Evaluating The Limit Of (1-x)^2 / (x^2 - X - X Ln X) As X Approaches 1

Introduction

In the realm of calculus, the evaluation of limits stands as a fundamental concept, serving as a gateway to understanding continuity, derivatives, and integrals. Among the diverse array of limit problems, those involving indeterminate forms often pose a unique challenge, requiring the application of specialized techniques to unravel their true value. In this article, we embark on a comprehensive exploration of the limit $\lim _{x \rightarrow 1} \frac{(1-x)^2}{x^2-x-x \ln x}$, delving into the intricacies of its evaluation and shedding light on the underlying principles that govern its behavior. This exploration will not only provide a step-by-step solution to the problem but also offer a deeper understanding of the techniques and concepts involved in limit calculations. Understanding limits is crucial not just for mathematicians, but also for engineers, physicists, and anyone working with dynamic systems. Limits help us to understand how functions behave near certain points, which is essential for modeling real-world phenomena. This article aims to equip you with the knowledge and skills to tackle similar limit problems with confidence. We will break down the problem into manageable steps, explaining the reasoning behind each step in detail. By the end of this article, you will not only know how to solve this particular limit but also how to approach other challenging limit problems. So, let's dive in and begin our exploration of this fascinating limit problem!

Identifying the Indeterminate Form

Before embarking on the evaluation of any limit, it is crucial to first ascertain its form. Direct substitution, a straightforward approach, involves substituting the value that x approaches into the expression and observing the result. In our case, as x approaches 1, direct substitution yields the expression $\frac{(1-1)^2}{1^2-1-1 \ln 1} = \frac{0}{0}$, an indeterminate form. This form signifies that the limit cannot be determined directly and necessitates the application of alternative techniques to unravel its value. The indeterminate form $\frac{0}{0}$ is a classic signal that further analysis is required. It tells us that both the numerator and the denominator are approaching zero, but it doesn't tell us how quickly they are approaching zero relative to each other. This is where more advanced techniques like L'Hôpital's Rule or algebraic manipulation become necessary. Recognizing this indeterminate form is the first critical step in solving the limit. Without this initial assessment, we might attempt incorrect approaches or arrive at a misleading conclusion. The indeterminate form is not a dead end; it's simply a signpost directing us to the appropriate path forward. It highlights the need for a more nuanced understanding of the function's behavior as x approaches 1. In the following sections, we will explore different methods for resolving this indeterminate form and ultimately finding the true value of the limit.

Employing L'Hôpital's Rule: A Powerful Technique

L'Hôpital's Rule, a cornerstone of calculus, provides a powerful tool for evaluating limits of indeterminate forms such as $\frac{0}{0}$ and $\frac{\infty}{\infty}$. This rule stipulates that if the limit of the ratio of two functions results in an indeterminate form, then the limit of the ratio of their derivatives, provided it exists, is equal to the original limit. Mathematically, if $\lim_{x \rightarrow c} f(x) = 0$ and $\lim_{x \rightarrow c} g(x) = 0$ (or both approach infinity), and $\lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}$ exists, then:

$\lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}$

In our case, we have $f(x) = (1-x)^2$ and $g(x) = x^2 - x - x \ln x$. To apply L'Hôpital's Rule, we first need to find the derivatives of f(x) and g(x). L'Hôpital's Rule is not a magic bullet, but it's an incredibly useful technique for dealing with indeterminate forms. However, it's essential to remember that it only applies when we have an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Applying it in other situations can lead to incorrect results. The rule works by comparing the rates at which the numerator and denominator are approaching zero (or infinity). By taking derivatives, we are essentially looking at the instantaneous rates of change, which can reveal the true behavior of the limit. It's also crucial to verify that the limit of the derivatives exists before concluding that it's equal to the original limit. If the limit of the derivatives is also an indeterminate form, we may need to apply L'Hôpital's Rule again. The power of L'Hôpital's Rule lies in its ability to transform complex limit problems into simpler ones. By repeatedly applying the rule, we can often simplify the expression until the limit can be evaluated directly.

Calculating the Derivatives

The derivative of $f(x) = (1-x)^2$ can be found using the chain rule:

$f'(x) = 2(1-x)(-1) = -2(1-x)$

For $g(x) = x^2 - x - x \ln x$, we need to apply the product rule to the term $x \ln x$:

$g'(x) = 2x - 1 - (1 \cdot \ln x + x \cdot \frac{1}{x}) = 2x - 1 - \ln x - 1 = 2x - 2 - \ln x$

The process of finding derivatives is a fundamental skill in calculus, and it's essential for applying L'Hôpital's Rule correctly. The chain rule is crucial for differentiating composite functions, while the product rule is necessary when dealing with the product of two functions. In this case, we had to apply both rules to find the derivatives of f(x) and g(x). A common mistake is to forget the chain rule when differentiating something like $(1-x)^2$. It's also important to be careful with the signs and to double-check your work. Once we have the derivatives, we can substitute them into L'Hôpital's Rule and see if the limit becomes easier to evaluate. If the resulting limit is still an indeterminate form, we may need to differentiate again. This iterative process can sometimes be necessary to fully resolve the limit. Accuracy in calculating derivatives is paramount, as any error here will propagate through the rest of the solution.

Applying L'Hôpital's Rule (First Application)

Now we apply L'Hôpital's Rule:

$\lim_{x \rightarrow 1} \frac{(1-x)^2}{x^2-x-x \ln x} = \lim_{x \rightarrow 1} \frac{-2(1-x)}{2x-2-\ln x}$

Substituting x = 1 into the new expression, we get $\frac{-2(1-1)}{2(1)-2-\ln 1} = \frac{0}{0}$, which is still an indeterminate form. This means we need to apply L'Hôpital's Rule again. The fact that we still have an indeterminate form after the first application of L'Hôpital's Rule is not uncommon. It simply means that the rates at which the numerator and denominator are approaching zero (or infinity) are still too similar to determine the limit directly. This is where the power of L'Hôpital's Rule truly shines – it allows us to repeatedly differentiate until the indeterminate form is resolved. Each application of the rule simplifies the expression further, bringing us closer to a form where the limit can be evaluated directly. It's like peeling back layers of an onion, gradually revealing the core value. The key is to be patient and persistent, and to carefully check your derivatives at each step. Sometimes, multiple applications of L'Hôpital's Rule are necessary to fully unravel the limit.

Applying L'Hôpital's Rule (Second Application)

Let's find the second derivatives:

$f''(x) = \frac{d}{dx} [-2(1-x)] = 2$

$g''(x) = \frac{d}{dx} [2x-2-\ln x] = 2 - \frac{1}{x}$

Applying L'Hôpital's Rule again:

$\lim_{x \rightarrow 1} \frac{-2(1-x)}{2x-2-\ln x} = \lim_{x \rightarrow 1} \frac{2}{2-\frac{1}{x}}$

Now, substituting x = 1, we get $\frac{2}{2-\frac{1}{1}} = \frac{2}{1} = 2$. Therefore, the limit is 2.

The second application of L'Hôpital's Rule has finally yielded a determinate form. By differentiating the numerator and denominator again, we have simplified the expression to the point where we can directly substitute x = 1 and obtain the limit. This highlights the iterative nature of L'Hôpital's Rule and its ability to resolve complex indeterminate forms. It's important to note that each application of the rule is valid only if the previous limit was an indeterminate form. Applying the rule unnecessarily can lead to incorrect results. The key is to carefully assess the form of the limit at each step and apply L'Hôpital's Rule only when it is appropriate. In this case, the second derivatives were much simpler than the first derivatives, making the final evaluation straightforward. The successful application of L'Hôpital's Rule demonstrates its effectiveness in solving challenging limit problems. It's a powerful tool in the calculus arsenal, and mastering its use is essential for anyone working with limits and derivatives.

Conclusion: The Limit Evaluated

Through the application of L'Hôpital's Rule, we have successfully determined that $\lim _{x \rightarrow 1} \frac{(1-x)^2}{x^2-x-x \ln x} = 2$. This journey involved identifying the indeterminate form, calculating derivatives, and applying the rule iteratively until a determinate form was achieved. This problem serves as a testament to the power and elegance of calculus in resolving seemingly complex mathematical challenges. The process of solving this limit problem has provided valuable insights into the application of L'Hôpital's Rule and the importance of careful and methodical problem-solving. By breaking down the problem into smaller steps, we were able to navigate the complexities and arrive at the correct answer. The techniques and concepts learned in this exercise can be applied to a wide range of limit problems, making this a valuable learning experience. Understanding limits is crucial for further studies in calculus and related fields, and mastering techniques like L'Hôpital's Rule is essential for success. The journey of solving this limit problem underscores the beauty and power of mathematics in unraveling the intricacies of the world around us. The final result, 2, represents the value that the function approaches as x gets closer and closer to 1. This understanding of function behavior is fundamental to many applications in science and engineering.