| 4 − 2x | 2(2−x) | 2 | ||||
lim x→2 | = lim x→2 | = | ||||
| sin3(2−x) | sin3(2 − x) | sin3 |
| sinx | ||
Trzeba skorzystać z granicy: limx→0 | =1 | |
| x |
| 2*(2−x) | ||
limx→2 | = | |
| sin[3*(2−x)] |
| 2*[3*(2−x)] | 2 | |||
=limx→2 | = | |||
| 3*sin[3*(2−x)] | 3 |
| [3(2−x)] | |
→1 | |
| sin[3*(2−x)] |
| 2(2−x) | 2 | |||
limx−>2 | = | |||
| 3sin(2−x)(2−x) *(2−x) | 3 |
| tgx | ||
a wzór | = 1 wykorzystam tutaj?: | |
| x |
| π | ||
x→1 (1−x)tg do potęgi | ||
| 2 |