Title | Calcular el límite: lim(x→0) (1-cos3x)/(1-cos4x ) |
---|---|
Author | Damian Reinoso |
Course | Calculo Diferencial I |
Institution | Universidad Central del Ecuador |
Pages | 1 |
File Size | 108.2 KB |
File Type | |
Total Downloads | 31 |
Total Views | 143 |
Calcular el límite:
lim(x→0) (1-cos3x)/(1-cos4x )...
Calcular el límite: lim
𝑥→0
=
1 − cos 3𝑥 1 − cos 4𝑥
1 − cos(3 ∙ 0) 1 − cos 0 1 − 1 0 → 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑐𝑖ó𝑛 = = = 1 − cos(4 ∙ 0) 1 − cos 0 1 − 1 0
Para levantar la indeterminación: 1 − cos 3𝑥 1 + cos 3𝑥 1 + cos 4𝑥 ∙ ∙ 1 − cos 4𝑥 1 + cos 3𝑥 1 + cos 4𝑥
= Todo por:
1
(1 + cos 4𝑥)(1 − cos2 3𝑥 ) (1 + cos 4𝑥)(sin2 3𝑥) = (1 + cos 3𝑥)(1 − cos2 4𝑥) (1 + cos 3𝑥)(sin2 4𝑥)
𝑥2
2 sin2 3𝑥 1 (1 + cos 4𝑥 )(sin2 3𝑥) ∙ 2 (1 + cos 4𝑥) ( 2 ) (1 + cos 4𝑥) (sin 3𝑥) 𝑥 𝑥 = 𝑥 = 2 2 4𝑥 = 1 sin (1 + cos 3𝑥 )(sin2 4𝑥) ∙ 2 (1 + cos 3𝑥) ( 2 ) (1 + cos 3𝑥) (sin 4𝑥) 𝑥 𝑥 𝑥
Dentro del paréntesis al cuadrado en el numerador por
3
3
sin 3𝑥 2 3 sin 3𝑥 2 (1 + cos 4𝑥) (3 ∙ (1 + cos 4𝑥) ( ∙ ) 3 𝑥 3𝑥 ) = = 2 2 sin 4𝑥 3 sin 4𝑥 (1 + cos 3𝑥) ( 𝑥 ) (1 + cos 3𝑥) ( ∙ 𝑥 ) 3
2 sin 3𝑥 2 sin 3𝑥 (1 + cos 4𝑥)(3)2 ( 3𝑥 ) 9 ∙ (1 + cos 4𝑥) ( 3𝑥 ) = = sin 4𝑥 2 sin 4𝑥 2 (1 + cos 3𝑥) ( (1 + cos 3𝑥) ( ) ) 𝑥 𝑥
Dentro del paréntesis al cuadrado en el numerador por
4
4
sin 3𝑥 2 sin 3𝑥 2 9 ∙ (1 + cos 4𝑥) ( 3𝑥 ) 9 ∙ (1 + cos 4𝑥) ( 3𝑥 ) = sin 4𝑥 2 4 sin 4𝑥 2 (1 + cos 3𝑥) (4 ∙ (1 + cos 3𝑥) ( ∙ ) ) 4 4𝑥 𝑥
sin 3𝑥 2 sin 3𝑥 2 sin 3𝑥 2 (1 + cos 4𝑥) ( 9 ∙ (1 + cos 4𝑥) ( ) ) 9 3𝑥 ) 3𝑥 3𝑥 = = = ∙ 2 sin 4𝑥 2 sin 4𝑥 sin 4𝑥 2 16 (1 + cos 3𝑥) ( (1 + cos 3𝑥)(4)2 ( 4𝑥 ) ) 16 ∙ (1 + cos 3𝑥) ( 4𝑥 ) 4𝑥 9 ∙ (1 + cos 4𝑥) (
Se calcula el límite: 2 sin 3𝑥 2 sin 3𝑥 (1 + cos 4𝑥) ( 3𝑥 ) 9 9 (1 + cos 4𝑥) ( 3𝑥 ) ∙ = lim ∙ lim 𝑥→0 16 sin 4𝑥 2 sin 4𝑥 2 16 𝑥→0 (1 + cos 3𝑥) ( (1 + cos 3𝑥) ( ) ) 4𝑥 4𝑥
Se sabe que: sin 𝑘𝑥 =1 𝑥→0 𝑘𝑥 lim
sin 3𝑥 2 ) 9 2 1 9 9 1 + 1 (1)2 9 1 + cos(4 ∙ 0) (lim = ∙ ∙ = ∙ 𝑥→0 3𝑥 2 = ∙ ∙ ∙ = 2 16 2 1 16 1 + 1 (1) 16 16 1 + cos(3 ∙ 0) sin 4𝑥 (lim ) 𝑥→0 4𝑥...