Limite d'une fonction (11)
5.4 Les limites trigonométriques
5.4.1 Propriétés 1
lim 0 |
sinx | = 1 |
| x | ||
lim 0 |
sinax | = 1 |
| ax | ||
lim 0 |
tanx | = 1 |
| x | ||
lim 0 |
tanax | = 1 |
| ax |
5.4.2 Propriété 2
lim 0 |
1-cosx | = | 1 | |
| x² | 2 |
Exemples
| 1) | lim 0 |
sin2x | = 1 |
| 2x | |||
| 2) | lim 0 |
tan(-3x) | = 1 |
| -3x |
Exercice 1 tp
Calculer la limite suivante
lim 0 | sin3x |
| sin9x |
Correction
lim 0 |
sin3x | = | lim 0 |
sin3x | × lim 0 |
9x |
| sin9x | 3x | 3sin9x |
lim 0 |
sin3x | = 1 |
| 3x |
et on a
lim 0 |
9x | = | lim 0 |
( | sin9x | )-1 = 1 |
| sin9x | 9x |
donc
lim 0 |
sin3x | = | 1 | |
| sin9x | 3 |
Exercice 2 tp
Calculer la limite suivante
lim 0 | tan4x |
| sin2x |
Correction
lim 0 |
tan4x | = | lim 0 |
tan4x | × lim 0 |
2.2x |
| sin2x | 4x | sin2x |
| on a | lim 0 |
tan4x | = 1 |
| 4x |
On a aussi
lim 0 |
2x | = | lim 0 |
( | sin2x | )-1 = 1 |
| sin2x | 2x |
| donc | lim 0 |
tan4x | = 2 |
| sin2x |
Exercice 3 tp
Calculer la limite suivante
lim 0 | 1 + sin²x - cos²x |
| x² |
Correction
On a 1+sin²x-cos²x = 1+1-cos²x-cos²x
= 2 - 2cos²x
Donc
lim 0 |
1 + sin²x - cos²x | = | lim 0 |
2(1-cos²x) |
| x² | x² |
| = | lim 0 |
2(1-cosx)(1+cosx) | |
| x² | |||
| = | lim 0 |
1-cosx | ×2(1+cosx) |
| x² |
| On a | lim 0 |
1-cosx | = | 1 |
| x² | 2 |
| et | lim 0 |
2(1+cosx) = 4 |
donc
lim 0 |
1 + sin²x - cos²x | = 2 |
| x² |