Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each limit. $$ \lim _{\theta \rightarrow 0} \frac{\cot (\pi \theta) \sin \theta}{2 \sec \theta} $$

Step-by-Step Solution

Answer

The limit is \(-\frac{\pi}{2}\).

Step 1: Understand the Problem

We are tasked with finding the limit of the expression \( \frac{\cot(\pi \theta) \sin \theta}{2 \sec \theta} \) as \( \theta \) approaches 0.

Step 2: Simplify Trigonometric Expressions

Recall that \( \cot(x) = \frac{\cos(x)}{\sin(x)} \) and \( \sec(x) = \frac{1}{\cos(x)} \). So, the expression becomes \( \frac{\frac{\cos(\pi \theta)}{\sin(\pi \theta)} \sin \theta}{2 \times \frac{1}{\cos \theta}} \).

Step 3: Rewrite the Expression

Rewriting, we have \( \frac{\cos(\pi \theta) \sin \theta \cos \theta}{2 \sin(\pi \theta)} \). This simplifies further to \( \frac{\cos(\pi \theta) \sin(\theta) \cos(\theta)}{2 \sin(\pi \theta)} \).

Step 4: Apply Limit Using L'Hopital’s Rule

As \( \theta \to 0 \), substitute \( \theta = 0 \) in the derivatives because the initial form is \( \frac{0}{0} \). Calculate derivatives: numerator \( \cos(\pi \theta) (\sin \theta \cos \theta)' + (\cos \theta) (\sin \theta)' \) and denominator \( \cos(\pi \theta) \).

Step 5: Evaluate Limit

As \( \theta \to 0 \), the expression simplifies, and terms cancel out, resulting in \( -\frac{\pi}{2} \).

Key Concepts

Trigonometry

Trigonometry is the branch of mathematics that studies the relationships between the angles and sides of triangles. It is built upon ratios of the sides in right-angled triangles, with the most common functions being sine, cosine, tangent, cotangent, secant, and cosecant.

In the given problem, we utilize three trigonometric functions: cotangent, sine, and secant. To better understand the limit, it’s crucial to recall their definitions:

• Cotangent (cot): This is the reciprocal of the tangent function, defined as \( \cot(x) = \frac{\cos(x)}{\sin(x)} \).
• Secant (sec): Known as the reciprocal of the cosine function, given by \( \sec(x) = \frac{1}{\cos(x)} \).

Applying these identities simplifies the original equation, turning it into terms that can be further processed. Recognizing these functions allows us to substitute and rewrite expressions in a form suitable for limit evaluation.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful method in calculus for evaluating limits of indeterminate forms, specifically \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms arise when directly substituting a value into a limit-operation expression.

In the exercise, when substituting \( \theta = 0 \), the expression simplifies to \( \frac{0}{0} \), necessitating the application of L'Hôpital's Rule. This process involves:

• Taking the derivative of the numerator and the derivative of the denominator independently.
• Re-evaluating the limit with the new fraction of derivatives.

For our problem, the derivatives are calculated as follows:

• The new numerator from derivatives of \( \cos(\pi \theta) (\sin \theta \cos \theta)' + (\cos \theta) (\sin \theta)' \).
• The denominator's derivative is \( \cos(\pi \theta) \).

Once derivatives are found, the limit is re-evaluated, revealing the simplified form and ultimately helping to find the solution. Understanding L'Hôpital's Rule is essential for dealing with complex limits in calculus.

Calculus

Calculus is a branch of mathematics focused on concepts of change and motion. It encompasses two main areas: differential calculus and integral calculus. Differential calculus involves finding derivatives, which represent rates of change and slopes of curves.

In our limit problem, calculus plays a crucial role in both simplifying the expression and evaluating the limit. By breaking down trigonometric expressions into simpler derivatives and then applying techniques like L'Hôpital's Rule, calculus allows us to navigate and find these challenging limits.

• Limit Evaluation: Calculus helps evaluate how functions behave as inputs approach certain values. Understanding limits is fundamental as they are the groundwork for continuity and differentiability.
• Derivative Calculation: As seen, deriving trigonometric functions reduces complex ratios into solvable formats, assisting in extracting meaningful results from indeterminate forms.

The principles of calculus ensure limits like the example given are solved accurately and efficiently, illustrating its importance in mathematical problem-solving.