Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each limit. $$ \lim _{t \rightarrow 0} \frac{\sin 3 t+4 t}{t \sec t} $$

Step-by-Step Solution

Answer

The limit evaluates to 7.

Step 1: Identify the Form of the Limit

Examine the limit \( \lim_{t \to 0} \frac{\sin 3t + 4t}{t \sec t} \). As \( t \rightarrow 0 \), both the numerator \( \sin 3t + 4t \rightarrow 0 \) and the denominator \( t \sec t \rightarrow 0 \), indicating an indeterminate form \( \frac{0}{0} \).

Step 2: Simplify the Denominator

Recall the definition of secant: \( \sec t = \frac{1}{\cos t} \). Thus, rewrite the denominator: \( t \sec t = \frac{t}{\cos t} \). Therefore, the limit becomes \( \lim_{t \to 0} \frac{(\sin 3t + 4t) \cos t}{t} \).

Step 3: Use L'Hôpital's Rule

Given the \( \frac{0}{0} \) indeterminate form, apply L'Hôpital's Rule by differentiating the numerator and denominator separately. The derivative of the numerator \( \sin 3t + 4t \cos t \) is \( 3 \cos 3t - 4t \sin t + 4 \cos t \). The derivative of the denominator \( t \) is 1.

Step 4: Substitute \( t \rightarrow 0 \)

Substitute \( t = 0 \) into the expression \( \frac{3 \cos 3t - 4t \sin t + 4 \cos t}{1} \). This simplifies to \( 3 \cos 0 + 4 \cos 0 = 3 \times 1 + 4 \times 1 = 7 \).

Step 5: Evaluate the Limit

The value of the limit \( \lim _{t \rightarrow 0} \frac{\sin 3 t+4 t}{t \sec t} \) is therefore 7.

Key Concepts

Indeterminate Form

When working with limits, sometimes you will encounter situations where both the numerator and denominator tend towards zero as the variable approaches a certain value. This is known as an indeterminate form. Specifically, the form \( \frac{0}{0} \) arises, which means you cannot directly evaluate the limit by simple substitution. Instead, you need to find a different approach to manipulate the expression or apply a special rule to determine the limit. This is where understanding indeterminate forms becomes crucial. In our exercise, both the top part (numerator) \( \sin 3t + 4t \) and the bottom part (denominator) \( t \sec t \) approach zero when \( t \to 0 \). It's typical to rewrite or transform the expression to resolve this uncertainty. Recognizing indeterminate forms helps you know when to use advanced techniques, such as L'Hôpital's Rule, to find a precise solution.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus for evaluating limits that give rise to the indeterminate form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When a limit is stuck in one of these forms, L'Hôpital's Rule says you can take the derivative of the numerator and the derivative of the denominator separately and then evaluate the limit of this new fraction. It's as if you're helping the limit "see through" the indeterminacy by looking at the rates of change of the top and bottom expressions. In our exercise, after identifying the \( \frac{0}{0} \) indeterminate form, we applied L'Hôpital's Rule. Differentiating the numerator \( \sin 3t + 4t \cos t \), we had to carefully apply derivative rules for sine and cosine functions. The resulting derivative was \( 3 \cos 3t - 4t \sin t + 4 \cos t \). Since the derivative of \( t \) is simply 1, we had a much simpler task to evaluate at \( t = 0 \). L'Hôpital's Rule clears up the indeterminate form, allowing us to substitute and solve easily.

Trigonometric Limits

Trigonometric limits often appear tricky at first glance, but understanding some basic properties and limits can simplify the process greatly. Key among these are the limits \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \) and \( \lim_{x \to 0} \cos x = 1 \). These are crucial to solving many trigonometric limit problems as they help simplify expressions involving sine and cosine as the variable approaches zero. In our limit problem, \( \frac{\sin 3t}{t} \) and \( \cos t \) played crucial roles. Recognizing that \( \sec t = \frac{1}{\cos t} \) allowed us to rewrite the expression, turning it into a form that was easier to work with using L'Hôpital's Rule. The critical step using trigonometric identities and limits allowed the transformation from a complex expression to a simpler one where substitution could easily provide the solution. Understanding these foundational limits is essential for solving problems involving trigonometric functions smoothly.