Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \lim _{x \rightarrow 0} \frac{e^{x}-\ln (1+x)-1}{x^{2}} $$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Identify the Indeterminate Form

The given limit is \( \lim _{x \rightarrow 0} \frac{e^{x}-\ln (1+x)-1}{x^{2}} \). As \( x \rightarrow 0 \), both the numerator and the denominator approach 0, which is an indeterminate form \( \frac{0}{0} \). This suggests that L'Hopital's Rule or series expansion might be useful here.

Step 2: Use Taylor Series Expansion

To simplify, we expand \( e^x \) and \( \ln(1+x) \) using their Taylor series around \( x = 0 \). We have \( e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \) and \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \).

Step 3: Expand the Numerator

Replace \( e^x \) and \( \ln(1+x) \) with their series in the expression for the numerator: \[ e^x - \ln(1+x) - 1 = (1 + x + \frac{x^2}{2} + \cdots) - (x - \frac{x^2}{2} + \cdots) - 1. \] Simplify this expression to get:\[ = 1 + x + \frac{x^2}{2} - x + \frac{x^2}{2} - 1 = x^2. \]

Step 4: Substitute Back into the Limit

Now substitute the simplified numerator back into the limit expression: \[ \lim _{x \rightarrow 0} \frac{x^2}{x^2} = \lim _{x \rightarrow 0} 1. \]

Step 5: Evaluate the Limit

As \( x \rightarrow 0 \), \( 1 \) remains constant. Therefore, the limit evaluates to 1.

Key Concepts

Taylor Series

Taylor Series are a powerful tool in calculus for approximating complex functions using infinite sums of their derivatives. When dealing with limit problems, such as the exercise given, Taylor Series help us break down functions into simpler components.

• These series expansions are written as the sum of derivatives evaluated at a specific point, typically around zero.
• The Taylor Series for a function helps us better analyze behavior near that point, often simplifying complicated expressions.

Let's consider the two functions used in the exercise:

• The Taylor Series for \( e^x \) is \( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \cdots \).
• For \( \ln(1+x) \), it is \( x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \).

As observed in the example, these series expansions simplify the difference in the numerator, making it possible to cancel terms in the limit expression.

Indeterminate Forms

Indeterminate Forms are situations where a limit seems to take on a form like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or other ambiguous expressions. These forms signal that more information or manipulation is needed to evaluate the limit.

• In this exercise, as \( x \rightarrow 0 \), the numerator and the denominator both approach 0, forming a \( \frac{0}{0} \) indeterminate form.
• This is crucial because it indicates the need for techniques like L'Hopital's Rule or Taylor Series expansion to resolve the ambiguity.

Identifying an indeterminate form allows mathematicians to apply the right tool and transform the limit into a form that can be evaluated with confidence. In this example, using Taylor Series helps clarify the behavior of the function near \( x = 0 \).

Limit Evaluation

Limit Evaluation is a fundamental concept in calculus that describes finding the value a function approaches as the input approaches a specific point. In the exercise, the limit of the function as \( x \rightarrow 0 \) is sought, with an initially problematic form.

• After simplifying the expression using Taylor Series, the numerator becomes \( x^2 \).
• Substituting back into the original limit gives \( \lim _{x \rightarrow 0} \frac{x^2}{x^2} = 1 \).

The cancelling of \( x^2 \) in the numerator and denominator reveals that the limit simplifies to a constant value of 1. Thus, the function approaches 1 as \( x \) moves towards 0. Being adept with techniques like series expansion allows accurate and straightforward limit evaluation, eliminating confusion over complex expressions.