Limite d'une fonction (2)
Exercice 1 tp
Soit f une fonction numérique définie par
| f(x) = | x²-1 |
| x³ - 1 |
Calculer
lim 1 | f(x) | lim -∞ | f(x) |
Correction
x²-1 = (x-1)(x+1)
et x³-1=x³ - 1³=(x-1)(x²+x+1).
lim 1 |
f(x) = | lim 1 |
(x-1)(x+1) |
| (x-1)(x²+x+1) | |||
| = | lim 1 |
x+1 | |
| x²+x+1 |
ainsi
lim 1 |
f(x) = | 2 |
| 3 |
lim -∞ |
f(x) = | lim -∞ |
x² | = | lim -∞ |
1 | = 0 |
| x³ | x |
Exercice 2 tp
Calculer
lim -5 |
x²+4x-5 | et | lim +∞ |
x²+4x-5 |
| 2x²-50 | 2x²-50 |
Exercice 3 tp
Calculer
lim 7 |
x-7 | + | x²-7x |
| x²-49 | 2x-14 |
Exercice 4 tp
Calculer
lim 3+ |
x-1 |
| x-3 |
Correction
Signe de x-3.
(x-3 s'annule en 3)
| x | -∞ | 3 | +∞ | |||
| x-3 | - | 0 | + |
Donc
lim 3+ |
x-1 | = | 2 |
| x-3 | 0+ |
ainsi
lim 3+ |
x-1 | = +∞ |
| x-3 |
Exercice 5 tp
Calculer
lim 2+ |
x-3 | et | lim 2- |
x-3 |
| x-2 | x-2 |
Correction
Signe de x-2.
(x-2 s'annule en 2)
| x | -∞ | 2 | +∞ | |||
| x-2 | - | 0 | + |
Donc
lim 2+ |
x-3 | = | -1 |
| x-2 | 0+ |
ainsi
lim 2+ |
x-1 | = -∞ |
| x-3 |
lim 2- |
x-3 | = | -1 |
| x-2 | 0- |
donc
lim 2- |
x-1 | = +∞ |
| x-3 |