Problem 52 - Calculus

Question

Compute \(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\)

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Indeterminate Form

First, substitute \( x = 0 \) into the function to check if it results in an indeterminate form. Substituting gives \( \frac{e^0 - 1}{0} = \frac{0}{0} \), which is an indeterminate form. This allows us to use L'Hôpital's Rule.

Step 2: Apply L'Hôpital's Rule

L'Hôpital's Rule states that if the limit produces an indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then \( \lim_{{x \to c}} \frac{f(x)}{g(x)} = \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \) if the latter limit exists. Compute the derivatives of the numerator and the denominator.

Step 3: Differentiate the Numerator and the Denominator

The derivative of the numerator, \( e^x - 1 \), is \( e^x \). The derivative of the denominator, \( x \), is \( 1 \). Thus, the limit \( \lim_{{x \to 0}} \frac{e^{x}-1}{x} = \lim_{{x \to 0}} \frac{e^x}{1} \).

Step 4: Evaluate the New Limit

Evaluate \( \lim_{{x \to 0}} e^x \). Since \( e^x \) is continuous, substituting \( x = 0 \) gives \( e^0 = 1 \).

Step 5: Conclusion

Therefore, the limit \( \lim_{{x \to 0}} \frac{e^x - 1}{x} \) is equal to \( 1 \).

Key Concepts

Understanding Indeterminate Forms

In calculus, when evaluating limits, an expression may sometimes result in what is called an "indeterminate form." These are forms where simple substitution results in ambiguous expressions, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms do not immediately indicate a limit but instead suggest the need for further analysis. Other indeterminate forms include \( 0^0 \), \( \infty - \infty \), and \( 1^\infty \).
To manage these forms, mathematicians implement additional techniques, such as algebraic manipulation or L'Hôpital's Rule, to find meaningful limits. Identifying indeterminate forms is the first step in applying these more advanced tools. Let's take the expression \( \frac{e^x - 1}{x} \) as an example. When substituting \( x = 0 \), this becomes \( \frac{0}{0} \), which is an indeterminate form. Hence, we need to explore other methods to find the limit.

Limits in Calculus: The Basics

Limits are foundational in calculus, providing a way to understand the behavior of functions as they approach certain points. They offer insights into function values when direct evaluation is not possible due to factors like division by zero. The notation \( \lim_{x \to c} f(x) \) denotes the value that function \( f(x) \) approaches as \( x \) gets closer to \( c \).
These concepts come in handy when evaluating functions that seem undefined at a particular point. In our example \( \lim_{x \to 0} \frac{e^x - 1}{x} \), the function isn’t directly evaluable at \( x = 0 \) due to division by zero. By identifying the limit, we can understand the function's behavior near that point. Calculating such limits often involves using approaches like L'Hôpital's Rule, which facilitates the evaluation when indeterminate forms are present.

Differentiation: Derivatives and Their Role

Differentiation is a critical operation in calculus that finds the rate at which a function is changing at any given point. This rate is known as the derivative. In calculus problems involving limits and indeterminate forms, derivatives play a vital role.
L'Hôpital's Rule, for instance, relies on derivatives to evaluate limits of indeterminate forms. If a limit expression results in \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), derivatives enable us to transform the expression to a solvable form. For the function \( \frac{e^x - 1}{x} \), the derivative of the numerator \( e^x - 1 \) is \( e^x \), and the derivative of the denominator \( x \) is 1. Thus, applying L'Hôpital's Rule, the limit \( \lim_{x \to 0} \frac{e^x - 1}{x} \) simplifies to \( \lim_{x \to 0} e^x \), which solves to 1. In this way, derivatives help in evaluating limits and in understanding how functions behave around specific points.