o -1^-$ , and then curves back down towards the x-axis as $x o - ext{infinity}$ .

This detailed visualization confirms our earlier finding about the horizontal asymptote. No matter how far left or right you go on the x-axis, the graph of $f(x)= rac{2 x}{1-x^2}$ will eventually get closer and closer to the x-axis ( $y=0$ ).

Conclusion: The Power of Limits

So, there you have it, folks! We've thoroughly analyzed the function $f(x)= rac{2 x}{1-x^2}$ . By applying the concept of limits at infinity, we were able to determine that as $x$ approaches infinity (both positive and negative), the value of the function $f(x)$ approaches 0. This means the correct statement describing the behavior of the function is B. The graph approaches 0 as $x$ approaches infinity.

Understanding these limits is a foundational skill in mathematics, particularly when you start graphing functions and analyzing their behavior. It helps us identify key features like horizontal asymptotes, which provide crucial information about the function's end behavior. Keep practicing, keep exploring, and don't be afraid to plug in those big numbers (or use algebraic tricks!) to see where your functions are headed. Happy calculating!