Hey guys! Today, we're diving headfirst into a fascinating limit problem that might seem a bit daunting at first glance. But don't worry, we'll break it down step-by-step, making sure everyone understands the underlying concepts and techniques involved. We're going to explore the limit as x approaches 0 of the function 7x / (√(3x+4) - 2). This isn't just about getting the answer; it's about understanding the process and the mathematical reasoning behind it. So, grab your thinking caps, and let's get started!
Understanding the Problem: A First Look
Before we jump into solving, let's take a moment to really understand what this limit problem is asking. The expression lim (x→0) 7x / (√(3x+4) - 2) is telling us to investigate what happens to the value of the function f(x) = 7x / (√(3x+4) - 2) as x gets incredibly close to 0. It's crucial to understand that we're not necessarily interested in what happens when x actually equals 0, but rather its behavior as it approaches 0 from both sides – the positive side and the negative side.
Now, if we were to simply plug in x = 0 directly into the function, we'd run into a bit of a snag. We'd get 7(0) / (√(3(0)+4) - 2) which simplifies to 0 / (√4 - 2), and further to 0 / (2 - 2), leaving us with 0/0. This is what we call an indeterminate form. An indeterminate form doesn't mean the limit doesn't exist; it just means we can't determine the limit directly by substitution. It's a signal that we need to use some algebraic manipulation or other techniques to figure out what's really going on. This is where the fun begins! We need to find a way to rewrite the function in a way that allows us to evaluate the limit without running into this 0/0 issue. This usually involves techniques like factoring, simplifying, or, in this case, a clever application of conjugates. Understanding why we can't just substitute and the meaning of an indeterminate form is a fundamental concept in calculus, and it's the key to unlocking this problem. We are essentially looking for a way to “unmask” the true behavior of the function near x = 0. So, let's dive into how we can do just that!
The Conjugate Trick: Our Key to Success
Alright, so we've established that direct substitution leads us to an indeterminate form. What's our next move? This is where a little algebraic wizardry comes into play – specifically, the technique of multiplying by the conjugate. This is a common and powerful method for dealing with limits involving square roots, and it's exactly what we need here. The conjugate of the denominator, √(3x+4) - 2, is simply √(3x+4) + 2. We create the conjugate by changing the minus sign between the square root and the constant to a plus sign. Why do we do this? Because when we multiply an expression by its conjugate, we eliminate the square root in the denominator, which often helps us simplify the expression and get rid of the indeterminate form.
Think of it like this: multiplying by the conjugate is like using a special tool to reshape the function into a more manageable form. But there's a golden rule in mathematics: we can't just change an expression willy-nilly. To keep things mathematically sound, we must multiply both the numerator and the denominator by the same thing. This is equivalent to multiplying the entire function by 1, which doesn't change its value but does change its appearance. So, we multiply both the top and bottom of our fraction by √(3x+4) + 2. This sets up a beautiful scenario. When we multiply the denominators (√(3x+4) - 2) and (√(3x+4) + 2), we'll use the difference of squares pattern: (a - b)(a + b) = a² - b². This will get rid of the square root, leaving us with a simpler expression in the denominator. Remember, the goal here is to manipulate the function algebraically so that we can avoid the 0/0 situation and directly evaluate the limit. This conjugate trick is a fundamental technique for handling limits involving radicals, and mastering it will take you a long way in calculus. It’s not just about memorizing the steps; it’s about understanding why it works and when to apply it. So, let’s see how this multiplication unfolds and how it helps us solve our limit.
Performing the Multiplication: A Step-by-Step Guide
Okay, guys, let's roll up our sleeves and get into the nitty-gritty of the multiplication. We've established that we need to multiply both the numerator and the denominator of our function by the conjugate, which is √(3x+4) + 2. So, we're going to take our original expression, 7x / (√(3x+4) - 2), and multiply it by (√(3x+4) + 2) / (√(3x+4) + 2). Remember, this is the same as multiplying by 1, so we're not changing the value of the expression, just its form. Let's start with the numerator. We have 7x multiplied by (√(3x+4) + 2). We can simply write this as 7x(√(3x+4) + 2) for now; we'll see if we need to distribute later. Now for the denominator, this is where the magic happens. We're multiplying (√(3x+4) - 2) by its conjugate (√(3x+4) + 2). As we discussed earlier, this is in the form (a - b)(a + b), which equals a² - b². So, in our case, a is √(3x+4) and b is 2. Applying the formula, we get (√(3x+4))² - 2². This simplifies to (3x+4) - 4. Notice how the square root is gone! This was the whole point of using the conjugate. Now, let's simplify the denominator further. We have (3x+4) - 4, which simply becomes 3x. This is a huge step forward. Our function is starting to look much cleaner and more manageable. We've successfully eliminated the square root from the denominator, and we're on our way to getting rid of that indeterminate form. It's important to take your time with these steps, ensuring that each multiplication and simplification is done carefully. A small error here can throw off the entire solution. This step-by-step approach allows us to carefully track how the function is transformed, and it sets the stage for the next crucial step: simplifying the fraction.
Simplifying the Fraction: Spotting the Cancellation
Fantastic! We've successfully multiplied by the conjugate and simplified the denominator. Now, let's take a look at our transformed expression. We have 7x(√(3x+4) + 2) in the numerator and 3x in the denominator. Do you see anything we can simplify? Bingo! We have an 'x' in both the numerator and the denominator. This is a crucial observation because it allows us to cancel out a common factor, which is often the key to resolving indeterminate forms. We can divide both the numerator and the denominator by 'x'. This gives us 7(√(3x+4) + 2) in the numerator and 3 in the denominator. Notice how the 'x' is completely gone from the denominator! This is exactly what we wanted. We've effectively eliminated the term that was causing the 0/0 indeterminate form. Remember, the reason we couldn't directly substitute x = 0 at the beginning was because it made the denominator zero. By canceling out the 'x', we've removed this obstruction. This simplification step is a powerful technique in evaluating limits. It highlights the importance of looking for common factors that can be canceled out. Often, algebraic manipulations like multiplying by the conjugate are just a means to an end – the end being the ability to simplify the expression and make it directly evaluable. This step also emphasizes the difference between a function and its simplified form. While the original function and the simplified function are equal everywhere except potentially at x = 0, their limits as x approaches 0 might be very different. Now that we've simplified our fraction, we're in a prime position to finally evaluate the limit. The path is clear; let's proceed to the final step.
Evaluating the Limit: The Final Step
We've reached the home stretch! After our algebraic maneuvering, we're left with the simplified expression 7(√(3x+4) + 2) / 3. Now, we can try substituting x = 0 directly into this expression and see what happens. Let's plug it in: 7(√(3(0)+4) + 2) / 3. This simplifies to 7(√4 + 2) / 3, which further simplifies to 7(2 + 2) / 3. Continuing the simplification, we have 7(4) / 3, which gives us 28/3. And there you have it! The limit as x approaches 0 of our original function, 7x / (√(3x+4) - 2), is 28/3. This is a concrete value, and it tells us precisely what happens to the function as x gets closer and closer to 0. It's important to recognize the significance of this result. We started with an indeterminate form, 0/0, which gave us no immediate information about the limit. Through a series of strategic algebraic steps – multiplying by the conjugate and simplifying – we transformed the function into a form where we could directly evaluate the limit. This process highlights the power of algebraic manipulation in calculus. It's not just about blindly applying formulas; it's about understanding the underlying structure of the function and using the right tools to reveal its behavior. Evaluating the limit is the final payoff, the culmination of all our hard work. It's the answer to our original question, and it gives us valuable insight into the function's behavior near a specific point. So, we've successfully navigated this limit problem, and hopefully, you've gained a deeper understanding of the techniques involved.
Conclusion: Mastering the Art of Limits
Alright, guys, we've tackled a challenging limit problem head-on, and hopefully, you've gained some valuable insights along the way. We started with the expression lim (x→0) 7x / (√(3x+4) - 2), encountered an indeterminate form, and used the powerful technique of multiplying by the conjugate to simplify the expression. Through careful algebraic manipulation, we were able to cancel out common factors, eliminate the indeterminate form, and finally, evaluate the limit to be 28/3. This journey illustrates several key concepts in calculus. First, it highlights the importance of understanding the concept of limits – what it means for a function to approach a certain value as its input gets closer to a specific point. Second, it demonstrates the significance of recognizing and dealing with indeterminate forms. Indeterminate forms are not roadblocks; they are opportunities to apply our algebraic skills and uncover the true behavior of the function. Third, it showcases the power of algebraic manipulation in calculus. Techniques like multiplying by the conjugate, factoring, and simplifying are essential tools in our problem-solving arsenal. But most importantly, this exercise underscores the importance of a step-by-step approach. Breaking down a complex problem into smaller, manageable steps allows us to tackle each challenge with clarity and precision. It also helps us avoid errors and maintain a clear understanding of the process. So, the next time you encounter a limit problem, remember the lessons we've learned here. Don't be intimidated by the complexity; embrace the challenge, and remember the power of algebraic manipulation. And remember, practice makes perfect! The more you work with limits, the more comfortable and confident you'll become in solving them. Keep exploring, keep learning, and keep pushing your mathematical boundaries. You've got this!
