| 8(x+2)−8 | x*arcctg(3−x) | |||
lim x→∞ | + | |||
| 3(2x−1)+2(3x+4)−5 | sin(πx) |
| an+1 | ||
=22 dla n ∊{1,2,3....} to prawdą jest że: | ||
| an |
| an+2 | ||
A: | =4n | |
| an |
| an+2 | ||
B: | =2n+1; | |
| an |
| an+2 | ||
c: | =2*4n; | |
| an |
| an+2 | ||
D: | =8*4n | |
| an |
| −3*π | ||
arcctg(ctg( | )). Jak coś takiego policzyć? Muszę operować na przedziale x∊[0,π] | |
| 2 |
| −3*π | 3π | π | π | |||||
ctg( | )=−ctg( | )=−ctg(π+ | )=−ctg( | ) | ||||
| 2 | 2 | 2 | 2 |
| (k2+3k−1)n2+n−4 | 1 | |||
an= | jest równa | ? | ||
| kn2−4 | 3 |
| 1 | ||
Obliczyć granice: lim x−>0 arcctg | ||
| x |