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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
Capitolo 2, Problema 2.4.28

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limt→0 2t / tan t

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1
Recognize that the limit involves a trigonometric function, specifically tan(t), which can be expressed in terms of sin(t) and cos(t). Recall that tan(t) = sin(t) / cos(t).
Rewrite the expression 2t / tan(t) as 2t / (sin(t) / cos(t)), which simplifies to 2t * (cos(t) / sin(t)).
This can be further simplified to (2t * cos(t)) / sin(t).
To apply the known limit lim(θ→0) sin(θ) / θ = 1, rewrite the expression as (2 * cos(t)) * (t / sin(t)).
Recognize that as t approaches 0, t / sin(t) approaches 1, and cos(t) approaches cos(0) = 1. Therefore, the limit can be evaluated by considering the product of these limits: 2 * 1 * 1.

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Limit of a Function

The limit of a function describes the value that a function approaches as the input approaches a certain point. In calculus, limits are fundamental for understanding continuity, derivatives, and integrals. The notation lim(x→a) f(x) indicates the limit of f(x) as x approaches a. Evaluating limits often involves techniques such as substitution, factoring, or using special limit properties.
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