Limit Calculation: Solving Lim (x->2) [(6-x)/(x^2-4) - 1/(x-2)]

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Hey guys! Let's dive into this interesting limit problem. We've got: lim (x->2) [(6-x)/(x^2-4) - 1/(x-2)]. Limits can seem tricky, but with a bit of algebraic manipulation, we can crack this one. The key here is to simplify the expression inside the limit before we directly substitute the value of x. So, grab your thinking caps, and let's get started!

Understanding the Problem

Before we jump into solving, let's quickly understand what this problem is asking. We need to find the value that the expression rac{6 - x}{x^2 - 4} - rac{1}{x - 2} approaches as x gets closer and closer to 2. Direct substitution might lead to an indeterminate form (like 0/0), so we need to simplify first.

When you first look at a limit problem like this, it's super tempting to just plug in the value that x is approaching. But hold up! That's a classic rookie mistake that can lead you down the wrong path, especially when you're dealing with rational functions. Plugging in x = 2 directly into our expression gives us rac{6 - 2}{2^2 - 4} - rac{1}{2 - 2} = rac{4}{0} - rac{1}{0} , which is undefined. We've got division by zero happening, and that's a big no-no in the math world. So, we need a different strategy. We need to massage this expression, clean it up, and make it so that we can plug in x = 2 without causing a mathematical catastrophe.

The first thing that should catch your eye is the denominator x^2 - 4 . This isn't just some random quadratic; it's a difference of squares! Recognizing these patterns is crucial in calculus. We can factor x^2 - 4 into (x - 2)(x + 2) . This is super helpful because we also have a term with (x - 2) in the denominator. Factoring is like unlocking a secret level in a video game; it reveals hidden possibilities for simplification. Once we factor, we can see a common factor emerging, and that's our ticket to making this limit solvable.

Next up, we need to deal with those fractions. We've got two fractions being subtracted, and to combine them, we need a common denominator. Think back to your algebra days – adding and subtracting fractions is all about that common denominator. In our case, we've got denominators of (x - 2)(x + 2) and (x - 2) . The least common denominator is going to be (x - 2)(x + 2) . This means we'll need to multiply the second fraction by rac{x + 2}{x + 2} to get that common denominator. This might seem like a bit of a detour, but trust me, it's a necessary step. Once we have that common denominator, we can combine the fractions and start simplifying the numerator.

After we combine the fractions, the real magic happens. We'll have a single fraction, and hopefully, some terms in the numerator will cancel out or simplify. This is where all our hard work pays off. We're aiming to eliminate the problematic term that was causing the division by zero. Often, in limit problems, there's a factor that's causing the issue, and by simplifying, we can get rid of it. It's like performing surgery on the expression, carefully removing the part that's making it sick. And once that problematic factor is gone, we're in the clear to finally substitute x = 2 and find our limit.

Step-by-Step Solution

  1. Factor the denominator: rac{6 - x}{x^2 - 4} - rac{1}{x - 2} = rac{6 - x}{(x - 2)(x + 2)} - rac{1}{x - 2}
  2. Find a common denominator: rac{6 - x}{(x - 2)(x + 2)} - rac{1}{x - 2} imes rac{x + 2}{x + 2} = rac{6 - x}{(x - 2)(x + 2)} - rac{x + 2}{(x - 2)(x + 2)}
  3. Combine the fractions: rac{6 - x - (x + 2)}{(x - 2)(x + 2)} = rac{6 - x - x - 2}{(x - 2)(x + 2)} = rac{4 - 2x}{(x - 2)(x + 2)}
  4. Factor out -2 from the numerator: rac{-2(x - 2)}{(x - 2)(x + 2)}
  5. Cancel the common factor (x - 2): rac{-2}{x + 2}
  6. Now, substitute x = 2: rac{-2}{2 + 2} = rac{-2}{4} = - rac{1}{2}

Detailed Breakdown of Each Step

Let's break down each step in detail so you can see exactly what's going on and why we're doing it. This isn't just about getting the right answer; it's about understanding the process and building a strong foundation for more complex problems.

Step 1: Factor the denominator

The first move we made was factoring the denominator x^2 - 4 . As we mentioned earlier, recognizing the difference of squares pattern is super important. x^2 - 4 fits this pattern perfectly: it's x^2 minus 2^2 . The difference of squares factorization is a classic: a^2 - b^2 = (a - b)(a + b) . So, x^2 - 4 becomes (x - 2)(x + 2) . Why is this so crucial? Well, we immediately see that we're going to have a common factor of (x - 2) with the other denominator. This is like finding a matching puzzle piece – it tells us we're on the right track for simplification.

Step 2: Find a common denominator

Next up, we needed a common denominator to combine those fractions. We had denominators of (x - 2)(x + 2) and (x - 2) . The least common denominator is the smallest expression that both denominators can divide into evenly. In this case, it's (x - 2)(x + 2) . So, we needed to multiply the second fraction, rac{1}{x - 2} , by rac{x + 2}{x + 2} . Remember, multiplying by rac{x + 2}{x + 2} is just like multiplying by 1, so we're not changing the value of the expression, just its appearance. This step is all about getting the fractions ready to be combined.

Step 3: Combine the fractions

Now we're ready to combine the fractions. We've got rac{6 - x}{(x - 2)(x + 2)} - rac{x + 2}{(x - 2)(x + 2)} . Since they have the same denominator, we can just subtract the numerators: rac{6 - x - (x + 2)}{(x - 2)(x + 2)} . Be super careful with that minus sign! It applies to the entire expression (x + 2) , so we need to distribute it. This gives us rac{6 - x - x - 2}{(x - 2)(x + 2)} . Now we simplify the numerator by combining like terms: rac{4 - 2x}{(x - 2)(x + 2)} . We're getting closer and closer to that simplified form!

Step 4: Factor out -2 from the numerator

Looking at the numerator, 4 - 2x , we can see a common factor of 2. But to make things even cleaner, we're going to factor out a -2. This gives us -2(x - 2) . Why factor out a negative? Because we have an (x - 2) in the denominator, and we want to see if we can cancel it out. Factoring out the -2 makes that cancellation much clearer.

Step 5: Cancel the common factor (x - 2)

This is the moment we've been waiting for! We have rac{-2(x - 2)}{(x - 2)(x + 2)} . The (x - 2) term appears in both the numerator and the denominator, so we can cancel it out. This is like hitting the jackpot in a math problem! Canceling this factor is what allows us to finally evaluate the limit without running into that division by zero issue. After canceling, we're left with the much simpler expression rac{-2}{x + 2} .

Step 6: Now, substitute x = 2

We've done the hard work, and now it's time for the payoff. We've simplified our expression to rac{-2}{x + 2} , and now we can safely substitute x = 2. Plugging in x = 2, we get rac{-2}{2 + 2} = rac{-2}{4} = - rac{1}{2} . And there it is! The limit of our original expression as x approaches 2 is - rac{1}{2} .

Final Answer

So, guys, the final answer is - rac{1}{2}. This problem is a classic example of how algebraic manipulation can help us solve seemingly tricky limit problems. Remember, always look for ways to simplify before you substitute! Keep practicing, and you'll become a limit-solving pro in no time! Analyzing each step meticulously ensures a solid grasp of the concepts, paving the way for tackling more complex problems with confidence.