Problem 52 - University Calculus: Early Transcendentals

Question

Find the limits. $$\lim _{x \rightarrow 1^{+}} x^{1 /(x-1)}$$

Step-by-Step Solution

Answer

The limit is \(e\).

Step 1: Substitute for Simplification

To handle this limit, recognize the form it takes when we substitute values close to 1. Direct substitution gives us \(x^{1/(x-1)} = 1^{1/(1-1)} = 1^{1/0}\), which is an indeterminate form.

Step 2: Use the Natural Logarithm

Introduce a logarithm to simplify the expression. Let \(y = x^{1/(x-1)}\), then take the natural logarithm on both sides: \(\ln(y) = \frac{\ln(x)}{x-1}\). Our task now transforms into computing \(\lim_{x \rightarrow 1^{+}} \ln(y)\).

Step 3: Transform to Evaluate the Limit

The expression \(\frac{\ln(x)}{x-1}\) is another indeterminate form \(\frac{0}{0}\) as \(x \rightarrow 1^{+}\). Apply L'Hôpital's Rule, which is applicable for indeterminacies like \(\frac{0}{0}\).

Step 4: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator independently: the derivative of \(\ln(x)\) is \(\frac{1}{x}\), and the derivative of \(x-1\) is 1. L'Hôpital's gives us \(\lim_{x \rightarrow 1^{+}} \frac{1/x}{1} = \lim_{x \rightarrow 1^{+}} \frac{1}{x}\).

Step 5: Compute the Result of the Limit

Evaluate the limit: \(\lim_{x \rightarrow 1^{+}} \frac{1}{x} = \frac{1}{1} = 1\). Thus, \(\ln(y) = 1\).

Step 6: Convert Back from the Logarithm

Since \(\ln(y) = 1\), solve for \(y\) by exponentiating both sides. Therefore, \(y = e\). This means \(\lim_{x \rightarrow 1^{+}} x^{1/(x-1)} = e\).

Key Concepts

Limit of a function

When we talk about the limit of a function, we're discussing the behavior of that function as its input approaches a particular point. Here, if we're given the expression \( \lim_{x \rightarrow 1^{+}} x^{1/(x-1)} \), our goal is to understand what happens to this expression as \( x \) approaches 1 from the right. To compute a limit, you might try direct substitution. However, this can sometimes lead to undefined or indeterminate expressions, like \( 1^{1/0} \) in this problem. Therefore, it's essential to use other mathematical techniques, such as natural logarithms and more sophisticated tools like L'Hôpital's Rule, to manage these complications and unveil the true behavior of the function near the point of interest. Limits form the foundation for understanding continuity, derivatives, and integrals in calculus.

The primary idea is to understand the destination of the function as \( x \) closes in on the chosen value, rather than what happens exactly at that point. It helps predict the pattern and behavior of functions, making it a core concept of calculus.

Indeterminate forms

Indeterminate forms are expressions that don't have a clear numerical value in the limit process. Commonly encountered types include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), 0⋅∞, ∞−∞, 1⁰⁰⁰, and 0⁰. These forms arise due to the limitations in direct substitution, creating ambiguity in evaluating limits. In the original exercise, you encountered \( 1^{1/0} \) which is an indeterminate form. Indeterminate forms require special methods to simplify them, such as logarithmic transformations or advanced rules like L'Hôpital's Rule.

Using L'Hôpital's Rule is a common technique for resolving indeterminate forms of \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It involves differentiating the numerator and denominator separately and finding the limit of the resulting equation. By resolving these forms, we gain clearer insights into the function's behavior near specific points.

Natural logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm with base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. It's especially useful in calculus due to its simple derivative: \( \frac{d}{dx} \ln(x) = \frac{1}{x} \). In the context of limits and indeterminate forms, the natural logarithm provides a powerful tool to simplify complex expressions.

In the given exercise, transforming \( x^{1/(x-1)} \) into a more manageable form using \( \ln(y) = \frac{\ln(x)}{x-1} \) was crucial. This transformation allowed the application of L'Hôpital's Rule by converting the complex power expression into a fraction whose limit could be more easily evaluated. By working with the logarithmic form, complex exponentials become simpler linear forms, aiding in easier differentiation and limit assessment.
Ultimately, using the properties of natural logarithms helps us solve intricate calculus problems more systematically and efficiently.