Problem 27 - Thomas Calculus in SI Units

Question

Use l'Hôpital's rule to find the limits. $$\lim _{\theta \rightarrow 0} \frac{3^{\sin \theta}-1}{\theta}$$

Step-by-Step Solution

Answer

The limit is \( \ln(3) \).

Step 1: Identify the Indeterminate Form

The given limit is \( \lim _{\theta \rightarrow 0} \frac{3^{\sin \theta}-1}{\theta} \). As \( \theta \rightarrow 0 \), \( \sin \theta \rightarrow 0 \) and \( 3^0 - 1 = 0 \). Thus, the limit takes the form \( \frac{0}{0} \), which is indeterminate. This applies l'Hôpital's Rule.

Step 2: Differentiate the Numerator and Denominator

Differentiate the numerator \( f(\theta) = 3^{\sin \theta} - 1 \). Using the chain rule, \( \frac{d}{d\theta} 3^{\sin \theta} = 3^{\sin \theta} \ln(3) \cos \theta \). The derivative of \( g(\theta) = \theta \) is \( 1 \).

Step 3: Apply l'Hôpital's Rule

Apply l'Hôpital's Rule: Take the derivative of the numerator and denominator and compute the new limit: \[ \lim_{\theta \rightarrow 0} \frac{3^{\sin \theta} \ln(3) \cos \theta}{1} \].

Step 4: Evaluate the Limit

Substitute \( \theta = 0 \) into the expression: \[ 3^{\sin 0} \ln(3) \cos 0 = 3^0 \ln(3) \cdot 1 = 1 \cdot \ln(3) \]. Hence, the limit is \( \ln(3) \).

Key Concepts

Understanding Limits

Limits are a fundamental concept in calculus, describing how a function behaves as it approaches a particular point. In our exercise, we're trying to find the limit of the expression \( \lim_{\theta \rightarrow 0} \frac{3^{\sin \theta} - 1}{\theta} \). This tells us about the behavior of the function close to \( \theta = 0 \).

• A limit can help predict function behavior without directly substituting the point, which might lead to an undefined form.
• The notation \( \lim_{x \rightarrow a} f(x) \) states that we're looking at the behavior of \( f(x) \) as \( x \) approaches \( a \).

In many cases, directly substituting \( \theta = 0 \) doesn't provide a straightforward result due to indeterminate forms, which will be discussed next.

Exploring Indeterminate Forms

Indeterminate forms like \( \frac{0}{0} \) occur frequently in limit problems. They imply that the limit isn't straightforward and often require tools like l'Hôpital's Rule. In our exercise, plugging \( \theta = 0 \) into \( 3^{\sin \theta} - 1 \) and \( \theta \) gives us \( 0 \) in both the numerator and denominator, forming \( \frac{0}{0} \).

• Indeterminate forms need special methods for resolution, as they indicate uncertainty.
• l'Hôpital's Rule is a common method to resolve these by using differentiation.

Recognizing an indeterminate form is the first step in employing l'Hôpital's Rule to calculate limits.

Diving into Differentiation

Differentiation is the process of finding a derivative, which indicates the rate of change. For using l'Hôpital's Rule, both the numerator and denominator must be differentiated. Let's explore how it's applied in our exercise:

• For \( f(\theta) = 3^{\sin \theta} - 1 \), the chain rule is used for differentiation: \( \frac{d}{d\theta}(3^{\sin\theta}) = 3^{\sin\theta} \ln(3) \cos\theta \).
• The denominator \( g(\theta) = \theta \) is straightforward with a derivative of \( 1 \).

Once the derivatives are obtained, we can apply l'Hôpital's Rule to find the new limit as \( \theta \rightarrow 0 \). Differentiation simplifies problems involving rates of change into manageable calculations.

The World of Exponential Functions

Exponential functions like \( 3^{\sin \theta} \) show up frequently in calculus due to their rapid growth properties. They form the crux of many calculus problems, including our limit problem. Here are some key points:

• Exponential functions have constant bases raised to variable powers, influencing the nature of a function's growth.
• The natural exponential function \( e^x \) is particularly vital, but functions like \( 3^x \) use similar principles.

In our limit, \( 3^{\sin \theta} \) manipulates the constant base \( 3 \) with a variable exponent, affecting the form of the function as \( \theta \) approaches different values. Understanding these functions helps anticipate how a limit behaves around indeterminate forms and through differentiation.