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Magazine Chapter 7: Problem 27
Use l'H么pital's rule to find the limits. $$\lim _{\theta \rightarrow 0} \frac{3^{\sin \theta}-1}{\theta}$$
Short Answer
Expert verified
The limit is \( \ln(3) \).
Step by step solution
01
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'H么pital's Rule.
02
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 \( 1 \).
03
Apply l'H么pital's Rule
Apply l'H么pital's Rule: Take the derivative of the numerator and denominator and compute the new limit: \[ \lim_{\theta \rightarrow 0} \frac{3^{\sin \theta} \ln(3) \cos \theta}{1} \].
04
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 \( \ln(3) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( a \).
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'H么pital'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'H么pital's Rule is a common method to resolve these by using differentiation.
Diving into Differentiation
Differentiation is the process of finding a derivative, which indicates the rate of change. For using l'H么pital'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 \( 1 \).
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.
