Problem 13 - Essential Calculus Early Transcendentals

Question

\(1-38=\) Find the limit. Use l'Hospital's Rule where appropriate. If there is a more elementary method, consider using it. If l'Hospital's Rule doesn't apply, explain why. $$\lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}}$$

Step-by-Step Solution

Answer

The limit is \( \frac{1}{2} \).

Step 1: Determine the Limit Form

First, substitute \( x = 0 \) into the function to check the form:\[\lim _{x \rightarrow 0} \frac{e^{x}-1-x}{x^{2}} = \frac{e^{0}-1-0}{0^{2}} = \frac{0}{0}\]This is an indeterminate form \( \frac{0}{0} \), so we can use l'Hospital's Rule.

Step 2: Apply l'Hospital's Rule - First Derivative

Since the limit gives an indeterminate form, apply l'Hospital's Rule by differentiating the numerator and the denominator.The first derivative of the numerator \( e^x - 1 - x \) is \( e^x - 1 \), and the derivative of the denominator \( x^2 \) is \( 2x \):\[\lim _{x \rightarrow 0} \frac{e^{x} - 1 - x}{x^{2}} = \lim _{x \rightarrow 0} \frac{e^{x} - 1}{2x}\]

Step 3: Evaluate the Limit - Check Form Again

Substitute \( x = 0 \) into the new expression:\[\lim _{x \rightarrow 0} \frac{e^{x} - 1}{2x} = \frac{e^{0} - 1}{2 \cdot 0} = \frac{0}{0}\]This is still \( \frac{0}{0} \), so we apply l'Hospital's Rule again.

Step 4: Apply l'Hospital's Rule - Second Derivative

Differentiate the numerator \( e^x - 1 \) to get \( e^x \), and differentiate the denominator \( 2x \) to get \( 2 \):\[\lim _{x \rightarrow 0} \frac{e^{x} - 1}{2x} = \lim _{x \rightarrow 0} \frac{e^{x}}{2}\]This simplifies further upon evaluating at \( x = 0 \).

Step 5: Evaluate the Limit

Substitute \( x = 0 \) again:\[\lim _{x \rightarrow 0} \frac{e^{x}}{2} = \frac{e^{0}}{2} = \frac{1}{2}\]So the limit is \( \frac{1}{2} \).

Key Concepts

Indeterminate Form

When you're dealing with limits in calculus, you often encounter situations that seem confusing at first. One of these is an 'indeterminate form.' The exercise gives us the expression \( \frac{e^x - 1 - x}{x^2} \) and asks us to find its limit as \( x \) approaches 0. To do this, we substitute \( x = 0 \) into the expression, ending up with \( \frac{0}{0} \). This result is undefined and doesn't give us any information about the limit. That's why it's called an 'indeterminate form.' Indeterminate forms are like unsolved puzzles, and special techniques such as l'Hospital's Rule help solve them. They emerge when both the numerator and the denominator of a fraction approach zero or infinity at the same time. It's a signal that more work is needed to find the true behavior of the limit.

Differentiation

Differentiation is the mathematical process of finding the derivative of a function. Derivatives represent how a function changes, providing insight into its behavior and trends. In our exercise, because we face an indeterminate form, we apply l'Hospital's Rule, which requires differentiating the numerator and the denominator until the indeterminate form is resolved.

• The derivative of \( e^x - 1 - x \) is \( e^x - 1 \).
• The derivative of \( x^2 \) is \( 2x \).

By differentiating the initial indeterminate function, we transform it into a new expression \( \frac{e^x - 1}{2x} \), then again repeat differentiation when \( \frac{0}{0} \) still appears. Using differentiation this way helps us slowly unravel the function's behavior near the point of interest, ultimately aiding in finding the limit.

Exponential Function

The exponential function, typically written as \( e^x \), is a vital concept in calculus and mathematics more widely. Exponential functions exhibit unique properties of continuous or compounded growth, often showing up in real-world scenarios like finance or population studies.In our exercise, \( e^x \) is a key player in the numerator. The reason for its significance is in its derivative properties:

• The derivative of \( e^x \) is \( e^x \).
• It maintains its form no matter how many times you differentiate it.

Understanding \( e^x \) allows us to simplify problems involving exponential expressions quickly, as seen in this exercise. Even after multiple differentiations, \( e^x \)'s simple behavior helps keep calculations straightforward, ultimately assisting in evaluating limits accurately.

Limit Evaluation

Evaluating a limit, simply put, involves discovering what value a function approaches as the input gets closer to a specific point.Initially, substituting \( x = 0 \) led us to an indeterminate form. By employing l'Hospital's Rule, the process of limit evaluation continues:1. Differentiate both the numerator and denominator. 2. Plug the value (here, \( x = 0 \)) back into the simplified function to evaluate the limit.3. If the result is still an indeterminate form like \( \frac{0}{0} \), differentiate again.After the second round of differentiation, we get \( \frac{e^x}{2} \). Substituting \( x = 0 \) into this function simplifies it to \( \frac{1}{2} \). Through this process, we've evaluated the limit accurately as \( \frac{1}{2} \), wrapping up the exercise by finding the exact behavior of the function around the point of interest.