Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each limit. $$ \lim _{\theta \rightarrow 0} \frac{\sin 3 \theta}{2 \theta} $$

Step-by-Step Solution

Answer

The limit is \( \frac{3}{2} \).

Step 1: Recognize the Standard Limit

Recall the standard limit: \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). This property will be very useful in solving the given problem.

Step 2: Rewrite the Expression

Rearrange the original limit \( \lim _{\theta \rightarrow 0} \frac{\sin 3 \theta}{2 \theta} \) into a form that resembles the standard limit. Note that the argument of sine is \( 3\theta \), so express it as \( \frac{\sin 3\theta}{3\theta} \) times \( \frac{3\theta}{2\theta} \), which simplifies the problem.

Step 3: Split and Simplify the Limit

Split the limit into two parts: \( \lim _{\theta \rightarrow 0} \frac{\sin 3 \theta}{3 \theta} \) and another constant factor \( \frac{3}{2} \). Hence, the expression becomes: \( \lim _{\theta \rightarrow 0} \frac{\sin 3\theta}{3\theta} \times \frac{3}{2} \).

Step 4: Apply Standard Limit Result

Apply the standard limit to the first part: \( \lim _{\theta \rightarrow 0} \frac{\sin 3 \theta}{3\theta} = 1 \). Combine this with the constant factor, \( \frac{3}{2} \).

Step 5: Evaluate the Final Limit

Multiply these together to get the final result: \( 1 \times \frac{3}{2} = \frac{3}{2} \). This is the value of the original limit.

Key Concepts

Trigonometric Limits

Understanding trigonometric limits is crucial when working with limits involving sine, cosine, and other trigonometric functions. These limits often occur in calculus, especially when evaluating limits as a variable approaches 0.
Trigonometric limits rely heavily on the inherent properties of trigonometric functions.
For example, one of the most used standard identities is for sine:

• \(\lim_{x \to 0} \frac{\sin x}{x} = 1\), a fundamental limit in calculus.

Applying this identity can simplify the evaluation of any limits involving sine. When dealing with these functions, remember the periodic and oscillatory nature of sine and cosine, which can initially seem daunting but remain predictable under limits.
For practice, break down more complicated expressions into forms that utilize these known trigonometric limits. This method makes handling definite squeeze-theorem problems more efficient. It allows students to recognize more complex expressions as variations of basic standard limits.

Standard Limit Theorem

The Standard Limit Theorem is a powerful tool in calculus.
This theorem gives us precise results for evaluating limits when variables approach a particular point. It forms the basis for understanding how functions behave around specific inputs.
One of the most vital standard limit theorems involves the sine function, where

• \(\lim_{x \to 0} \frac{\sin x}{x} = 1\).

Using this theorem streamlines solving problems by reducing complex expressions into known, solvable limits. This theorem highlights the behavior of sine and cosine around zero, simplifying calculations with these functions.
It is essential to recognize when you can apply this theorem. It allows for transforming more complex expressions into a suitable form that takes advantage of this rule. Understanding this theorem's application can significantly remove difficulties in solving trigonometric limits.
Always look for opportunities to use the standard limit to evaluate limits more efficiently.

Limit Properties

Limit properties are essential for all calculus students to grasp.
These properties give guidelines about how limits interact with the arithmetic of functions.
Basics include:

• If \(\lim_{x \to a} f(x)\) and \(\lim_{x \to a} g(x)\) exist, then:
• \(\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)\)
• \(\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)\)
• \(\lim_{x \to a} [f(x) \, g(x)] = \lim_{x \to a} f(x) \, \lim_{x \to a} g(x)\)
• For division, \(\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}\) if \(\lim_{x \to a} g(x) eq 0\).

When using these properties, it's essential to verify that all individual limits exist and are finite. Understanding these properties allows students to deconstruct complex limit problems effectively. By splitting expressions into manageable parts, you solve what might first appear complicated, step by step.
This toolkit approach is about being alert to rewrite and adjust expressions actively, making the use of limit properties a flexible part of problem-solving in calculus.