Prove The Limit: Step-by-Step Solution

Hey guys! Ever find yourself staring at a limit problem and feeling totally lost? Don't worry, we've all been there. Today, we're going to break down a classic limit problem step-by-step, so you can see exactly how it's done. We'll tackle the problem:

(a) Prove that $\lim _{x \rightarrow 5}\left(\frac{1}{x-5}-\frac{25}{x^3-5 x^2}\right)=\frac{2}{5}$

This might look intimidating at first, but I promise, it's totally manageable. We'll use some basic algebra and a little bit of limit theory to get to the solution. So, grab your thinking caps, and let's dive in!

Understanding Limits

Before we jump into the problem, let's quickly recap what a limit actually means. In simple terms, a limit tells us what value a function approaches as its input gets closer and closer to a specific value. In our case, we want to know what happens to the expression $\frac{1}{x-5}-\frac{25}{x^3-5 x^2}$ as x gets closer and closer to 5.

Now, you might be tempted to just plug in x = 5 directly. But hold on! If we do that, we get a division by zero situation, which is a big no-no in the math world. That's where the magic of limits comes in. We need to manipulate the expression algebraically before we try to evaluate the limit.

The key concept here is that we're interested in what happens near 5, not at 5 itself. This subtle distinction allows us to perform some clever algebraic tricks to simplify the expression and get rid of the problematic division by zero.

Think of it like this: you're trying to drive your car as close as possible to a wall without actually crashing into it. The limit is like figuring out how close you can get. You might not be able to touch the wall (just like we can't directly substitute x = 5), but you can get arbitrarily close.

Step-by-Step Solution

Okay, let's get down to business. Here's how we can prove that $\lim _{x \rightarrow 5}\left(\frac{1}{x-5}-\frac{25}{x^3-5 x^2}\right)=\frac{2}{5}$:

1. Simplify the Expression

The first step is to simplify the expression inside the limit. This usually involves finding a common denominator and combining the fractions. Focusing on algebraic manipulation is key here. We have:

$\frac{1}{x-5}-\frac{25}{x^3-5 x^2}$

Notice that we can factor an $x^2$ out of the denominator in the second term:

$\frac{1}{x-5}-\frac{25}{x^2(x-5)}$

Now we can see that the common denominator is $x^2(x-5)$. Let's rewrite the first fraction with this denominator:

$\frac{x^2}{x^2(x-5)}-\frac{25}{x^2(x-5)}$

Now we can combine the fractions:

$\frac{x^2 - 25}{x^2(x-5)}$

2. Factor and Cancel

The next step is to factor the numerator. We recognize that $x^2 - 25$ is a difference of squares, which factors as $(x-5)(x+5)$. So we have:

$\frac{(x-5)(x+5)}{x^2(x-5)}$

Now comes the crucial part: we can cancel the $(x-5)$ terms in the numerator and denominator. Remember, we can do this because we're considering the limit as x approaches 5, not when x is actually equal to 5. So, $(x-5)$ is close to zero, but not zero itself. This gives us:

$\frac{x+5}{x^2}$

3. Evaluate the Limit

Now that we've simplified the expression, we can finally evaluate the limit. We substitute x = 5 into the simplified expression:

$\lim_{x \rightarrow 5} \frac{x+5}{x^2} = \frac{5+5}{5^2} = \frac{10}{25} = \frac{2}{5}$

And there you have it! We've shown that $\lim _{x \rightarrow 5}\left(\frac{1}{x-5}-\frac{25}{x^3-5 x^2}\right)=\frac{2}{5}$. This confirms our initial goal using algebraic manipulation and limit evaluation.

Why Does This Work?

It's important to understand why we can cancel the $(x-5)$ term. We're not saying that $(x-5)$ is zero. Instead, we're saying that as x gets arbitrarily close to 5, the ratio $\frac{x-5}{x-5}$ gets arbitrarily close to 1. This is a fundamental concept in limit theory.

If we were to graph the original function and the simplified function, we would see that they are identical everywhere except at x = 5. The original function has a "hole" at x = 5, while the simplified function is continuous at x = 5. The limit allows us to "fill in" that hole and find the value the function approaches.

Common Mistakes to Avoid

When working with limits, there are a few common mistakes to watch out for:

• Plugging in directly without simplifying: As we saw, plugging in x = 5 directly into the original expression leads to division by zero. Always try to simplify the expression first.
• Incorrectly factoring: Make sure you factor the expression correctly. A small factoring error can throw off the entire solution.
• Forgetting to cancel: Canceling common factors is a crucial step in many limit problems. Don't forget to do it!
• Ignoring the limit notation: Remember to keep writing "$\lim_{x \rightarrow 5}$" until you actually substitute x = 5 into the expression. This reminds you that you're working with a limit and haven't yet evaluated it.

Practice Makes Perfect

The best way to master limits is to practice, practice, practice! Try working through similar problems, and don't be afraid to make mistakes. Every mistake is a learning opportunity. Regular practice strengthens understanding of the fundamental limit principles.

Conclusion

So, there you have it! We've successfully proven that $\lim _{x \rightarrow 5}\left(\frac{1}{x-5}-\frac{25}{x^3-5 x^2}\right)=\frac{2}{5}$. By simplifying the expression, factoring, canceling, and then evaluating the limit, we were able to arrive at the answer. Remember, the key to solving limit problems is to manipulate the expression algebraically until you can eliminate any problematic divisions by zero.

I hope this step-by-step guide was helpful. Keep practicing, and you'll be a limit-solving pro in no time! Understanding limits is crucial for calculus and further mathematical studies. Good luck, guys!