Oblicz granice
qwerty: lim x→0+ (ctgx)sin(x)
13 gru 20:12
ford:
(ctgx)
sin(x) = e
ln(ctgx)sin(x) = e
sin(x)*ln(ctgx)
liczę granicę wykładnika
| | ln(ctgx) | | ln(ctg0) | | ∞ | |
lim sin(x)*ln(ctgx) = lim |
| = |
| = [ |
| ] = H = |
| | | | | | ∞ | |
| | | | sinx | | sin0 | |
= lim |
| = lim |
| = |
| = |
| | | | cos2x | | (cos0)2 | |
granica wykładnika to 0
więc
lim e
sin(x)*ln(ctgx) = e
0 =
1
13 gru 20:42