Asymptotes & Limits: Solving F(x) = X^2/(x^2-1)

Hey everyone! Let's dive into a cool math problem involving limits and asymptotes. We're going to explore the function $f(x) = \frac{x^2}{x^2-1}$. Our mission is to find the limits as x approaches 1 from both sides and to identify any asymptotes this function might have. Buckle up, because we're about to get mathematical!

Understanding Limits and Asymptotes

Before we jump into the specifics, let’s make sure we're all on the same page about what limits and asymptotes actually are. It's super important to grasp these concepts before tackling the problem.

What are Limits?

In simple terms, a limit tells us what value a function approaches as the input (in our case, x) gets closer and closer to a specific value. Think of it like this: imagine you're walking towards a destination. The limit is the place you're heading, even if you never quite reach it. We write $\lim_{x \to a} f(x) = L$ to say that as x approaches a, the function f(x) approaches L. Limits are crucial for understanding the behavior of functions, especially around points where they might be undefined.

What are Asymptotes?

Asymptotes are like invisible guide rails for a function. They are lines that the graph of a function approaches but never actually touches (or, in some cases, touches only a few times). There are three main types of asymptotes:

• Vertical Asymptotes: These occur where the function approaches infinity (or negative infinity) as x approaches a certain value. They happen when the denominator of a rational function (like ours) becomes zero.
• Horizontal Asymptotes: These describe the behavior of the function as x approaches infinity or negative infinity. They tell us where the function settles down in the long run.
• Oblique (or Slant) Asymptotes: These are diagonal lines that the function approaches as x goes to infinity or negative infinity. They occur when the degree of the numerator is exactly one more than the degree of the denominator.

Now that we've refreshed our understanding of limits and asymptotes, we're ready to tackle the problem at hand. Let's get back to our function, $f(x) = \frac{x^2}{x^2-1}$, and see what we can discover!

Part A: Calculating the Limits

Okay, let's get started with the first part of the problem: finding the limits. We need to determine the values of $\lim_{x \to 1^-} f(x)$ and $\lim_{x \to 1^+} f(x)$. This means we're looking at what happens to our function, $f(x) = \frac{x^2}{x^2-1}$, as x approaches 1 from the left (values less than 1) and from the right (values greater than 1).

Limit as x Approaches 1 from the Left ($\lim_{x \to 1^-} f(x)$)

When we say x approaches 1 from the left, we're considering values like 0.9, 0.99, 0.999, and so on. These values are getting closer and closer to 1, but they're always a little bit less than 1. Let's think about what happens to our function as x gets closer to 1 from this direction.

Our function is $f(x) = \frac{x^2}{x^2-1}$. The key part to focus on here is the denominator, $x^2 - 1$. As x approaches 1 from the left, $x^2$ will also approach 1, but it will be slightly less than 1 (since we're squaring a number slightly less than 1). This means that $x^2 - 1$ will be a small negative number.

So, we have:

• The numerator, $x^2$, is approaching 1 (a positive number).
• The denominator, $x^2 - 1$, is approaching 0 from the negative side (a very small negative number).

When we divide a positive number by a very small negative number, we get a very large negative number. In mathematical terms, this means the limit is negative infinity. Therefore,

$\lim_{x \to 1^-} f(x) = -\infty$

Limit as x Approaches 1 from the Right ($\lim_{x \to 1^+} f(x)$)

Now, let's consider what happens as x approaches 1 from the right. This means we're looking at values like 1.1, 1.01, 1.001, and so on. These values are also getting closer and closer to 1, but they're always a little bit greater than 1.

Again, we focus on the denominator, $x^2 - 1$. As x approaches 1 from the right, $x^2$ will approach 1, but it will be slightly greater than 1 (since we're squaring a number slightly greater than 1). This means that $x^2 - 1$ will be a small positive number.

So, we have:

• The numerator, $x^2$, is approaching 1 (a positive number).
• The denominator, $x^2 - 1$, is approaching 0 from the positive side (a very small positive number).

When we divide a positive number by a very small positive number, we get a very large positive number. This means the limit is positive infinity. Therefore,

$\lim_{x \to 1^+} f(x) = +\infty$

Conclusion for Part A

We've successfully found the limits! As x approaches 1 from the left, the function goes to negative infinity, and as x approaches 1 from the right, the function goes to positive infinity. This tells us something important about the behavior of the function near x = 1, which will be crucial when we look for asymptotes.

Part B: Finding the Asymptotes

Now, let's move on to the second part of our problem: finding all the asymptotes of the function $f(x) = \frac{x^2}{x^2-1}$. As we discussed earlier, asymptotes are like guide rails for the function's graph. They help us understand how the function behaves as x approaches certain values or infinity.

Vertical Asymptotes

Vertical asymptotes occur where the denominator of a rational function equals zero (and the numerator doesn't). So, to find the vertical asymptotes, we need to solve the equation $x^2 - 1 = 0$.

This is a simple quadratic equation that we can factor:

$x^2 - 1 = (x - 1)(x + 1) = 0$

This gives us two solutions: $x = 1$ and $x = -1$. These are the locations of our vertical asymptotes. We already saw in Part A that the function approaches infinity as x approaches 1 from either side, which confirms that x = 1 is indeed a vertical asymptote. We would find similar behavior near x = -1.

So, we have two vertical asymptotes:

• $x = 1$
• $x = -1$

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of the function as x approaches infinity or negative infinity. To find them, we need to look at the limit of the function as x goes to $\infty$ and $-\infty$.

Let's start with $\lim_{x \to \infty} \frac{x^2}{x^2-1}$. To evaluate this limit, we can divide both the numerator and the denominator by the highest power of x present, which in this case is $x^2$:

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

As x approaches infinity, $1/x^2$ approaches 0. So, we have:

$\lim_{x \to \infty} \frac{1}{1 - 1/x^2} = \frac{1}{1 - 0} = 1$

This tells us that there is a horizontal asymptote at $y = 1$ as x approaches infinity.

Now, let's consider $\lim_{x \to -\infty} \frac{x^2}{x^2-1}$. We can use the same trick and divide both the numerator and the denominator by $x^2$:

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

As x approaches negative infinity, $1/x^2$ still approaches 0. So, we have:

$\lim_{x \to -\infty} \frac{1}{1 - 1/x^2} = \frac{1}{1 - 0} = 1$

This tells us that there is also a horizontal asymptote at $y = 1$ as x approaches negative infinity.

So, we have one horizontal asymptote:

• $y = 1$

Oblique Asymptotes

Oblique asymptotes occur when the degree of the numerator is exactly one more than the degree of the denominator. In our case, the degree of the numerator ($x^2$) is the same as the degree of the denominator ($x^2 - 1$), so there are no oblique asymptotes.

Conclusion for Part B

We've found all the asymptotes for the function $f(x) = \frac{x^2}{x^2-1}$:

• Vertical Asymptotes: $x = 1$ and $x = -1$
• Horizontal Asymptote: $y = 1$
• Oblique Asymptotes: None

Final Thoughts

We've successfully tackled this problem! We found the limits of the function as x approaches 1 from both sides, and we identified all the asymptotes. By understanding these concepts, we can get a really good idea of how the function behaves. This is super useful in many areas of math and science. Keep practicing, guys, and you'll become asymptote and limit masters in no time!