Question

Evaluate the limits with either L'Hôpital's rule or previously learned methods. $$ \lim _{x \rightarrow \infty}\left(1-\frac{1}{x}\right)^{x} $$

Short answer

The limit is \( \frac{1}{e} \).

Step-by-step solution

Recognize the Indeterminate Form

First, let's recognize the expression \( \lim_{x \rightarrow \infty} \left(1-\frac{1}{x}\right)^{x} \). As \( x \to \infty \), the base \( 1 - \frac{1}{x} \to 1 \) and the exponent \( x \to \infty \). The expression is of the form \( 1^{\infty} \), an indeterminate form.

Transform the Expression Using Logarithms

To evaluate the limit, let's take the natural logarithm: Define \( y = \left(1-\frac{1}{x}\right)^{x} \), then \( \ln y = x \ln\left(1-\frac{1}{x}\right) \). Now, we'll find \( \lim_{x \to \infty} \ln y \).

Apply Expansion for Simplification

Inspect the logarithmic expression \( \ln\left(1-\frac{1}{x}\right) \). For small \( t \), \( \ln (1-t) \approx -t \), so we have \( \ln\left(1-\frac{1}{x}\right) \approx -\frac{1}{x} \) for large \( x \).

Evaluate the Limit of the Logarithmic Expression

Substitute the approximation into the logarithm: \( x \cdot \ln\left(1-\frac{1}{x}\right) \approx x \cdot \left(-\frac{1}{x}\right) \approx -1 \). Hence, \( \lim_{x \to \infty} x \ln\left(1-\frac{1}{x}\right) = -1 \).

Compute the Final Limit

Since \( \lim_{x \to \infty} \ln y = -1 \), it follows \( \lim_{x \to \infty} y = \lim_{x \to \infty} e^{\ln y} = e^{-1} = \frac{1}{e} \). Therefore, \( \lim_{x \to \infty}\left(1-\frac{1}{x}\right)^{x} = \frac{1}{e} \).

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule allows us to differentiate the numerator and the denominator separately to find the limit of a rational function, as long as the derivatives exist and are continuous in a neighborhood around the point of interest. However, it's important to note that in some cases, like the one in the exercise, L'Hôpital's Rule does not apply directly because it is not a rational expression. To use L'Hôpital's Rule, make sure:

• The limit you are evaluating forms an indeterminate type (e.g., \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)).
• The derivatives of the numerator and denominator must exist and be continuous near the point you're evaluating.

If both these conditions are met, you can differentiate the top and bottom until the indeterminate form is resolved or the limit can be evaluated by substitution. Remember, not all indeterminate forms can be solved using this rule, so understanding the context of your problem is key.

Indeterminate Forms

In calculus, indeterminate forms are expressions like \( 0^0 \), \( 1^\infty \), \( \infty - \infty \), and others where direct substitution would make the expression seemingly unsolvable because of their undefined nature. They signal that more work needs to be done to uncover what the actual limit might be, often requiring transformation or re-evaluation. For example, if you directly plug infinity into the expression \( \left(1-\frac{1}{x}\right)^{x} \), you would encounter an indeterminate form of \( 1^\infty \). This is indeterminate because the base approaches 1 while the exponent skyrockets to infinity, creating a tension between stagnation and explosive growth.To resolve an indeterminate form, you might need to:

• Rewrite the expression using known algebraic identities.
• Utilize logarithmic transformations, as was done in the solution to the original exercise, to bring the expression into a form that can be simplified and solved.
• Apply L'Hôpital's Rule where appropriate, though not for all types of indeterminate forms.

Recognizing these forms early on helps in choosing the best strategy for further evaluation.

Limits

Limits are a fundamental concept in calculus, acting as the building blocks for understanding continuity, derivatives, and integrals. A limit seeks to find what a function's value approaches as the input gets closer to a certain point, often leading to more insight into the behavior of complex functions.In cases like \( \lim_{x \rightarrow \infty} \left(1-\frac{1}{x}\right)^{x} \), the limit examines how the expression behaves as \( x \) grows indefinitely. To solve this, we often manipulate the expression to facilitate easier evaluation, either through simplifying techniques, transformations, or using rules like L'Hôpital's.Here are some key points about limits:

• Limits can describe behavior as \( x \) approaches finite values, infinity, or negative infinity.
• They help in defining derivatives and integrals, which are cornerstone concepts in calculus and analysis.
• Understanding limits improves problem-solving techniques and enhances our comprehension of function behaviors at critical points.

Mastering limits involves practice and familiarity with a variety of approaches to handle different scenarios, ultimately strengthening your grasp on calculus as a whole.