Solving Limits: Step-by-Step Guide For (x→2)

Hey everyone! Today, we're diving into a classic calculus problem: finding the limit of a rational function. Specifically, we're tackling ${\lim _{x \rightarrow 2} \frac{x^2-4 x+4}{x^2-7 x+10}}$. This might look intimidating at first, but don't worry, we'll break it down step by step. Understanding limits is crucial in calculus, as they form the foundation for concepts like derivatives and integrals. So, let's roll up our sleeves and get started!

Understanding the Limit Problem

So, what does ${\lim _{x \rightarrow 2} \frac{x^2-4 x+4}{x^2-7 x+10}}$ actually mean? In layman's terms, we want to know what value the function ${f(x) = \frac{x^2-4 x+4}{x^2-7 x+10}}$ approaches as x gets closer and closer to 2, without actually being equal to 2. This is a fundamental concept in calculus, and it's used to define continuity, derivatives, and integrals. Now, the first thing you might think to do is just plug in x = 2 directly into the function. However, if we do that, we get:

${f(2) = \frac{2^2 - 4(2) + 4}{2^2 - 7(2) + 10} = \frac{4 - 8 + 4}{4 - 14 + 10} = \frac{0}{0}}$

Uh oh! We've got an 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. Indeterminate forms are like puzzles in calculus – they tell us there's more work to be done. They often indicate the presence of a common factor in the numerator and denominator that we can cancel out. So, what's our next move? This is where our algebra skills come into play. Factoring is our key tool here. By factoring the numerator and the denominator, we can often simplify the expression and eliminate the indeterminate form. Remember, the goal is to rewrite the function in a form where we can directly substitute x = 2 and get a meaningful answer.

Factoring the Quadratic Expressions

The key to cracking this limit problem lies in factoring. Let's take a look at the numerator and the denominator separately. First, we have the numerator: ${x^2 - 4x + 4}$. This is a quadratic expression, and it looks like it might be a perfect square trinomial. A perfect square trinomial is a quadratic that can be factored into the form ${(ax + b)^2}$ or ${(ax - b)^2}$. To see if it fits this pattern, we can look for two numbers that multiply to give the constant term (4) and add up to give the coefficient of the linear term (-4). In this case, those numbers are -2 and -2. So, we can factor the numerator as follows:

${x^2 - 4x + 4 = (x - 2)(x - 2) = (x - 2)^2}$

Great! Now let's move on to the denominator: ${x^2 - 7x + 10}$. This is another quadratic expression, and we need to find two numbers that multiply to give 10 and add up to -7. After a bit of thought, we can see that -2 and -5 fit the bill. So, we can factor the denominator as:

${x^2 - 7x + 10 = (x - 2)(x - 5)}$

Now that we've factored both the numerator and the denominator, we can rewrite the original limit expression. Factoring is such a powerful technique in algebra and calculus. It allows us to simplify complex expressions and reveal hidden structures. In this case, it's going to help us get rid of that pesky indeterminate form and find the limit. By factoring, we're essentially uncovering the underlying behavior of the function near x = 2. We're isolating the factor that's causing the problem (the factor that makes both the numerator and denominator zero) so we can deal with it appropriately.

Simplifying the Rational Function

Alright, we've factored the numerator and denominator, and now we're ready for the next step: simplification! Remember, our original limit was ${\lim _{x \rightarrow 2} \frac{x^2-4 x+4}{x^2-7 x+10}}$. After factoring, we have:

${\lim _{x \rightarrow 2} \frac{(x - 2)(x - 2)}{(x - 2)(x - 5)}}$

Now, here's where the magic happens. We notice that we have a common factor of ${(x - 2)}$ in both the numerator and the denominator. As long as x is not equal to 2 (and remember, in a limit, we're looking at values close to 2, not equal to 2), we can cancel out this common factor. This is a crucial step, as it eliminates the indeterminate form we encountered earlier. Cancelling the common factor gives us:

${\lim _{x \rightarrow 2} \frac{(x - 2)}{(x - 5)}}$

Look at that! The expression has been significantly simplified. We've gone from a potentially messy quadratic rational function to a much cleaner linear rational function. This simplification is what allows us to finally evaluate the limit. By canceling the common factor, we've essentially removed the discontinuity at x = 2. The original function had a "hole" at x = 2 because both the numerator and denominator were zero there. But by factoring and canceling, we've filled in that hole, and we can now see what value the function is approaching as x gets close to 2. Simplifying rational functions is a fundamental skill in calculus and algebra. It allows us to work with expressions that are easier to handle and understand. In this case, it's the key to finding the limit.

Evaluating the Limit

We've simplified our expression down to ${\lim _{x \rightarrow 2} \frac{(x - 2)}{(x - 5)}}$. Now comes the moment of truth: evaluating the limit. Remember, after simplifying, we can often just plug in the value that x is approaching. In this case, x is approaching 2, so let's substitute x = 2 into our simplified expression:

${\frac{(2 - 2)}{(2 - 5)} = \frac{0}{-3} = 0}$

And there we have it! The limit of the function as x approaches 2 is 0. This means that as x gets closer and closer to 2, the value of the function ${f(x) = \frac{x^2-4 x+4}{x^2-7 x+10}}$ gets closer and closer to 0. This result makes sense when we consider the simplification we performed earlier. We canceled out the common factor of ${(x - 2)}$, which was the source of the indeterminate form. After canceling, the numerator became ${(x - 2)}$, which clearly approaches 0 as x approaches 2. The denominator, ${(x - 5)}$, approaches -3 as x approaches 2, so the overall fraction approaches ${\frac{0}{-3} = 0}$. Evaluating limits is a core skill in calculus. It allows us to understand the behavior of functions near specific points, even if the function is not defined at those points. In this case, we were able to find the limit by using a combination of factoring, simplifying, and direct substitution. This process is a common strategy for evaluating limits of rational functions. The fact that the limit exists and is equal to 0 tells us something important about the function's behavior near x = 2. It suggests that the function is well-behaved in that neighborhood, even though it had an indeterminate form at x = 2 initially.

Conclusion: Mastering Limit Calculations

So, guys, we've successfully navigated the limit problem ${\lim _{x \rightarrow 2} \frac{x^2-4 x+4}{x^2-7 x+10}}$! We started with an indeterminate form, used factoring to simplify the expression, and then evaluated the limit by direct substitution. This whole process highlights the power of algebraic manipulation in calculus. Factoring, simplifying, and canceling common factors are essential techniques for handling limits, especially those involving rational functions. Remember, when you encounter an indeterminate form like ${\frac{0}{0}}$, don't panic! It's a clue that you need to do some algebraic work to reveal the true behavior of the function. Limits are the bedrock of calculus. They allow us to define continuity, derivatives, and integrals, which are the fundamental tools for analyzing change and motion. By mastering limit calculations, you're building a strong foundation for your calculus journey. Keep practicing, keep exploring, and you'll become a limit-solving pro in no time! This specific example demonstrates a common strategy for evaluating limits of rational functions: try direct substitution first, and if you get an indeterminate form, factor and simplify. This approach can be applied to a wide range of limit problems, so it's a valuable tool to have in your calculus toolkit. And remember, the more you practice, the more comfortable you'll become with these techniques, and the more confident you'll be in your ability to tackle even the most challenging limit problems. So, keep up the great work, and happy calculating!