Ch. 2 - Limits and Continuity

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 2 - Limits and Continuity

Problem 2.4.40

Chapter 2, Problem 2.4.40

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limθ→0 sin θ cot 2θ

Verified step by step guidance

First, recognize that cotangent is the reciprocal of tangent. Therefore, \( \cot 2\theta = \frac{1}{\tan 2\theta} \).

Rewrite the expression \( \sin \theta \cot 2\theta \) as \( \sin \theta \cdot \frac{1}{\tan 2\theta} = \frac{\sin \theta}{\tan 2\theta} \).

Recall that \( \tan 2\theta = \frac{\sin 2\theta}{\cos 2\theta} \), so \( \frac{1}{\tan 2\theta} = \frac{\cos 2\theta}{\sin 2\theta} \). Substitute this into the expression to get \( \frac{\sin \theta \cdot \cos 2\theta}{\sin 2\theta} \).

Use the double angle identity for sine: \( \sin 2\theta = 2\sin \theta \cos \theta \). Substitute this into the expression to get \( \frac{\sin \theta \cdot \cos 2\theta}{2\sin \theta \cos \theta} \).

Simplify the expression by canceling \( \sin \theta \) from the numerator and the denominator, resulting in \( \frac{\cos 2\theta}{2\cos \theta} \). Now, evaluate the limit as \( \theta \to 0 \) using the fact that \( \cos 0 = 1 \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limit of a Function

The limit of a function describes the behavior of the function as the input approaches a particular value. In calculus, understanding limits is crucial for analyzing the continuity and differentiability of functions. For the given problem, evaluating the limit as θ approaches 0 is essential to determine the behavior of the expression sin θ cot 2θ.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables. In this problem, recognizing that cot 2θ is equivalent to cos 2θ/sin 2θ helps simplify the expression. This simplification is necessary to apply the limit property limθ→0 sin θ / θ = 1 effectively.

Squeeze Theorem

The Squeeze Theorem is a method for finding the limit of a function by comparing it to two other functions whose limits are known and 'squeeze' the function of interest. In this context, the theorem can be used to justify the limit of sin θ / θ as θ approaches 0, which is a foundational result in calculus and helps in evaluating the given limit problem.

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Related Practice

Textbook Question

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

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Textbook Question

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a. It can be shown that the inequalities 1 − x²/ 6 < (x sin x) / (2−2cos x) < 1 hold for all values of x close to zero (except for x = 0). What, if anything, does this tell you about limx→0 (x sin x) / (2 − 2cos x)?

Give reasons for your answer.

[Technology Exercise] b. Graph y = 1 − (x²/6), y=(x sinx)/(2 − 2cos x), and y = 1 together for −2 ≤ x ≤2. Comment on the behavior of the graphs as x→0.

Textbook Question

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

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Textbook Question

Limits and Infinity

Find the limits in Exercises 37–46.

x⁴ + x³

lim -----------------

x→∞ 12x³ + 128

Textbook Question

Using the Formal Definition

Prove the limit statements in Exercises 37–50.

lim x→0 √(4 − x) = 2

Textbook Question

Find the limits in Exercises 31–40. Are the functions continuous at the point being approached?

lim x → π/6 √(csc² x + 5√3 tan x)

Using limθ→0 sin θ / θ = 1Find the limits in Exercises 23–46.limθ→0 sin θ cot 2θ

Verified video answer for a similar problem:

Key Concepts

Limit of a Function

Trigonometric Identities

Squeeze Theorem

Using limθ→0 sin θ / θ = 1

Find the limits in Exercises 23–46.

limθ→0 sin θ cot 2θ