Granica
Pomoc: Jak obliczyć granicę ((6/π)arcsint))(1/(2t−1) przy t→1/2.
1 lut 00:13
piotr: ((6/π)arcsint))
1/(2t−1) = e
ln((6/π)arcsint)(1/(2t−1)
| | ln((6/π)arcsint) | |
limt→1/2 |
| = |
| | 2t−1 | |
| | 1/(arcsint)*1/√1−t2 | | 6/π*2/√3 | | 2√3 | |
limt→1/2 |
| = |
| = |
| |
| | 2 | | 2 | | π | |
lim
t→1/2((6/π)arcsint))
1/(2t−1) = e
2√3/π
1 lut 09:17
piotr: zastosowano regułę l'Hospitala:
| | f(x) | | f'(x) | |
lim |
| = lim |
| |
| | g(x) | | g'(x) | |
1 lut 09:19
Pomoc: Dzięki wielkie
1 lut 13:58