Are you looking for a way to tackle tricky limits, specifically those that involve L’Hôpital’s Rule? This post dives into a practical example of finding the limit using L’Hôpital’s Rule Limit, focusing on the function x ln(1 + 1/x). We’ll walk through the process step-by-step, making it easy to understand.

Initially, the limit presents a seemingly complex scenario. However, we’ll show you how to transform the problem into a more manageable form. By applying L’Hôpital’s Rule Limit, we can effectively evaluate the limit. This method proves incredibly useful when dealing with indeterminate forms, like the one encountered in this example. Therefore, understanding this technique is crucial for success in calculus.

“Success is not final, failure is not fatal: it is the courage to continue that counts” – Winston Churchill

Evaluating a Limit Using L’Hôpital’s Rule

This example demonstrates the application of L’Hôpital’s rule to find the limit of a function as ##x## approaches infinity.

Problem Statement

Find the limit:

### \lim_{x \to \infty} \left( x \cdot \ln\left(1 + \frac{1}{x}\right) \right) ###

This problem involves a limit that is in an indeterminate form of the type ##\infty \cdot 0##. L’Hôpital’s rule is a powerful tool for evaluating such limits.

Solution

Understanding the Problem

The given limit has the indeterminate form ##\infty \cdot 0##. To apply L’Hôpital’s rule, we need to rewrite the expression in a form where the limit is in the indeterminate form ##\frac{0}{0}## or ##\frac{\infty}{\infty}##.

Solving the Problem

We rewrite the expression as:

### L = \lim_{x \to \infty} x \ln\left(1 + \frac{1}{x}\right) = \lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{\frac{1}{x}} ###

Now, as ##x## approaches infinity, the numerator approaches ##\ln(1) = 0## and the denominator approaches 0. This is in the indeterminate form ##\frac{0}{0}##, so we can apply L’Hôpital’s rule.

Applying L’Hôpital’s Rule

Differentiating the numerator and denominator with respect to ##x##, we get:

### L = \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{1}{x}}}{1} ###

Simplifying the expression:

### L = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} ###

Evaluating the Limit

As ##x## approaches infinity, ##\frac{1}{x}## approaches 0. Thus:

### L = \frac{1}{1 + 0} = 1 ###

Final Solution

Therefore, the limit is:

##1##

This example clearly demonstrates the use of L’Hôpital’s rule to evaluate a limit involving logarithms and rational functions. This technique is frequently employed in Calculus to evaluate limits that are in indeterminate forms.

Problem Type	Equation/Expression	Method/Concept
Limit Evaluation	### \lim_{x \to \infty} \left( x \cdot \ln\left(1 + \frac{1}{x}\right) \right) ###	L’Hôpital’s Rule, Limit at Infinity
Indeterminate Form	##\infty \cdot 0##, then rewritten as ##\frac{0}{0}##	Rewrite the limit to fit L’Hôpital’s Rule format
L’Hôpital’s Rule Application	### L = \lim_{x \to \infty} \frac{\ln\left(1 + \frac{1}{x}\right)}{\frac{1}{x}} ###	Applying L’Hôpital’s Rule to evaluate the limit
Differentiation	### \frac{d}{dx} (\ln(1 + \frac{1}{x})) = \frac{1}{1 + \frac{1}{x}} \cdot (-\frac{1}{x^2}) ###, ### \frac{d}{dx} (\frac{1}{x}) = -\frac{1}{x^2} ###	Differentiating the numerator and denominator
Simplification	### L = \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{1}{x}}}{1} ###, ### L = \lim_{x \to \infty} \frac{1}{1 + \frac{1}{x}} ###	Simplifying the expression after applying L’Hôpital’s Rule
Limit Evaluation (Final)	### L = \frac{1}{1 + 0} = 1 ###	Evaluating the limit as ##x## approaches infinity
Result	1	Final Answer

This example provides a practical demonstration of L’Hôpital’s Rule Limit, a powerful technique for evaluating limits that result in indeterminate forms like ∞ ⋅ 0 or 0/0. By rewriting the limit in a suitable form, we can apply L’Hôpital’s rule and find the solution efficiently.

Understanding the underlying principles behind L’Hôpital’s Rule is crucial. The rule states that if the limit of the ratio of two functions results in an indeterminate form (0/0 or ∞/∞), then the limit of the ratio is equal to the limit of the ratio of their derivatives. This allows us to transform a seemingly complex limit problem into a simpler one, facilitating its solution.

• Indeterminate Forms: Recognizing indeterminate forms like ∞ ⋅ 0, 0/0, and ∞/∞ is the first step in applying L’Hôpital’s Rule Limit. The example showcased how to manipulate the limit to achieve a suitable indeterminate form.
• Derivative Application: The core of L’Hôpital’s rule involves finding the derivatives of the numerator and denominator functions. This step requires a strong understanding of differentiation techniques.
• Repeated Application (if necessary): In some cases, applying L’Hôpital’s rule multiple times might be needed to obtain a determinate form. This is a common occurrence, especially with more complex functions.
• Limit Evaluation: Once the derivatives are found, evaluating the limit of the ratio of the derivatives provides the final solution.
• Limit Evaluation: Once the derivatives are found, evaluating the limit of the ratio of the derivatives provides the final solution.

L’Hôpital’s Rule Limit is a valuable tool for tackling a wide range of limit problems in calculus. By mastering this technique, students can efficiently evaluate challenging limits and gain a deeper understanding of the underlying concepts of calculus.

This example serves as a foundational exercise in applying L’Hôpital’s Rule. Practice with various examples will further enhance your understanding and proficiency in using this crucial mathematical tool.