Problem 31 - Thomas Calculus

Question

Find the limits in Exercises \(21-42\) $$\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta}$$

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Identify the limit structure

We're given the limit expression \( \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\sin 2 \theta} \). This is an indeterminate form \( \frac{0}{0} \) since both numerator and denominator approach 0 as \( \theta \rightarrow 0 \). Thus, it requires algebraic manipulation or L'Hôpital's Rule to evaluate.

Step 2: Apply Trigonometric Identity

Use the trigonometric identity \( \sin 2\theta = 2 \sin \theta \cos \theta \) to rewrite the denominator of the limit expression. Therefore, the limit becomes:\[ \lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{2 \sin \theta \cos \theta} \]

Step 3: Use L'Hôpital's Rule

Since the current form is still \( \frac{0}{0} \), apply L'Hôpital's Rule, which involves differentiating the numerator and the denominator with respect to \( \theta \). The derivative of the numerator \( 1 - \cos \theta \) is \( \sin \theta \), and the derivative of the denominator \( 2 \sin \theta \cos \theta \) is \( 2(\cos^2 \theta - \sin^2 \theta) \). Then the limit becomes:\[ \lim _{\theta \rightarrow 0} \frac{\sin \theta}{2(\cos^2 \theta - \sin^2 \theta)} \]

Step 4: Evaluate the derivative limit at \( \theta = 0 \)

Substitute \( \theta = 0 \) into the differentiated expression. \( \sin(0) = 0 \), \( \cos(0) = 1 \), so the denominator becomes \( 2(1 - 0) = 2 \). Hence, the limit evaluates to:\[ \lim _{\theta \rightarrow 0} \frac{0}{2} = 0 \]

Step 5: Conclude the limit result

Thus, the final evaluated limit is 0. This concludes that as \( \theta \) approaches 0, the expression \( \frac{1-\cos \theta}{\sin 2 \theta} \) approaches 0.

Key Concepts

Trigonometric Limits

Trigonometric limits are an essential part of calculus, especially when dealing with oscillating functions like sine and cosine. Understanding these limits helps in analyzing behavior of trigonometric functions as variables approach specific values. One common situation involves expressions where angles approach zero. Such limits exploit known trigonometric identities and behaviors such as:

• As \(\theta \rightarrow 0\), \(\sin \theta \approx \theta\) and \(1 - \cos \theta \approx \frac{1}{2}\theta^2\).
• For double angles, like \(\sin 2\theta\), we can rewrite it using the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\).

For the given exercise, using the identity for \(\sin 2\theta\) simplifies the problem by breaking it down into smaller, manageable parts. Recognizing these types of limits makes it easier to apply calculus principles consistently.

Indeterminate Forms

When evaluating limits, we often encounter indeterminate forms like \(\frac{0}{0}\). These forms indicate uncertainty because direct substitution doesn't provide a clear answer. They occur frequently in calculus, especially when functions approach a common point.In indeterminate cases, simple substitution won't work, so we need additional algebraic manipulation or calculus methods like L'Hôpital's Rule. For the limit \(\lim_{\theta \rightarrow 0} \frac{1 - \cos \theta}{\sin 2\theta}\), both the numerator and denominator approach 0, creating a \(\frac{0}{0}\) form symbolizing an indeterminate limit. Recognizing indeterminate forms is crucial as it directs us to use more sophisticated techniques to find the limit's actual value.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used for solving indeterminate limits, such as \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\). This rule states that if a limit produces an indeterminate form, it can often be resolved by differentiating the numerator and denominator separately.The rule can be applied if:

• Both the original numerator and denominator converge to 0 or \(\infty\) at the limit point.
• The derivatives of both parts exist and are continuous near that point.

For example, in the exercise \(\lim_{\theta \rightarrow 0} \frac{1 - \cos \theta}{\sin 2\theta}\), after applying L'Hôpital's Rule, we differentiate the numerator \(1 - \cos\theta\) to get \(\sin\theta\), and the denominator \(2\sin\theta \cos\theta\) to obtain \(2(\cos^2\theta - \sin^2\theta)\). Evaluating these new expressions can lead to a simpler form that provides the limit value.Therefore, L'Hôpital's Rule transforms complex indeterminate forms into solvable problems, making it an invaluable method in calculus.