Problem 16 - Calculus - AP Edition

Question

Find the limits. $$\lim _{x \rightarrow 0^{+}} \frac{1-\ln x}{e^{1 / x}}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the Function

We are given the function \( f(x) = \frac{1-\ln x}{e^{1/x}} \). Our task is to find the limit as \( x \to 0^{+} \).

Step 2: Explore the Behavior of Numerator

As \( x \to 0^{+} \), \( \ln(x) \to -\infty \). Therefore, the numerator \( 1 - \ln(x) \to \infty \).

Step 3: Explore the Behavior of Denominator

As \( x \to 0^{+} \), \( \frac{1}{x} \to \infty \), leading \( e^{1/x} \to \infty \). So the denominator becomes infinitely large as well.

Step 4: Determine Type of Indeterminate Form

Both the numerator and denominator approach infinity, indicating an indeterminate form of type \( \frac{\infty}{\infty} \). This suggests using L'Hôpital's Rule could be helpful.

Step 5: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if \( \lim_{x \to a} \frac{f(x)}{g(x)} \) is of the form \( \frac{\infty}{\infty} \), then \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]. First, we find the derivatives: \( f'(x) = \frac{-1}{x} \) and \( g'(x) = -\frac{e^{1/x}}{x^2} \).

Step 6: Simplify the New Expression

Now, substitute back the derivatives: \[ \lim_{x \to 0^{+}} \frac{-\frac{1}{x}}{-\frac{e^{1/x}}{x^2}} = \lim_{x \to 0^{+}} \frac{x^2}{x e^{1/x}} = \lim_{x \to 0^{+}} xe^{-1/x}. \]

Step 7: Evaluate the Simplified Limit

Let \( u = \frac{1}{x} \), then as \( x \to 0^{+} \), \( u \to \infty \). The expression becomes \( \lim_{u \to \infty} \frac{1}{u} e^{-u} \). As \( u \to \infty \), \( \frac{1}{u} \to 0 \) faster than \( e^{-u} \) which also approaches 0, so the overall limit is \( 0 \).

Step 8: Conclusion

Thus, the limit of the original function is \( 0 \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus, designed to help evaluate limits of indeterminate forms, specifically when they take the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When confronted with such forms, L'Hôpital's Rule allows us to differentiate the numerator and the denominator separately and then take the limit of their quotient instead.
In this exercise, we encounter an indeterminate form of type \( \frac{\infty}{\infty} \), indicating that both the numerator and the denominator are approaching infinity as \( x \to 0^+ \). By using L'Hôpital's Rule, we simplify the process to evaluate the limit by finding the derivatives:

• The derivative of \( 1 - \ln x \) is \( -\frac{1}{x} \).
• The derivative of \( e^{1/x} \) is \( -\frac{e^{1/x}}{x^2} \).

Substituting these derivatives back into our limit expression enables us to simplify complex functions and reach the solution efficiently.

Indeterminate Forms

Indeterminate forms arise in calculus when the limit of a function does not initially appear to provide enough information to determine a clear answer. These forms, like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0 \times \infty \), and others, can often result in confusion about how to proceed.
In this exercise, the form \( \frac{\infty}{\infty} \) means both the numerator and denominator are becoming infinitely large as \( x \to 0^{+} \). This doesn't immediately resolve to a single number, so additional analysis is required.
L'Hôpital's Rule is particularly useful in these situations as it provides a method to delve further into the problem. By examining the derivatives of the numerator and denominator, you can often transform the indeterminate form into one that can be evaluated directly or with other limit techniques. Thus, understanding and identifying indeterminate forms is crucial for solving complex limit problems.

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to a variable exponent, typically written as \( e^x \). In this exercise, the component \( e^{1/x} \) represents an exponential function where the exponent \( 1/x \) changes as \( x \to 0^+ \).
Understanding how exponential functions behave is vital. As \( x \) approaches zero from the positive side, \( 1/x \) becomes very large, causing \( e^{1/x} \) to increase dramatically towards infinity.
Exponential functions are crucial because they model a variety of real-world behaviors such as population growth, radioactive decay, and, here, the rate of change of another variable. They often appear in calculus problems due to their unique properties and behavior under limits. In this context, understanding how the exponential function \( e^{1/x} \) behaves directly influences the outcome of applying L'Hôpital's Rule to find the limit of the original problem.