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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 2.4.45

Using limθ→0 sin θ / θ = 1


Find the limits in Exercises 23–46.


limx→0 (1 − cos 3x) / 2x

Guida verificata passo dopo passo
1
Recognize that the limit involves a trigonometric function and can be approached using trigonometric identities and the known limit lim(θ→0) (sin θ / θ) = 1.
Rewrite the expression (1 - cos 3x) using the trigonometric identity: 1 - cos A = 2sin²(A/2). In this case, A = 3x, so 1 - cos 3x = 2sin²(3x/2).
Substitute the identity into the limit expression: lim(x→0) (2sin²(3x/2) / 2x). This simplifies to lim(x→0) (sin²(3x/2) / x).
To further simplify, use the substitution u = 3x/2, which implies that as x approaches 0, u also approaches 0. Therefore, x = (2/3)u, and the limit becomes lim(u→0) (sin²(u) / ((2/3)u)).
Apply the limit property lim(u→0) (sin u / u) = 1 to the expression: lim(u→0) (sin²(u) / u) = (lim(u→0) (sin u / u)) * (lim(u→0) sin u) = 1 * 0 = 0. Thus, the original limit evaluates to 0.

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