Fonction Logarithme (6)
Exercice 1 tp
Calculer
| lim +∞ |
1 - lnx |
| x | |
| lim +∞ | x - ln(x²-1) |
Correction
| lim +∞ |
1-lnx | = | lim +∞ |
1 | - | ln(x) |
| x | x | x |
On a
| lim +∞ |
1 | = 0 | lim +∞ |
ln(x) | = 0 | |
| x | x |
| Donc | lim +∞ |
1-lnx | = 0 |
| x |
| lim +∞ |
x - ln(x²-1) | = | lim +∞ |
x - ln(x²(1 - | 1 | ) |
| x² |
| = | lim +∞ |
x - ln(x²) - ln(1 - | 1 | ) |
| x² |
| = | lim +∞ |
x (1 - 2 | ln(x) | ) - ln(1 - | 1 | ) |
| x | x² |
| On a | lim +∞ |
ln(x) | = 0 |
| x |
| Et | lim +∞ |
ln(1 - | 1 | ) = 0 |
| x² |
Donc
| lim +∞ |
x-ln(x²-1) = | lim +∞ |
x = +∞ |