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Ch. 2 - Limits and Continuity
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 2 - Limits and ContinuityProblema 2.4.40
Capitolo 2, Problema 2.4.40

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limθ→0 sin θ cot 2θ

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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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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
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In Exercises 77–80, find a function that satisfies the given conditions and sketch its graph. (The answers here are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.)


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Horizontal and Vertical Asymptotes


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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)

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Using the Formal Definition


Prove the limit statements in Exercises 37–50.


lim x→0 √(4 − x) = 2

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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)

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Centering Intervals About a Point


In Exercises 1–6, sketch the interval (a,b), on the x-axis with the point c inside. Then find a value of δ>0 such that a < x < b whenever 0 < |x−c| < δ.


a=4/9, b=4/7, c=1/2

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