Problem 26 - Calculus Early Transcendentals

Question

Evaluate the following limits. $$\lim _{x \rightarrow 0} \frac{e^{x}-x-1}{5 x^{2}}$$

Step-by-Step Solution

Answer

Question: Evaluate the limit of the following expression: \(\lim _{x \rightarrow 0} \frac{e^{x}-x-1}{5 x^{2}}\) Answer: \(\frac{1}{10}\)

Step 1: Identify the indeterminate form

As x approaches 0, the function becomes \(\frac{e^0-0-1}{5 \cdot 0^2} = \frac{1-1}{0}\). This is an indeterminate form of the type 0/0.

Step 2: Apply L'Hôpital's Rule

Since the limit is of the indeterminate form 0/0, we can apply L'Hôpital's Rule which states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) is an indeterminate form of 0/0, then \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\), provided that the latter limit exists.

Step 3: Differentiate numerator and denominator

Calculate the derivatives of the numerator and denominator function: Numerator derivative: \(f'(x) = \frac{d}{dx} \left(e^{x} - x - 1\right) = e^{x} - 1\) Denominator derivative: \(g'(x) = \frac{d}{dx} \left(5x^2\right) = 10x\)

Step 4: Apply L'Hôpital's Rule again

Now we have a new fraction: \(\frac{e^{x} - 1}{10x}\). As x approaches 0, the expression becomes \(\frac{e^0 - 1}{0} = \frac{0}{0}\), which is also an indeterminate form of the type 0/0. Therefore, we apply L'Hôpital's Rule for a second time: Numerator derivative: \(f''(x) = \frac{d}{dx} \left(e^{x} - 1\right) = e^{x}\) Denominator derivative: \(g''(x) = \frac{d}{dx} \left(10x\right) = 10\)

Step 5: Evaluate the limit

We now have the new fraction \(\frac{e^{x}}{10}\), which is not an indeterminate form. As x approaches 0, the expression becomes \(\frac{e^0}{10} = \frac{1}{10}\). Therefore: $$\lim _{x \rightarrow 0} \frac{e^{x}-x-1}{5 x^{2}} = \frac{1}{10}$$

Key Concepts

Indeterminate Form

When evaluating limits, we sometimes encounter expressions that don't have a straightforward answer. These are called "indeterminate forms," and they typically arise in calculus when evaluating

• Functions that result in 0/0, like in our example
• Expressing other combinations like ∞/∞, 0 • ∞, and more

When facing an indeterminate form, direct substitution into the limit won't work. In our example, as x approaches 0, both the numerator and the denominator approach 0, creating a 0/0 form.
This means we need a special technique, like L'Hôpital's Rule, to resolve the form and find the correct limit. Since just plugging in the numbers doesn't suffice, applying calculus tools helps solve these tricky cases.

Limit Evaluation

Finding the limit of a function as the input approaches certain values is a crucial part of calculus. Evaluating limits helps us understand the behavior of functions near specific points, even if the function isn't defined there.
In the given example, we aim to evaluate the limit of \(\lim_{x \to 0} \frac{e^x - x - 1}{5x^2}\). Initially, substituting x = 0 gives an indeterminate form 0/0. This signals the need for further evaluation specialized methods.
L'Hôpital's Rule becomes handy here, allowing us to replace the original limit with a limit involving the derivatives of the numerator and the denominator. If the new limit formed is still indeterminate, we repeat the process until we achieve a determinate form. After applying L'Hôpital's Rule twice, the limit becomes easy to evaluate, as direct substitution can finally yield a determinate number.

Differentiation

Differentiation is a fundamental concept in calculus involving the computation of the derivative of a function. Differentiation helps us understand how a function changes at any given point. It is particularly useful in evaluating limits involving indeterminate forms.
In the problem, after identifying the indeterminate form, differentiation is used step-by-step. First, we apply the derivative to both

• The numerator \(e^x - x - 1\), and
• The denominator \(5x^2\)

This results in a new fraction \(\frac{e^x - 1}{10x}\), which also is an indeterminate form. Differentiation is repeated, leading us to \(\frac{e^x}{10}\), which is determinate.
This concise application of differentiation enables us to progressively reduce the complexity of the fractions involved to finally evaluate the limit successfully.