Evaluate The Limit: Lim (x→a) [e^x Ln(a) - E^a Ln(x)] / (x^2 - A^2), Where A Is A Strictly Positive Real Number.

Jun 17, 2025 by ADMIN 113 views

$Exploring the Limit of a Function: A Deep Dive into ${\lim_{x \to a} \frac{e^x \ln(a) - e^a \ln(x)}{x^2 - a^2}}$$

Introduction

In the realm of calculus, evaluating limits is a fundamental skill. Limits help us understand the behavior of functions as their input approaches a particular value. This article delves into the intricate evaluation of a specific limit: ${\lim_{x \to a} \frac{e^x \ln(a) - e^a \ln(x)}{x^2 - a^2}}$ , where a is a strictly positive real number. This limit is not straightforward and requires careful application of techniques such as L'Hôpital's Rule. Understanding this limit provides valuable insights into the interplay between exponential and logarithmic functions, and how they behave near specific points.

When we first encounter this limit, a direct substitution of x = a leads to an indeterminate form of 0/0, since both the numerator and the denominator become zero. This indeterminacy signals that we need a more sophisticated approach. L'Hôpital's Rule, a powerful tool in calculus, is designed to handle such situations. L'Hôpital's Rule allows us to take the derivatives of the numerator and the denominator separately and then re-evaluate the limit. This process can simplify the expression and reveal the true behavior of the function as x approaches a. The application of L'Hôpital's Rule isn't merely a mechanical step; it requires a deep understanding of the conditions under which it can be applied and the implications of the resulting limit. Furthermore, the problem underscores the importance of recognizing indeterminate forms and the various techniques available to resolve them, such as factoring, algebraic manipulation, and, as in this case, L'Hôpital's Rule.

Navigating through such complex limits enhances our analytical skills and sharpens our understanding of how different types of functions interact. This exploration isn't just about finding a numerical answer; it's about gaining a more profound appreciation for the underlying concepts of calculus and their applications in various mathematical and scientific contexts. Understanding the behavior of this limit not only solidifies our grasp on calculus but also builds a strong foundation for tackling more advanced problems in mathematical analysis. The steps involved in solving this limit—recognizing the indeterminate form, applying L'Hôpital's Rule, and simplifying the resulting expression—are essential techniques that are frequently used in various areas of mathematics and engineering.

Problem Statement

We are tasked with evaluating the limit: ${\lim_{x \to a} \frac{e^x \ln(a) - e^a \ln(x)}{x^2 - a^2}}$ , where a is a strictly positive real number. This problem is intriguing because it combines exponential and logarithmic functions in a way that initially results in an indeterminate form. The presence of both e^x and ln(x) complicates the evaluation, necessitating the use of advanced techniques such as L'Hôpital's Rule. The limit's structure hints at the delicate balance between exponential growth and logarithmic decay as x approaches a, making it an excellent example for illustrating the power of calculus in resolving such complexities.

To solve this, our initial step is to recognize the indeterminate form. When x approaches a, the expression becomes:

Numerator: e^a ln(a) - e^a ln(a) = 0

Denominator: a^2 - a^2 = 0

This 0/0 form indicates that we cannot directly substitute x = a to find the limit. Instead, we must employ a method that allows us to "break" the indeterminacy. This is where L'Hôpital's Rule comes into play. This rule is a cornerstone of advanced calculus, enabling us to handle limits of quotients where both the numerator and denominator approach zero or infinity. However, before we can apply the rule, it is crucial to ensure that all the necessary conditions are met. The functions in the numerator and denominator must be differentiable, and the limit of their derivatives must exist. Meeting these conditions guarantees that applying L'Hôpital's Rule will lead us to a valid solution.

The challenge of this problem lies not only in the application of L'Hôpital's Rule but also in the subsequent simplification of the resulting expression. After differentiating the numerator and denominator, we obtain a new limit that might still appear complex. Successfully navigating this stage requires a strong grasp of differentiation techniques and algebraic manipulation. The ability to simplify the expression effectively is key to ultimately finding the limit's value. This process underscores the interconnectedness of different mathematical concepts and the importance of a holistic approach to problem-solving in calculus.

Applying L'Hôpital's Rule

Since we have an indeterminate form of 0/0, we can apply L'Hôpital's Rule. This rule states that if ${\lim_{x \to c} \frac{f(x)}{g(x)}}$ is of the form 0/0 or ∞/∞, then:

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

provided the limit on the right side exists.

In our case, let f(x) = e^x ln(a) - e^a ln(x) and g(x) = x^2 - a^2. We need to find the derivatives f'(x) and g'(x).

The derivative of f(x) with respect to x is:

${f'(x) = \frac{d}{dx} [e^x \ln(a) - e^a \ln(x)] = e^x \ln(a) - e^a \frac{1}{x}}$

Here, the derivative of e^x ln(a) is e^x ln(a) because ln(a) is a constant, and the derivative of e^a ln(x) is e^a/x because e^a is a constant. This step highlights the importance of recognizing constant terms and applying the appropriate differentiation rules. The ability to correctly differentiate such expressions is crucial for the accurate application of L'Hôpital's Rule.

The derivative of g(x) with respect to x is:

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

This derivative is straightforward, utilizing the power rule of differentiation. The simplicity of this derivative contrasts with the slightly more complex derivative of f(x), emphasizing the need for a flexible approach to differentiation.

Now, we apply L'Hôpital's Rule:

${\lim_{x \to a} \frac{e^x \ln(a) - e^a \ln(x)}{x^2 - a^2} = \lim_{x \to a} \frac{e^x \ln(a) - e^a / x}{2x}}$

This step transforms the original limit into a new limit involving the derivatives. However, directly substituting x = a into this new limit still yields an indeterminate form, albeit one that is slightly different from the original. This situation often occurs when applying L'Hôpital's Rule, and it may be necessary to apply the rule multiple times to fully resolve the indeterminacy. The persistence of the indeterminate form underscores the complexity of the problem and the need for a systematic approach.

Evaluating the New Limit

If we substitute x = a into the new limit, we get:

${\frac{e^a \ln(a) - e^a / a}{2a}}$

This expression is not immediately revealing, and we still need to simplify it further. To do this, we can try to combine the terms in the numerator by finding a common denominator. This algebraic manipulation is a crucial step in making the limit more manageable and revealing its true behavior.

Let's rewrite the numerator:

${e^a \ln(a) - \frac{e^a}{a} = \frac{a e^a \ln(a) - e^a}{a}}$

Now, substitute this back into the limit:

${\lim_{x \to a} \frac{\frac{a e^a \ln(a) - e^a}{a}}{2a} = \frac{a e^a \ln(a) - e^a}{2a^2}}$

This expression is now in a more simplified form. However, to ensure we have correctly evaluated the limit, it's essential to double-check our steps and consider whether any further simplifications are possible. This critical evaluation is a hallmark of rigorous mathematical problem-solving.

We can factor out e^a from the numerator:

${\frac{e^a (a \ln(a) - 1)}{2a^2}}$

This is the simplified form of the limit. It represents the final value of the limit as x approaches a. The expression highlights the interplay between the exponential function e^a, the logarithmic function ln(a), and the algebraic term a. The fact that the limit is expressed in terms of a underscores that its value depends directly on the specific value of a. This result is not just a numerical answer; it's a function of a, providing a more complete understanding of how the original limit behaves for different values of a.

Final Result

The limit ${\lim_{x \to a} \frac{e^x \ln(a) - e^a \ln(x)}{x^2 - a^2}}$ evaluates to:

${\frac{e^a (a \ln(a) - 1)}{2a^2}}$

This final result encapsulates the entire process of evaluating the limit. It shows how the initial indeterminate form was resolved through the application of L'Hôpital's Rule, followed by careful algebraic manipulation and simplification. The result is not just a number; it's an expression that depends on the value of a, providing a comprehensive understanding of the limit's behavior for various values of a. The expression highlights the intricate relationship between exponential and logarithmic functions and the role of calculus in unraveling their complexities.

The process of finding this limit involved several key steps, each requiring a specific set of skills and techniques. Recognizing the indeterminate form was the crucial first step, signaling the need for a more advanced method. Applying L'Hôpital's Rule allowed us to transform the limit into a more manageable form. The subsequent algebraic simplification was essential in revealing the final result. This entire process underscores the importance of a multifaceted approach to problem-solving in calculus, combining analytical skills with algebraic proficiency. Moreover, the final result is a testament to the power of calculus in handling complex mathematical expressions and providing insights into the behavior of functions near specific points.

Conclusion

In conclusion, the evaluation of the limit ${\lim_{x \to a} \frac{e^x \ln(a) - e^a \ln(x)}{x^2 - a^2}}$ demonstrates the power and elegance of calculus techniques. Starting from an indeterminate form, we successfully applied L'Hôpital's Rule and algebraic manipulation to arrive at the final result: ${\frac{e^a (a \ln(a) - 1)}{2a^2}}$ . This process highlights the importance of recognizing indeterminate forms, applying appropriate rules, and simplifying expressions to uncover the underlying behavior of functions.

The journey through this problem is not just about finding a solution; it’s about understanding the process. Each step, from identifying the indeterminate form to applying L'Hôpital's Rule and simplifying the expression, contributes to a deeper understanding of calculus concepts. The final result is a testament to the effectiveness of these techniques and the power of mathematical analysis. This exploration serves as a valuable exercise in honing problem-solving skills and reinforcing fundamental principles of calculus. Furthermore, the result itself offers insights into how exponential and logarithmic functions interact, emphasizing the importance of such functions in mathematical modeling and analysis.

The skills acquired in solving this limit problem are transferable to a wide range of mathematical and scientific contexts. The ability to handle indeterminate forms, apply differentiation rules, and simplify complex expressions are essential tools in various fields, including physics, engineering, and economics. This underscores the practical significance of mastering calculus techniques and the value of engaging with challenging problems that push the boundaries of our mathematical understanding. The satisfaction of arriving at a solution through careful and methodical application of mathematical principles is a reward in itself, encouraging further exploration and discovery in the world of mathematics.