Limit Calculation: Solving $\lim_{x \to 1} (rac{1}{x+1} - rac{x-1}{x^2-1})$

Hey guys! Today, we're diving into a super interesting limit problem that might seem a bit tricky at first glance. We've got to find the value of the limit: $\lim_{x \to 1} (\frac{1}{x+1} - \frac{x-1}{x^2-1})$. Sounds like fun, right? Don't worry, we'll break it down step by step so it's crystal clear. So, grab your thinking caps, and let's get started!

Understanding the Problem

So, let's talk about what we're actually trying to do here. We're dealing with a limit, which is a fundamental concept in calculus. Basically, a limit tells us what value a function approaches as the input (in this case, x) gets closer and closer to a specific value (which is 1 in our problem). The function we're working with is $(\frac{1}{x+1} - \frac{x-1}{x^2-1})$. This looks a bit complex, but don't sweat it – we'll simplify it soon enough. When we see $\lim_{x \to 1}$, it's like asking, "Hey, what happens to this function as x gets super close to 1?" This is super important in various fields, from physics to engineering, where we often need to understand how things behave under extreme conditions or near certain critical points. Remember, limits are all about understanding the behavior of a function, not necessarily the exact value at a single point. That's why we need to use some algebraic tricks to make it easier to evaluate. This limit problem is a classic example where direct substitution might lead to an indeterminate form, so we need to be clever about how we approach it.

Why Can't We Just Plug in x = 1?

You might be thinking, "Why can't we just plug in x = 1 and call it a day?" That's a great question! If we try that, we get: $(\frac{1}{1+1} - \frac{1-1}{1^2-1}) = (\frac{1}{2} - \frac{0}{0})$. Uh-oh! We've got a $\frac{0}{0}$ situation, which is what we call an indeterminate form. This basically means we can't determine the limit directly by substituting x = 1. It's like hitting a roadblock – we need to find another route. Indeterminate forms are common in limit problems, and they tell us that we need to do some more work, like simplifying the expression or using techniques like L'Hôpital's Rule (which we won't need for this particular problem, thankfully!). The fact that we get an indeterminate form doesn't mean the limit doesn't exist; it just means we need to use some algebraic manipulation to find out what it is. So, let's roll up our sleeves and get to the fun part – simplifying!

Simplifying the Expression

Okay, so we can't just plug in x = 1. What's the next move? We need to simplify the expression inside the limit to get rid of that troublesome $\frac{0}{0}$ situation. The key here is to focus on the denominator $x^2 - 1$. Does that look familiar to anyone? It should! It's a classic difference of squares. We can factorize $x^2 - 1$ as $(x - 1)(x + 1)$. Factoring is a super powerful tool in algebra, and it's going to help us a lot here. By recognizing this pattern, we can rewrite our expression and hopefully cancel out some terms. This is a common strategy in limit problems: look for opportunities to factor, simplify, and cancel. So, let's rewrite our limit with the factored denominator:

$\lim_{x \to 1} (\frac{1}{x+1} - \frac{x-1}{(x-1)(x+1)})$.

See anything we can cancel? You bet!

Cancelling Common Factors

Now we're talking! Look closely, guys. We've got $(x - 1)$ in both the numerator and the denominator of the second term. Time to do some canceling! Remember, we can cancel common factors because $\frac{x-1}{x-1}$ is equal to 1 (as long as $x$ isn't 1, which is fine because we're taking the limit as x approaches 1, not at x = 1). This is a crucial step. By canceling out the $(x - 1)$ terms, we're getting rid of the part of the expression that was causing the $\frac{0}{0}$ indeterminate form. After canceling, our expression looks much simpler:

$\lim_{x \to 1} (\frac{1}{x+1} - \frac{1}{x+1})$

Wow, that's a lot cleaner, isn't it? What happens when you subtract something from itself? You get zero, of course!

Evaluating the Limit

Alright, we've done the hard work of simplifying the expression. Now comes the easy part: evaluating the limit. After canceling those common factors, we're left with:

$\lim_{x \to 1} (\frac{1}{x+1} - \frac{1}{x+1})$

As we pointed out, anything minus itself is zero, so we have:

$\lim_{x \to 1} (0)$

This is pretty straightforward. The limit of a constant (in this case, 0) is just the constant itself. So, the value of the limit is:

0

That's it! We've found the answer. The limit of the given expression as x approaches 1 is 0. This might have seemed like a tough problem at first, but by breaking it down into smaller steps – understanding the problem, simplifying the expression, and then evaluating the limit – we were able to solve it without too much trouble. Remember, guys, practice makes perfect. The more limit problems you solve, the better you'll get at spotting these tricks and techniques.

Conclusion

So, to recap, we started with the limit $\lim_{x \to 1} (\frac{1}{x+1} - \frac{x-1}{x^2-1})$ and walked through the steps to find its value. We learned why we couldn't just plug in x = 1 (hello, indeterminate form!), and how to use factoring and cancellation to simplify the expression. The key takeaway here is that when you're faced with a limit problem, especially one that gives you an indeterminate form, don't panic! Look for ways to simplify the expression using algebraic techniques. Factoring, canceling common factors, and combining fractions are your best friends in these situations. Once we simplified the expression, evaluating the limit was a piece of cake. We found that the value of the limit is 0. Remember, understanding limits is crucial for calculus and beyond, so mastering these techniques will pay off big time. Keep practicing, keep exploring, and you'll become limit-solving pros in no time! And most importantly, have fun with it! Math can be like a puzzle, and solving it is super satisfying.

Final Answer

The final answer is A. 0. We successfully navigated this limit problem by simplifying the expression and evaluating the result. Great job, everyone! Keep up the awesome work!