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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 2 - Limits and ContinuityProblem 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
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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.
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Limits of Rational Functions: Denominator = 0

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.
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Verifying Trig Equations as Identities

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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Fundamental Theorem of Calculus Part 1
Related Practice
Textbook Question

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limt→0 sin(1 − cos t) / (1 − cos t)

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Textbook Question

Limits as x → ∞ or x → −∞


The process by which we determine limits of rational functions applies equally well to ratios containing noninteger or negative powers of x. Divide numerator and denominator by the highest power of x in the denominator and proceed from there. Find the limits in Exercises 23–36. Write ∞ or −∞ where appropriate.


lim x→∞ (x − 3) / √(4x² + 25)

Textbook Question

Calculating Limits


Find the limits in Exercises 11–22.


limx→−1/2 4x(3x+4)²

Textbook Question

At what points are the functions in Exercises 13–30 continuous?


y = √(2x + 3)

Textbook Question

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limh→0 sin(sin h) / sin h

Textbook Question

Limits and Infinity


Find the limits in Exercises 37–46.


x⁴ + x³

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

x→∞ 12x³ + 128