Problem 17 - Calculus: Early Transcendentals

Question

Compute the following limits. $$ \lim _{x \rightarrow 0} \frac{x^{2}}{e^{x}-x-1} $$

Step-by-Step Solution

Answer

The limit is 2.

Step 1: Assess the Limit Form

First, determine the form of the function as \(x\) approaches 0. Substitute \(x = 0\) into the numerator and the denominator: \(\frac{0^2}{e^0 - 0 - 1} = \frac{0}{0}\). This is an indeterminate form, so we need further analysis.

Step 2: Apply L'Hôpital's Rule

Since the limit is in the form \(\frac{0}{0}\), we apply L'Hôpital's Rule, which requires us to differentiate the numerator and denominator separately and then compute the limit again as \(x\) approaches 0.

Step 3: Differentiate the Numerator and Denominator

Differentiate the numerator \(x^2\) to get \(2x\). Differentiate the denominator \(e^x - x - 1\) to get \(e^x - 1\). Consequently, the limit becomes:\[\lim_{{x \to 0}} \frac{2x}{e^x - 1}\]

Step 4: Evaluate the New Limit

Substitute \(x = 0\) into the new expression: \(\frac{2 \cdot 0}{e^0 - 1} = \frac{0}{0}\). This again gives an indeterminate form, so we apply L'Hôpital's Rule one more time.

Step 5: Differentiate Again with L'Hôpital's Rule

Differentiate the numerator \(2x\) to get \(2\). Differentiate the denominator \(e^x - 1\) to get \(e^x\). Now, the limit is:\[\lim_{{x \to 0}} \frac{2}{e^x}\]

Step 6: Evaluate the Simplified Limit

Substitute \(x = 0\) into \( \frac{2}{e^x} \): \( \frac{2}{e^0} = \frac{2}{1} = 2\). Thus, the limit evaluates to 2.

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful technique in calculus, designed to resolve tricky limits that lead to indeterminate forms. It is particularly helpful when dealing with expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) yields an indeterminate form, then:

• We can differentiate the numerator \(f(x)\) and denominator \(g(x)\).
• We then attempt to find \( \lim_{x \to c} \frac{f'(x)}{g'(x)} \).

Repeat this process as needed if the new limit continues to be indeterminate.
The simplicity of L'Hôpital's Rule lies in its straightforward application—differentiate and try again. However, ensure that the conditions for using the rule are met, specifically that both \(f(x)\) and \(g(x)\) are differentiable around \(c\), except potentially at \(c\) itself.

Limits

In calculus, a limit helps us understand the behavior of a function as it approaches a particular point. The limit \( \lim_{x \to 0} \frac{x^2}{e^x - x - 1} \) asks what value the expression reaches as \(x\) gets closer to 0.
This concept is foundational because it forms the basis of more advanced topics like derivatives and integrals. To determine limits effectively, you can:

• Substitute the value directly if possible.
• Consider the right-hand and left-hand limits.
• Use algebraic manipulation to simplify the expression first.

Limits can become more challenging when they approach forms like \( \frac{0}{0} \). That's when tools such as L'Hôpital's Rule come into play.

Indeterminate Forms

Indeterminate forms are expressions where direct calculation of a limit results in an ambiguity, typically expressed as \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), or similar forms like \(0 \times \infty\). These scenarios mean that the limit

• Can not be directly found.
• Requires additional tools or manipulation to solve.

In this exercise, when \( x = 0 \), both the numerator \(x^2\) and the denominator \(e^x - x - 1\) result in 0, leading to the form \( \frac{0}{0} \).
These forms indicate that the function's behavior is not immediately clear, hence extra steps or rules, like L'Hôpital's, are needed to properly evaluate the limit.