Problem 57 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 54-59, use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. \(\lim _{n \rightarrow \infty}\left(\frac{n-1}{n+1}\right)^{n}\)

Step-by-Step Solution

Answer

The limit is \( e^{-2} \).

Step 1: Examine the Original Limit

We need to find \( \lim _{n \rightarrow \infty}\left(\frac{n-1}{n+1}\right)^{n} \). This is a classic indeterminate form of the type \( \infty^0 \).

Step 2: Use Logarithms to Simplify

Taking the natural logarithm of the expression allows us to simplify the limit: \[ \ln L = \lim_{n \to \infty} n \ln \left( \frac{n-1}{n+1} \right) \].

Step 3: Simplify the Logarithmic Expression

The expression inside the logarithm can be rewritten using common denominator rules: \(\ln \left( \frac{n-1}{n+1} \right) = \ln \left( 1 - \frac{2}{n+1} \right)\). For large \(n\), this becomes approximately \( -\frac{2}{n+1} \).

Step 4: Evaluate the Limit Using L'Hopital's Rule

Now, we have: \[ \ln L = \lim_{n \to \infty} n \cdot \left(-\frac{2}{n+1}\right) = -2 \lim_{n \to \infty} \frac{n}{n+1}.\] This can be simplified using L'Hopital's Rule: \[ -2 \lim_{n \to \infty} \frac{n}{n+1} = -2 \lim_{n \to \infty} \frac{1}{1} = -2. \]

Step 5: Interpret the Logarithmic Limit Result

From \( \ln L = -2 \), we know that \( L = e^{-2} \). This means that the original limit is \( e^{-2} \).

Key Concepts

Indeterminate Forms

When dealing with limits in calculus, you may often encounter a form that doesn't provide an obvious solution, known as an indeterminate form. Some common examples include \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), and powers like \( 0^0 \), \( 1^\infty \), and \( \infty^0 \). Each of these forms requires a different approach to determine the limit.

In the exercise provided, the form is \( \infty^0 \), which arises when trying to find the limit of \( \left(\frac{n-1}{n+1}\right)^{n} \) as \( n \) approaches infinity. This is tricky since direct computation isn't possible. Instead, we use additional mathematical techniques like logarithms or substitutions to transform the expression into a factorial form that is solvable. Understanding indeterminate forms is crucial, as it opens up the toolbox of techniques that include L'Hopital's Rule and the use of logarithmic identities.

L'Hopital's Rule

L'Hopital's Rule is a powerful tool in calculus used to resolve limits that present in indeterminate form, specifically \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule can be applied repeatedly until a determinate form is achieved. L'Hopital's Rule states: If \( \lim_{x \to c} f(x) = 0 \) and \( \lim_{x \to c} g(x) = 0 \), 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 exists.

In our original problem, after logarithmically simplifying the expression, we end up with the limit \( \lim_{n \to \infty} \frac{n}{n+1} \). This expression suggests the use of L'Hopital's Rule due to its indeterminate form \( \frac{\infty}{\infty} \). Differentiating the numerator and the denominator, the limit simplifies to \( \frac{1}{1} \), giving a direct solution for the problem. This demonstrates how L'Hopital's Rule aids in tackling challenging limits.

Exponential Limits

Exponential limits can be daunting due to their exponential nature, often resulting in forms such as \( \infty^0 \) or \( 1^\infty \). These require special handling, frequently involving the application of logarithms. By transforming the exponential expression into a logarithmic one, the problem can be made solvable using differential calculus techniques like L'Hopital's Rule.

In the given exercise, the problem involves evaluating an exponential expression of \( \left(\frac{n-1}{n+1}\right)^{n} \). By taking the natural logarithm, the problem becomes easier to handle: \( \ln L = \lim_{n \to \infty} n \ln \left( \frac{n-1}{n+1} \right) \). This logarithmic form provides a more tractable problem, enabling the use of established techniques to reach a solution.

Mastery of exponential limits involves practicing strategies that bridge between logarithmic and exponential forms.

Natural Logarithms

Natural logarithms are an essential mathematical concept, often used to simplify complex limit problems. The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). It simplifies multiplicative relationships into additive ones, turning powers and roots into products and fractions into sums and differences.

In the problem solved, the natural logarithm transforms the exponential expression \( \left( \frac{n-1}{n+1} \right)^{n} \) into a more manageable form: \( n \ln\left(1 - \frac{2}{n+1}\right) \). This transformation allows application of L'Hopital's Rule and easier evaluation of the limit. Understanding natural logarithms lets you unravel exponential problems and provides insights into scale transformations in calculus.

By incorporating \( \ln \) into problem-solving strategies, handling growth problems, compound interest, and complex limits becomes more straightforward, affording mathematical flexibility and precision.