Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the indicated limit. Make sure that you have an indeterminate form before you apply l'Hôpital's Rule. $$ \lim _{x \rightarrow 0} \frac{e^{x}-\ln (1+x)-1}{x^{2}} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Form

First, substitute \(x = 0\) into the given expression to check whether an indeterminate form, such as \(\frac{0}{0}\), occurs. Compute each part of the expression as \(x \rightarrow 0\). For the numerator, \(e^x - \ln(1+x) - 1\), it becomes \(1 - 0 - 1 = 0\). The denominator, \(x^2\), becomes \(0\). This confirms the indeterminate form \(\frac{0}{0}\).

Step 2: Apply L'Hôpital's Rule

Since we have a \(\frac{0}{0}\) form, apply L'Hôpital's Rule, which states that \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\) if the limit is indeterminate. Differentiate the numerator and the denominator separately. The derivative of the numerator \(e^x - \ln(1+x) - 1\) is \(e^x - \frac{1}{1+x}\), and the derivative of the denominator \(x^2\) is \(2x\).

Step 3: Evaluate the New Limit

Now evaluate the limit of the derivatives: \(\lim_{x \to 0} \frac{e^x - \frac{1}{1+x}}{2x}\). Substitute \(x = 0\) in the numerator \(e^x - \frac{1}{1+x}\), it becomes \(1 - 1 = 0\), and the denominator \(2x\) becomes \(0\), which is still indeterminate. Therefore, further application of L'Hôpital's Rule is necessary.

Step 4: Differentiate Again

Apply L'Hôpital's Rule once more. Differentiate the numerator \(e^x - \frac{1}{1+x}\) to get \(e^x + \frac{1}{(1+x)^2}\), and the denominator \(2x\) is differentiated to \(2\).

Step 5: Compute the Final Limit

Evaluate \(\lim_{x \to 0} \frac{e^x - \frac{1}{1+x}}{2x}\) again, using the new derivatives: \(\lim_{x \to 0} \frac{e^x + \frac{1}{(1+x)^2}}{2}\). Substitute \(x = 0\) to get \(\frac{1 + 1}{2} = \frac{2}{2} = 1\). Hence, the limit is \(\frac{1}{2}\).

Key Concepts

Indeterminate Forms

When evaluating limits, especially in calculus, you might encounter scenarios known as indeterminate forms. These forms arise when direct substitution in a limit results in expressions that do not provide direct numerical values, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These are termed indeterminate because they do not immediately lead to a definite answer. The limit \[ \lim_{x \rightarrow 0} \frac{e^{x}-\ln (1+x)-1}{x^{2}} \] exemplifies an indeterminate form. When \(x\) approaches 0, both the numerator \(e^{x} - \ln(1+x) - 1\) and the denominator \(x^{2}\) approach 0, resulting in the form \(\frac{0}{0}\). Indeterminate forms signal the need for special techniques, like l'Hôpital's Rule, to evaluate the limit effectively.

Derivatives

Derivatives play a crucial role in calculus for finding rates of change. In the context of limit evaluation, derivatives are used in l'Hôpital's Rule to simplify indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Let's consider the original problem's numerator and denominator. The numerator \(e^x - \ln(1+x) - 1\) entails derivatives of functions such as the exponential function \(e^x\) and the natural logarithm \(\ln(1+x)\). The derivative calculations give us - The derivative of \(e^x\) is \(e^x\). - The derivative of \(-\ln(1+x)\) is \(-\frac{1}{1+x}\) using the chain rule.Applying these, we obtain the total derivative for the numerator as \(e^x - \frac{1}{1+x}\).
The denominator \(x^2\) differentiates to \(2x\). These derivatives help reformulate the limit, aiding in further steps required for limit evaluation.

Limit Evaluation

Limit evaluation using l'Hôpital's Rule involves repeated differentiation, especially when initial attempts still yield indeterminate forms. In our example, after one cycle of applications leading to: \[ \lim_{x \to 0} \frac{e^x - \frac{1}{1+x}}{2x} \]substitution again results in \(\frac{0}{0}\), an indeterminate form. Thus, another round of differentiation was necessary. Derivatives calculated are revisited. Differentiating \( e^x - \frac{1}{1+x} \) gives \( e^x + \frac{1}{(1+x)^2} \). The denominator \(2x\) becomes \(2\). This turns our limit expression into:\[ \lim_{x \to 0} \frac{e^x + \frac{1}{(1+x)^2}}{2} \] Now, substituting \(x = 0\) directly gives:- Numerator: \(1 + 1 = 2\) because both term components \(e^x\) and \(\frac{1}{(1+x)^2}\) converge to 1. - Denominator: \(2\), a constant. Hence, the limit evaluates to \(\frac{2}{2} = 1\). This process highlights how iterative limit evaluations consolidate understanding and application of derivatives.