Evaluate The Limit: A Step-by-Step Guide

Hey guys! Today, we're diving into a fascinating problem from the realm of calculus: evaluating the limit of a rational function. Specifically, we'll be tackling the limit: $\lim _{x \rightarrow -3} \frac{x^2 - 2x - 15}{x^2 + 9x + 18}$. This type of problem is a classic example of what you might encounter in your calculus journey, and mastering it will equip you with essential skills for more complex challenges. Understanding limits is crucial in calculus as it forms the foundation for concepts like continuity, derivatives, and integrals. When we talk about the limit of a function as x approaches a certain value, we're essentially asking: what value does the function get closer and closer to as x gets closer and closer to that value? In this case, we want to know what happens to the expression $\frac{x^2 - 2x - 15}{x^2 + 9x + 18}$ as x gets closer and closer to -3. Directly substituting x = -3 into the expression gives us an indeterminate form, which means we need to do some algebraic manipulation to find the limit. This involves techniques like factoring, simplifying, and sometimes even rationalizing the numerator or denominator. So, let's roll up our sleeves and get started! We'll break down each step in detail, making sure you understand the why behind the how. By the end of this guide, you'll not only be able to solve this particular limit problem, but you'll also have a solid understanding of the general approach for evaluating limits of rational functions. Remember, practice makes perfect, so don't hesitate to try out similar problems on your own. Let's get this limit evaluated!

Initial Assessment: Why Can't We Just Plug In?

Okay, so our first instinct when faced with a limit problem might be to just plug in the value that x is approaching. Let's see what happens if we try that here with our limit, $\lim _{x \rightarrow -3} \frac{x^2 - 2x - 15}{x^2 + 9x + 18}$. If we substitute x = -3 directly into the expression, we get: $\frac{(-3)^2 - 2(-3) - 15}{(-3)^2 + 9(-3) + 18} = \frac{9 + 6 - 15}{9 - 27 + 18} = \frac{0}{0}$. Uh-oh! We've run into a problem – we have the indeterminate form $\frac{0}{0}$. This doesn't mean the limit doesn't exist; it just means we can't find it by direct substitution. Think of it like this: the $\frac{0}{0}$ form is a signal that there's more to the story, and we need to dig deeper to uncover the true behavior of the function as x approaches -3. The indeterminate form arises because both the numerator and the denominator are approaching zero simultaneously. This suggests that there might be a common factor that's causing both to become zero. Our goal now is to identify and eliminate this common factor. This is where our algebraic skills come into play! We need to manipulate the expression in a way that allows us to cancel out the problematic term and reveal the true limit. So, what's our next move? Well, given that we suspect a common factor, factoring the numerator and denominator seems like a promising strategy. Factoring will help us break down the polynomials into their constituent parts, making it easier to spot any shared factors. Remember, the $\frac{0}{0}$ form is a common hurdle in limit problems, and learning how to overcome it is a crucial step in mastering calculus. So, let's move on to factoring and see if we can crack this limit open!

Factoring the Numerator and Denominator

Alright guys, now that we know we can't just plug in x = -3, it's time to put on our factoring hats! Factoring is a powerful technique for simplifying expressions and revealing hidden structures. In our case, we need to factor both the numerator, $x^2 - 2x - 15$, and the denominator, $x^2 + 9x + 18$. Let's start with the numerator. We're looking for two numbers that multiply to -15 and add up to -2. After a bit of thought, we can see that -5 and 3 fit the bill perfectly. So, we can factor the numerator as: $x^2 - 2x - 15 = (x - 5)(x + 3)$. Great! Now, let's tackle the denominator. This time, we need two numbers that multiply to 18 and add up to 9. These numbers are 6 and 3. Therefore, we can factor the denominator as: $x^2 + 9x + 18 = (x + 6)(x + 3)$. Excellent! We've successfully factored both the numerator and the denominator. Now, let's rewrite our limit with these factored expressions: $\lim _{x \rightarrow -3} \frac{(x - 5)(x + 3)}{(x + 6)(x + 3)}$. Do you see anything interesting? Notice that we have a common factor of (x + 3) in both the numerator and the denominator. This is exactly what we were hoping for! Remember how the $\frac{0}{0}$ form suggested a common factor? Well, here it is. This common factor is what was causing both the numerator and the denominator to approach zero as x approached -3. Now, we can move on to the next step: simplifying the expression by canceling out this common factor. This will get us one step closer to evaluating the limit. Factoring is a fundamental skill in algebra and calculus, and it's essential for tackling limit problems like this one. By breaking down the polynomials into their factors, we've uncovered a crucial piece of the puzzle. Let's move on and simplify!

Simplifying the Expression: Canceling Common Factors

Okay, awesome! We've factored both the numerator and the denominator, and we've identified the common factor (x + 3). Now comes the satisfying part: canceling out that common factor! Remember, we can cancel factors because we're dealing with a limit, and we're interested in what happens as x approaches -3, not necessarily what happens exactly at x = -3. So, as long as x is not exactly -3, we can safely divide both the numerator and the denominator by (x + 3). This gives us: $\lim _{x \rightarrow -3} \frac{(x - 5)(x + 3)}{(x + 6)(x + 3)} = \lim _{x \rightarrow -3} \frac{x - 5}{x + 6}$. Look at that! Our expression has become much simpler. We've eliminated the factor that was causing the indeterminate form. Now, we have a rational function that's much easier to handle. This is a crucial step in evaluating limits of this type. By canceling out the common factor, we've essentially removed the "hole" in the function at x = -3. The original function had a discontinuity at x = -3 because the denominator was zero, but the simplified function is continuous at x = -3. The limit, however, describes the behavior of the function as x gets close to -3, not necessarily the value of the function at x = -3. So, even though the original function is undefined at x = -3, the limit can still exist. Now that we've simplified the expression, we're in a much better position to evaluate the limit. We can try direct substitution again and see if it works this time. Remember, our goal is to find the value that the function approaches as x approaches -3. Let's move on and try substituting again!

Evaluating the Limit: Direct Substitution

Alright, guys! We've done the hard work of factoring and simplifying. Now comes the moment of truth: let's see if we can finally evaluate the limit. We've arrived at the simplified expression: $\lim _{x \rightarrow -3} \frac{x - 5}{x + 6}$. Remember, we couldn't use direct substitution initially because we got the indeterminate form $\frac{0}{0}$. But now that we've canceled out the common factor, let's try substituting x = -3 again. Plugging in x = -3 into our simplified expression, we get: $\frac{-3 - 5}{-3 + 6} = \frac{-8}{3}$. Ta-da! We've got a value. This means that as x approaches -3, the function $\frac{x^2 - 2x - 15}{x^2 + 9x + 18}$ approaches $\frac{-8}{3}$. So, we can confidently say that: $\lim _{x \rightarrow -3} \frac{x^2 - 2x - 15}{x^2 + 9x + 18} = \frac{-8}{3}$. This is the limit we were looking for! It's important to note that the limit exists even though the original function is undefined at x = -3. This is a key concept in understanding limits. The limit describes the function's behavior near a point, not necessarily at the point. In this case, the function approaches $\frac{-8}{3}$ as x gets closer and closer to -3, even though the function itself doesn't have a value at x = -3. We've successfully navigated the indeterminate form, factored the polynomials, canceled the common factor, and finally, evaluated the limit using direct substitution. This is a classic example of how to handle limits of rational functions, and the techniques we've used here can be applied to a wide range of similar problems. Congratulations! You've just mastered a fundamental concept in calculus. But remember, practice makes perfect, so keep working on those limit problems!

Conclusion: Key Takeaways and Further Practice

Okay, guys, we've reached the end of our journey to evaluate the limit $\lim _{x \rightarrow -3} \frac{x^2 - 2x - 15}{x^2 + 9x + 18}$, and we've successfully found that it equals $\frac{-8}{3}$. Let's take a moment to recap the key steps we took and the important concepts we learned along the way. First, we encountered the indeterminate form $\frac{0}{0}$ when we tried direct substitution. This signaled that we needed to do some algebraic manipulation. Then, we employed the powerful technique of factoring to break down both the numerator and the denominator into their constituent factors. This allowed us to identify the common factor (x + 3), which was the culprit behind the indeterminate form. Next, we simplified the expression by canceling out the common factor. This is a crucial step because it removes the discontinuity at x = -3 and allows us to evaluate the limit. Finally, we used direct substitution on the simplified expression to find the limit. This worked because we had eliminated the problematic factor. So, the key takeaways from this problem are:

• Indeterminate forms like $\frac{0}{0}$ indicate that further algebraic manipulation is needed.
• Factoring is a powerful tool for simplifying rational expressions.
• Canceling common factors can remove discontinuities and allow for direct substitution.
• The limit describes the function's behavior near a point, not necessarily at the point.

These principles are fundamental to understanding limits and will serve you well in your calculus studies. Now, to solidify your understanding, it's essential to practice more problems. Here are a few suggestions for further practice:

1. Try evaluating limits of other rational functions that result in the $\frac{0}{0}$ form. Look for opportunities to factor and simplify.
2. Explore limits that involve other indeterminate forms, such as $\frac{\infty}{\infty}$.
3. Consider using graphing calculators or software to visualize the functions and their behavior near the limit point. This can provide valuable intuition.

Remember, mastering limits is crucial for success in calculus. Keep practicing, and you'll become a limit-evaluating pro in no time! Great job today, guys! Keep up the awesome work.