Evaluate Limit (x^2-4)/(x^2+x-2) As X→2

Hey guys! Today, we're diving into a classic calculus problem: evaluating the limit of a rational function. Specifically, we're going to tackle the limit of (x^2 - 4) / (x^2 + x - 2) as x approaches 2. This is a fundamental concept in calculus, and understanding how to solve these types of problems is crucial for mastering more advanced topics. So, let's break it down step by step and make sure we've got a solid grasp on the process. This isn't just about getting the right answer; it's about understanding why we do what we do. We'll be using factoring, simplification, and a bit of limit law magic to get to our solution. By the end of this, you'll be able to confidently approach similar limit problems. Think of this as a building block for your calculus journey, and we're going to make sure that foundation is strong. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into solving, let's really understand what the question is asking. We're looking at the function f(x) = (x^2 - 4) / (x^2 + x - 2) and we want to know what happens to the value of this function as x gets incredibly close to 2. It's super important to note that we're not actually plugging in x = 2 directly at first. That's because if we did, we'd get 0/0, which is an indeterminate form. Indeterminate forms are like little puzzles in calculus; they tell us we need to do some more work to find the actual limit. We're interested in the behavior of the function near 2, not necessarily at 2. Imagine zooming in closer and closer to the point where x is 2 on the graph of this function. What value does the function seem to be approaching? That's what we're trying to find. This concept of approaching a value is the heart of what limits are all about. They allow us to analyze functions even where they might not be perfectly defined. So, with that understanding, let's move on to the algebraic techniques we'll use to crack this limit problem.

Initial Substitution and Indeterminate Form

The first thing you should always do when evaluating a limit is to try direct substitution. It's the simplest approach, and sometimes it just works! So, let's plug x = 2 into our function: f(2) = (2^2 - 4) / (2^2 + 2 - 2) = (4 - 4) / (4 + 2 - 2) = 0 / 4 = 0. Oops! We got 0/4, which simplifies to just 0. But hang on, plugging in x = 2 gave us (2^2-4)/(22+2-2) = 0/4, which equals 0. However, plugging in x=2 gives us 0/0, an indeterminate form. This doesn't mean the limit doesn't exist; it just means we need to do more work to figure out what's going on. The indeterminate form is a sign that there might be a common factor in the numerator and denominator that we can cancel out. This is a crucial observation because it guides our next step: factoring. We've hit a roadblock with direct substitution, but that's perfectly normal. Indeterminate forms are common in limit problems, and they're a signal that we need to use algebraic manipulation to simplify the expression. So, let's move on to factoring and see if we can untangle this limit.

Factoring the Numerator and Denominator

Okay, guys, this is where our algebra skills come into play! We need to factor both the numerator and the denominator of our rational function. Factoring is like reverse multiplication; we're trying to find the expressions that, when multiplied together, give us the original expressions. Let's start with the numerator: x^2 - 4. This is a classic difference of squares pattern! Remember that a^2 - b^2 can be factored into (a - b)(a + b). In our case, a = x and b = 2, so x^2 - 4 factors into (x - 2)(x + 2). Awesome! Now, let's tackle the denominator: x^2 + x - 2. This is a quadratic expression, and we're looking for two numbers that multiply to -2 and add to 1 (the coefficient of the x term). Those numbers are 2 and -1. So, x^2 + x - 2 factors into (x + 2)(x - 1). Great! We've successfully factored both the numerator and the denominator. Now we have (x - 2)(x + 2) / (x + 2)(x - 1). Do you see anything exciting? We're setting ourselves up for a beautiful simplification.

Simplifying the Expression

Now comes the satisfying part: simplification! We've factored the numerator and denominator, and we can now see a common factor. Notice that both the numerator and denominator have a factor of (x + 2). Since we're considering the limit as x approaches 2 (but not actually equal to 2), we know that (x + 2) is not zero in the neighborhood of 2. This means we can safely cancel out the (x + 2) factor from both the top and bottom of our fraction. This simplification is the key to resolving the indeterminate form we encountered earlier. By canceling out this common factor, we're essentially removing the source of the 0/0 issue. After canceling, our expression becomes (x - 2) / (x - 1). This is a much simpler expression to work with! Remember, we can only cancel factors that are not equal to zero. That's why it's important that we're considering the limit as x approaches 2, not necessarily at x = 2. Now that we've simplified our expression, let's try plugging in x = 2 again and see what happens.

Evaluating the Limit After Simplification

Alright, we've done the hard work of factoring and simplifying. Now, let's see if our efforts have paid off. We have the simplified expression (x - 2) / (x - 1), and we want to find the limit as x approaches 2. Let's try direct substitution again. We plug in x = 2 into the simplified expression: (2 - 2) / (2 - 1) = 0 / 1 = 0. Woohoo! We got a definite value. The limit of our simplified expression as x approaches 2 is 0. This is a fantastic result! It means that as x gets closer and closer to 2, the value of our function (x^2 - 4) / (x^2 + x - 2) gets closer and closer to 0. This highlights the power of algebraic manipulation in evaluating limits. By factoring and simplifying, we were able to transform an indeterminate form into a clear, defined value. We can now confidently state that the limit of (x^2 - 4) / (x^2 + x - 2) as x approaches 2 is 0. This is a great example of how limits work, and it demonstrates a common technique used in calculus. But let's solidify our understanding by summarizing the steps we took and discussing the underlying principles.

Applying Limit Laws

While we essentially used direct substitution after simplifying, it's good to understand the limit laws that allow us to do this. The limit laws are like the rules of the road for evaluating limits, and they provide a rigorous justification for our steps. In this case, we're using the fact that the limit of a quotient is the quotient of the limits (provided the limit of the denominator is not zero). After simplifying, we had lim (x→2) [(x - 2) / (x - 1)]. We can think of this as [lim (x→2) (x - 2)] / [lim (x→2) (x - 1)]. The limit of (x - 2) as x approaches 2 is simply 2 - 2 = 0. The limit of (x - 1) as x approaches 2 is 2 - 1 = 1. So, we have 0 / 1 = 0, which confirms our result. Understanding these limit laws helps us to see why our algebraic manipulations are valid and how we can rigorously justify our answers. They also provide a framework for evaluating more complex limits in the future. So, while we got to the answer using a more intuitive approach, it's important to know the underlying mathematical principles at play.

Conclusion

So, guys, we've successfully navigated the limit of (x^2 - 4) / (x^2 + x - 2) as x approaches 2! We started by recognizing the indeterminate form, then used our factoring skills to simplify the expression. Finally, we applied direct substitution (or, more rigorously, limit laws) to arrive at our answer: 0. This problem is a fantastic illustration of the core concepts of limits and the importance of algebraic manipulation in calculus. By understanding how to factor, simplify, and apply limit laws, you're well on your way to mastering more complex calculus problems. Remember, the key is to practice and to understand why each step is taken. Limits are a fundamental building block for calculus, and with a solid understanding of these concepts, you'll be able to tackle a wide range of problems. Keep practicing, keep exploring, and you'll become a limit-solving pro in no time!

Key Takeaways:

• Always try direct substitution first.
• If you encounter an indeterminate form (like 0/0), try factoring.
• Simplify the expression by canceling common factors.
• Apply limit laws to rigorously justify your steps.
• Practice, practice, practice!