Limite d'une fonctions (3)
Exercice 1 tp
Calculer
lim (-5)+ |
x | et | lim 5- |
x |
| x²-25 | x²-25 |
Correction
Signe de x²-25=(x-5)(x+5)
(x²-25 s'annule en -5 et 5)
| x | -∞ | -5 | 5 | +∞ | |||
| x²-25 | + | 0 | - | 0 | + |
Donc
lim (-5)+ |
x | = | -5 |
| x²-25 | 0- |
ainsi
lim (-5)+ |
x | = +∞ |
| x²-25 |
Et on a
lim 5- |
x | = | 5 |
| x²-25 | 0- |
donc
lim 5- |
x | = -∞ |
| x²-25 |
Exercice 2 tp
Calculer
lim (-1)- |
x²-x-2 | lim (-1)+ |
x²-x-2 | |
| (x+1)² | (x+1)² |
Correction
On pose p(x)=x²-x-2 et q(x)=(x+1)²
Les deux polynômes p(x) et q(x) s'annulent en -1 donc les deux sont divisibles par x+1
on factorise p(x)
Δ=b²-4ac
=(-1)²-4.1.(-2)=9>0
Donc
| ou | x = | -b-√(Δ) | = | -(-1)-√(9) |
| 2a | 2.1 | |||
| x = | -b+√(Δ) | = | -(-1)+√(9) | |
| 2a | 2.1 |
ou encore x = -1 ou x = 2
ainsi p(x)=(x+1)(x-2).
lim (-1)- |
x²-x-2 | = | lim (-1)- |
(x+1)(x-2) |
| (x+1)² | (x+1)² |
| = | lim (-1)- |
x-2 |
| x+1 |
Signe de x+1.
(x+1 s'annule en -1)
| x | -∞ | -1 | +∞ | ||
| x+1 | - | 0 | + |
donc
lim (-1)- |
x-2 | = | -3 |
| x+1 | 0- |
lim (-1)- |
x²-x-2 | = +∞ |
| (x+1)² |
et on a
lim (-1)+ |
x-2 | = | -3 |
| x+1 | 0+ |
donc
lim (-1)+ |
x²-x-2 | = -∞ |
| (x+1)²) |
Exercice 3 tp
1) Calculer
lim (-10)- |
x-12 | lim (-10)+ |
x-12 | |
| -x²+2x+120 | -x²+2x+120 |
2) Calculer
lim 12- |
x-12 | lim 12+ |
x-12 | |
| -x²+2x+120 | -x²+2x+120 |