Limit Calculation: (1/(3x-1))/(9x^2-1) As X→1/3
Hey guys! Let's dive into a fascinating problem in calculus: finding the limit of a function. Specifically, we're going to tackle the limit of (1/(3x-1))/(9x²-1) as x approaches 1/3. This is a classic example that combines algebraic manipulation and limit concepts, and I’m excited to walk you through each step. Buckle up, because we’re about to explore how to solve this intriguing mathematical puzzle!
Understanding the Problem
Before we jump into solving, let’s make sure we understand exactly what we're trying to find. We are given the function:
f(x) = (1/(3x-1)) / (9x² - 1)
Our mission, should we choose to accept it (and we do!), is to determine what value this function approaches as x gets incredibly close to 1/3. This is written mathematically as:
lim (x→1/3) [ (1/(3x-1)) / (9x² - 1) ]
Why is this interesting? Well, if we try to directly substitute x = 1/3 into the function, we run into a problem. The denominator 3x - 1 becomes zero, and we all know you can't divide by zero! This means we have an indeterminate form, which basically tells us we need to do some more digging to find the limit. But don’t worry, we have the tools to handle this. We'll use some algebraic tricks to simplify the expression and then find the limit.
It's crucial to identify these issues early on because they dictate our approach. Simply plugging in the value sometimes works, but when it doesn't, that's where the real math magic begins! We need to understand why direct substitution fails and what strategies we can employ to overcome this hurdle. This initial assessment sets the stage for the rest of our solution.
Remember, limits are fundamental to calculus. They help us understand the behavior of functions, especially near points where the function might be undefined. By mastering the techniques to evaluate limits, we unlock the ability to analyze more complex functions and their properties. So, let's get started and see how we can unravel this limit problem!
Step 1: Simplifying the Expression
The heart of solving this limit lies in simplifying the given expression. Remember, our function is:
f(x) = (1/(3x-1)) / (9x² - 1)
This looks a bit clunky, so let's clean it up. The key here is to recognize that the denominator (9x² - 1) is a difference of squares. Ah, those trusty algebraic identities! We can factor it as:
9x² - 1 = (3x - 1)(3x + 1)
Now, substitute this factorization back into our function:
f(x) = (1/(3x-1)) / [ (3x - 1)(3x + 1) ]
To make things even clearer, let's rewrite the division as multiplication by the reciprocal:
f(x) = (1/(3x-1)) * [ 1 / ((3x - 1)(3x + 1)) ]
Now we can combine the terms in the denominator:
f(x) = 1 / [ (3x - 1)(3x - 1)(3x + 1) ]
f(x) = 1 / [ (3x - 1)²(3x + 1) ]
Wait a minute! Notice anything interesting? We have a (3x - 1) term in both the numerator and the denominator. This is our golden ticket! We can cancel one (3x - 1) term from the numerator and denominator, but ONLY if x ≠ 1/3. Remember, we're taking the limit as x approaches 1/3, so we're interested in values close to 1/3, not exactly 1/3. This is a crucial distinction in limit calculations. After canceling, our simplified function becomes:
f(x) = 1 / [ (3x - 1)(3x + 1) ]
This simplification is a game-changer! It transforms our function into something much easier to handle. By recognizing and applying the difference of squares factorization, we've cleared a major hurdle. This step highlights the power of algebraic manipulation in simplifying complex expressions and preparing them for limit evaluation. Now that we have a more manageable form, let's move on to the next step: evaluating the limit.
Step 2: Evaluating the Limit
Okay, guys, we've done the hard work of simplifying our expression. Now comes the fun part: actually finding the limit! We've transformed our original function into:
f(x) = 1 / [ (3x - 1)(3x + 1) ]
And we want to find:
lim (x→1/3) [ 1 / ( (3x - 1)(3x + 1) ) ]
Now that we've canceled the problematic term, we can try direct substitution again. This means we plug in x = 1/3 into our simplified function:
f(1/3) = 1 / [ (3(1/3) - 1)(3(1/3) + 1) ]
Let's break it down:
3 * (1/3) = 1
So, we have:
f(1/3) = 1 / [ (1 - 1)(1 + 1) ]
f(1/3) = 1 / [ (0)(2) ]
Oops! We still have a zero in the denominator! This tells us that we made a mistake in the previous step. Let's go back to where we simplified the function:
f(x) = (1/(3x-1)) * [ 1 / ((3x - 1)(3x + 1)) ]
Before cancelling, let's combine the fractions correctly:
f(x) = 1 / ((3x - 1) * (3x - 1)(3x + 1))
f(x) = 1 / ((3x - 1)² (3x + 1))
Now, let's try factoring 9x² - 1 correctly and rewrite the original function:
f(x) = (1 / (3x - 1)) / (9x² - 1)
f(x) = (1 / (3x - 1)) / ((3x - 1)(3x + 1))
To divide by a fraction, we multiply by its reciprocal:
f(x) = (1 / (3x - 1)) * (1 / ((3x - 1)(3x + 1)))
f(x) = 1 / ((3x - 1)²(3x + 1))
We made a mistake by cancelling (3x - 1) too early. We should not have canceled it because it leads to an incorrect simplification. We need to proceed with the correct simplified form:
f(x) = 1 / ((3x - 1)²(3x + 1))
Now, let's analyze the limit as x approaches 1/3:
lim (x→1/3) [1 / ((3x - 1)²(3x + 1))]
As x approaches 1/3, (3x - 1) approaches 0, so (3x - 1)² also approaches 0. At the same time, (3x + 1) approaches (3(1/3) + 1) = 2. Thus, we have:
lim (x→1/3) [1 / ((3x - 1)²(3x + 1))] = 1 / (0 * 2) = 1 / 0
This indicates that the limit is going to infinity. Since the denominator approaches 0 and is squared, it will always be positive. Therefore, the function will approach positive infinity.
So, we can conclude:
lim (x→1/3) [1 / ((3x - 1)²(3x + 1))] = ∞
Step 3: Discussing the Result
Alright, let's talk about what our result actually means. We've determined that:
lim (x→1/3) [ (1/(3x-1)) / (9x² - 1) ] = ∞
This tells us that as x gets closer and closer to 1/3, the value of our function grows without bound – it shoots off to infinity. This is a classic example of a function having a vertical asymptote at x = 1/3. In simpler terms, the function has a dramatic, unbounded behavior near this point.
Think about it graphically: if you were to plot this function, you'd see the curve getting closer and closer to the vertical line x = 1/3, but never actually touching it. Instead, the curve would skyrocket upwards (or plummet downwards, depending on which side you approach from). This vertical asymptote is a visual representation of the function's limit approaching infinity.
Why did this happen? It all boils down to the denominator. As x approaches 1/3, the term (3x - 1)² gets incredibly small, approaching zero. And we know that dividing by a very small number results in a very large number. Since we have a square in the denominator, the term is always positive, so the function goes to positive infinity.
This kind of behavior is super important to understand in calculus. Functions with infinite limits can model many real-world phenomena, from the concentration of a substance in a chemical reaction to the behavior of electrical circuits. Recognizing and interpreting these limits is a key skill in higher-level math and its applications.
What did we learn? We learned that not all limits are finite numbers. Sometimes, functions can grow without bound, leading to infinite limits. We also saw how crucial it is to correctly simplify expressions and to be careful when canceling terms. A small mistake in algebra can lead to a completely different result. This problem also reinforced the idea that limits tell us about the behavior of a function near a point, not necessarily at the point itself.
So, next time you encounter a limit problem, remember our journey here. Simplify, evaluate, and most importantly, think about what the result means in the bigger picture. Math isn't just about getting the right answer; it's about understanding the story the numbers are telling us.
Conclusion
Well, there you have it, guys! We've successfully navigated the tricky terrain of limits and determined that the limit of (1/(3x-1))/(9x²-1) as x approaches 1/3 is infinity. We started by understanding the problem, then used algebraic manipulation to simplify the expression. We had a little detour when we realized our initial simplification was incorrect, highlighting the importance of double-checking our steps. Finally, we correctly evaluated the limit and discussed what our result means in terms of the function's behavior.
This problem was a fantastic example of how limits work and how they can reveal fascinating aspects of a function's behavior. It also underscored the importance of careful algebraic manipulation and attention to detail. Remember, guys, math is like a puzzle – each piece must fit perfectly to reveal the final picture. Keep practicing, keep exploring, and you'll become limit-solving pros in no time!