Problem 29 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let $$ f(x)= \begin{cases}\frac{e^{x}-1}{x}, & \text { if } x \neq 0 \\ c, & \text { if } x=0\end{cases} $$ What value of \(c\) makes \(f(x)\) continuous at \(x=0\) ?

Step-by-Step Solution

Answer

The value of \(c\) that makes \(f(x)\) continuous at \(x=0\) is 1.

Step 1: Understand Continuity at a Point

For a function to be continuous at a point, the function's value at that point must be equal to the limit of the function as it approaches that point. Here, we need to ensure that \( f(0) = \lim_{{x \to 0}} f(x) \). Since \( f(0) = c \), we need \( c = \lim_{{x \to 0}} f(x) \).

Step 2: Express Limit for \(x \neq 0\)

For \(x eq 0\), the function is given by \( f(x) = \frac{e^x - 1}{x} \). We need to find \( \lim_{{x \to 0}} \frac{e^x - 1}{x} \).

Step 3: Simplify Using L'Hôpital's Rule

Since both the numerator \(e^x - 1\) and the denominator \(x\) tend to 0 as \(x\) tends to 0, we have an indeterminate form \(\frac{0}{0}\). Thus, we can apply L'Hôpital's Rule, which involves taking the derivatives of the numerator and the denominator.

Step 4: Differentiate Numerator and Denominator

The derivative of the numerator \(e^x - 1\) with respect to \(x\) is \(e^x\), and the derivative of the denominator \(x\) is 1. So the limit becomes \( \lim_{{x \to 0}} \frac{e^x}{1} = \lim_{{x \to 0}} e^x \).

Step 5: Compute the Limit

Evaluate the limit \( \lim_{{x \to 0}} e^x \), which simplifies to \( e^0 = 1 \). Therefore, \( \lim_{{x \to 0}} f(x) = 1 \). This is the value we need for continuity.

Step 6: Determine the Value of \(c\)

For the function to be continuous at \(x = 0\), we need \( c = 1 \), as \( \lim_{{x \to 0}} f(x) = 1 \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used to find limits of indeterminate forms. Specifically, it helps us tackle the \({0/0}\) and \(\infty/\infty\) scenarios. When you find a limit that seems stuck due to these forms, L'Hôpital's Rule comes to rescue. The idea is simple: instead of working directly with the troublesome function, you differentiate the numerator and denominator separately. But remember! You can only apply this rule when both the numerator and the denominator approach zero or infinity simultaneously. Here’s a quick step-by-step:

• Identify if the limit is an indeterminate form.
• Differentiate the numerator.
• Differentiate the denominator.
• Evaluate the new limit. If it's still indeterminate, apply L'Hôpital's Rule again.

Keep in mind that coming across an indeterminate form is a clue that L'Hôpital's Rule might be useful.

Limit of a Function

Understanding the limit of a function is central to calculus. When you hear that, think of the value that a function approaches as the input gets closer to some number.For example, in our exercise with \( f(x) = \frac{e^x - 1}{x}\), we're looking for what happens to \(f(x)\) as \(x\) approaches zero.To tackle this, approach it steadily:

• Substitute values increasingly close to the target (like zero), if direct substitution fails, as it does in \( \frac{e^x - 1}{x} \).
• Use methods like factoring, multiplying by conjugates, or tools like L'Hôpital's Rule if the direct approach doesn't work.

For continuous functions, the limit at a point matches the function’s value there. Limits lead to a smooth transition over these points.

Indeterminate Forms

Indeterminate forms are situations in calculus where standard computations hit a dead end. They pop up in limits, making precise evaluation tough initially. These forms signal uncertainty primarily because they don't settle at an obvious value right away.In the limit \( \lim_{{x \to 0}} \frac{e^x - 1}{x} \), the indeterminate form is \( \frac{0}{0} \). Here's the list of common indeterminate forms:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \(0 \cdot \infty \)
• \(1^\infty \)
• \(0^0\)
• \(\infty - \infty\)

When you recognize these forms, you know that traditional arithmetic fails here. Using calculus techniques, like L'Hôpital's Rule, helps unravel these tricky expressions to find meaningful limits.

Differentiation

Differentiation is the core method used in calculus to determine the rate at which one quantity changes with respect to another. It's the process of calculating a derivative. A derivative gives us a snapshot of a function's slope or rate of change at any point.When using L'Hôpital's Rule, differentiation is crucial. Take our function \( f(x) = \frac{e^x - 1}{x} \):

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

By applying these derivatives, we transformed our original indeterminate form into a more manageable limit: \(\lim_{{x \to 0}} \frac{e^x}{1} = \lim_{{x \to 0}} e^x = 1 \).Here are a few basics of differentiation:

• The derivative of a constant is zero.
• The derivative of \(x^n\) is \(nx^{n-1}\).
• Basic rules like the product rule, quotient rule, and chain rule expand the differentiation toolkit.

These basic skills are invaluable in handling indeterminate forms efficiently.