Limite d'une fonction (1)
Exercice 1 tp
Calculer
lim -∞ |
-7x³+3x²-5 |
lim +∞ |
5x³+8x-7 |
Correction
lim -∞ |
-7x³+3x²-5 | = | lim -∞ |
-7x³= -7(-∞)=+∞ |
lim +∞ |
5x³+8x-7 | = | lim +∞ |
5x³= 5(+∞)=+∞ |
Exercice 2 tp
Calculer
lim -∞ |
4x+3 |
| 7x-2 | |
lim +∞ |
3x-1 |
| 4x²+5x |
Correction
On a
lim -∞ |
4x+3 | = | lim -∞ |
4x | = | 4 |
| 7x-2 | 7x | 7 |
et on a
lim +∞ |
3x-1 | = | lim +∞ |
3x |
| 4x²+5x | 4x² | |||
| = | lim +∞ |
3 | =0 | |
| 4x |
Exercice 3 tp
Calculer
lim -2 | x²+2x |
| x+2 |
Correction
| 0 | Forme indéterminée |
| 0 |
On a x²+2x = x(x+2) donc
lim -2 |
x²+2x | = | lim -2 |
x(x+2) |
| x+2 | x+2 |
| = | lim -2 |
x = -2 |
ainsi
lim -2 |
x²+2x | = -2 |
| x+2 |
Exercice 4 tp
Soit f une fonction numérique définie par
| f(x) = | 2x²-5x+2 |
| x-2 |
Calculer
lim 2 |
f(x) |
lim +∞ |
f(x) |
Correction
On pose p(x)=2x²-5x+2
p(2)=0 donc le polynôme p(x) est divisible par x-2.
| 2x² | -5x | +2 | x-2 | |
| -2x² | +4x | 2x-1 | ||
| 0 | -x | +2 | ||
| +x | -2 | |||
| 0 | 0 |
donc p(x)=(x-2)(2x-1) (on trouve le même résultat si on utilise Δ).
lim 2 |
f(x) = | lim 2 |
(x-2)(2x-1) |
| x-2 | |||
| = | lim 2 |
2x-1 | = 3 |
lim +∞ |
f(x) = | lim +∞ |
2x² |
| x | |||
| = | lim +∞ |
2x | = +∞ |