Problem 52 - Thomas Calculus

Question

Find the limits in Exercises \(51-66\) $$ \lim _{x \rightarrow 1^{+}} x^{1 /(x-1)} $$

Step-by-Step Solution

Answer

The limit is \( e \).

Step 1: Understand the Expression

The given limit is \( \lim _{x \rightarrow 1^{+}} x^{1 /(x-1)} \). This implies that \( x \) approaches 1 from the right. Hence, \( x > 1 \).

Step 2: Rewrite the Expression in Logarithmic Form

To solve this limit, first take the natural logarithm of the expression. Let \( y = x^{1/(x-1)} \). Then \( \ln y = \frac{1}{x-1} \ln x \). The limit now becomes \( \lim_{x \rightarrow 1^+} \ln y = \lim_{x \rightarrow 1^+} \frac{\ln x}{x-1} \).

Step 3: Apply L'Hôpital's Rule

The limit \( \lim_{x \rightarrow 1^+} \frac{\ln x}{x-1} \) is in the indeterminate form \( \frac{0}{0} \). Apply L'Hôpital's Rule to differentiate the numerator and the denominator: \( \lim_{x \rightarrow 1^+} \frac{\ln x}{x-1} = \lim_{x \rightarrow 1^+} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx}(x-1)} = \lim_{x \rightarrow 1^+} \frac{1/x}{1} = \lim_{x \rightarrow 1^+} \frac{1}{x}. \)

Step 4: Evaluate the Simplified Limit

Now, evaluate \( \lim_{x \rightarrow 1^+} \frac{1}{x} = \frac{1}{1} = 1 \). Thus, \( \lim_{x \rightarrow 1^+} \ln y = 1 \).

Step 5: Solve for Original Expression

Since \( \lim_{x \rightarrow 1^+} \ln y = 1 \), it follows that \( \ln y = 1 \Rightarrow y = e^1 = e \).

Step 6: Conclude the Problem

Conclude that the original limit \( \lim_{x \rightarrow 1^{+}} x^{1/(x-1)} = e \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is an essential tool in calculus for evaluating limits that result in indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule helps you differentiate both the numerator and the denominator separately and then take the limit again. It simplifies the process of solving limits by converting a complex expression into a simpler one.For mathematicians and students alike, it provides a systematic approach to solve problems quickly without tedious algebra. When you encounter a limit problem, first check if it's in an indeterminate form. If it is, check the conditions under which L'Hôpital's Rule can be applied:

• The limit must initially result in an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• The functions in the numerator and denominator must be differentiable at the point of interest.
• The limit of their derivatives must exist or be evaluated.

In the discussed problem, the rule was applied to the expression \( \frac{\ln x}{x-1} \) to convert it into \( \frac{1/x}{1} \), which is much easier to evaluate.

Natural Logarithm

The natural logarithm, denoted as \( \ln x \), is the logarithm with the base \( e \), where \( e \) is approximately equal to 2.718. It is a fundamental mathematical function showing up frequently in calculus and real-world applications. The natural logarithm is particularly useful in modeling exponential growth or decay and in solving various calculus problems.In the exercise, the natural logarithm is used to transform the initial complex limit into a manageable form. Often when dealing with exponents, taking logarithms makes the mathematics far simpler. The conversion \( y = x^{1/(x-1)} \) into \( \ln y = \frac{1}{x-1}\ln x \) allows the problematic limit to become something more approachable: \( \lim_{x \rightarrow 1^+} \frac{\ln x}{x-1} \). Hence, this transformation is a common technique to streamline challenging problems.

Indeterminate Forms

Indeterminate forms are expressions that do not initially provide enough information to determine a limit. Common types are \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \cdot \infty \), \( \infty - \infty \), \( 0^0 \), \( 1^\infty \), and \( \infty^0 \). These forms appear when evaluating limits at certain points leads to ambiguous results.In the problem at hand, when approaching the limit \( \lim_{x \rightarrow 1^+} \frac{\ln x}{x-1} \), the result initially appears as \( \frac{0}{0} \). Since direct substitution does not resolve the limit, recognizing it as an indeterminate form is necessary to decide to utilize L'Hôpital's Rule. Identifying indeterminate forms allows mathematicians to apply special strategies like L'Hôpital's Rule, Taylor series, or algebraic simplification to find the limits.

Calculus Problem Solving

Solving calculus problems often requires a deep understanding of concepts like derivatives, integrals, and limits. It is crucial to approach problems systematically:

• Identify what you need to find. In limit problems, confirm if it involves infinite behavior or finite approaches.
• Check if the expression leads to indeterminate forms; apply strategies like L'Hôpital's Rule if needed.
• Use algebraic transformations to simplify the expression, like converting exponentials into logarithms.
• Evaluate the limit using the simplified form or additional techniques (e.g., numerical estimation if required).

Focus on understanding each method and why it is applied, as this will guide you in finding the solution smoothly. In our exercise, rewriting the expression using logarithms and applying L'Hôpital's Rule was integral to reaching the conclusion that the limit \( \lim_{x \rightarrow 1^{+}} x^{1/(x-1)} = e \). Such structured steps are valuable not only for obtaining the right answer but also for enhancing overall calculus comprehension.