Problem 10 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The limit evaluates to \( \frac{9}{4} \).

Step 1: Identify the Form of the Limit

We need to evaluate \( \lim _{t \rightarrow 0} \frac{\sin ^{2} 3t}{2t} \). This limit is of the indeterminate form \( \frac{0}{0} \) as \( t \rightarrow 0 \).

Step 2: Use Trigonometric Limit Property

We use the known limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) to simplify \( \frac{\sin^2 3t}{2t} \). Note that \( \sin^2 3t = (\sin 3t)^2 \).

Step 3: Rewrite the Expression

Rewrite the limit to separate the squared sine function:\[ \frac{\sin^2 3t}{2t} = \frac{\sin 3t \cdot \sin 3t}{2t} = \frac{\sin 3t}{t} \cdot \frac{\sin 3t}{2}\]

Step 4: Apply Trigonometric Limit

Use \( \lim_{t \to 0} \frac{\sin 3t}{3t} = 1 \) by letting \( x = 3t \):\[ \lim_{t \to 0} \frac{\sin 3t}{2t} = \lim_{t \to 0} \left( \frac{3}{2} \cdot \frac{\sin 3t}{3t} \right) = \frac{3}{2} \cdot 1 = \frac{3}{2} \]

Step 5: Conclude the Evaluation

Hence, the limit evaluates to \( \frac{3}{2} \). Thus, \( \lim _{t \rightarrow 0} \frac{\sin ^{2} 3t}{2t} = \frac{9}{4} \).

Key Concepts

Trigonometric Limits

Trigonometric limits are crucial in calculus, especially when dealing with functions that involve sine and cosine. These limits help us understand behavior and trends as a variable approaches a specific point. In this problem, the limit of interest is represented as \( \lim _{t \rightarrow 0} \frac{\sin ^{2} 3 t}{2 t} \). Trigonometric limits, like the famous \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \), guide us in simplifying and evaluating complex expressions.

If you encounter sine or cosine functions approaching zero in the numerator or denominator, recalling these limits can provide valuable shortcuts to finding solutions quickly. Recognizing these patterns will become easier with practice, making trigonometric limits less daunting.

Indeterminate Forms

Indeterminate forms arise when evaluating limits. These forms, such as \( \frac{0}{0} \), seem unsolvable at first glance, but they signal that algebraic manipulation or special techniques may uncover the answer. In the given limit \( \lim _{t \rightarrow 0} \frac{\sin ^{2} 3 t}{2 t} \), we encounter an indeterminate form.

Upon replacing \( t \) by zero, both the numerator \( \sin^2 3t \) and the denominator \( 2t \) become zero, leading to \( \frac{0}{0} \). This tells us that direct substitution is not the way forward. Instead, we need to factor or simplify using known limits or identities to resolve the indeterminacy and evaluate the limit correctly.

Sine Function

The sine function is one of the fundamental trigonometric functions. In this exercise, we're working with the function \( \sin 3t \), specifically its square: \( \sin^2 3t \). Understanding how sine behaves, especially near zero, is important to effectively use trigonometric limits.

The interesting property about the sine function is its relationship with angles in radians. Around the point zero, \( \sin x \approx x \) holds true under the limit. This characteristic is utilized to simplify expressions like \( \frac{\sin x}{x} \). Recognizing how \( \sin 3t \) behaves when \( t \) is very small aids in resolving complicated limits.

Limit Evaluation Techniques

Limit evaluation techniques, including substitution and factoring, are vital when approaching indeterminate forms. This exercise uses the trigonometric limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) cleverly adapted for \( \sin 3t \).

We start by rewriting the limit \( \frac{\sin^2 3t}{2t} \) to separate the problems into more familiar forms: \( \frac{\sin 3t}{t} \cdot \frac{\sin 3t}{2} \). Then, the limit \( \lim_{t \to 0} \frac{\sin 3t}{3t} = 1 \) helps us adjust the expression so it's manageable. By methodically adjusting variables and equivalent expressions, complex limits simplify considerably. This systematic approach is crucial for solving intricate calculus problems efficiently.