lim(x2−3)*ex= mamy [∞*0]= lim (x2−3)/(1/ex)= [∞/∞] [H]
x→−∞
lim 2x/−e−x = lim 2/∞= 0
Dobrze I jakby ta granica dążyła do +∞ to wynik byłby ∞
| 1 | 1 | 1 | ||||
( | )' = − | *ex = − | ||||
| ex | (ex)2 | ex |
| 1 | ||
t(x) = ex f(t) = | ||
| t |
| 1 | 1 | 1 | 1 | |||||
f'(t) = − | *t'(x) = − | *(ex)' = − | *ex = − | |||||
| t2 | e2x | e2x | ex |
| 2x | 2x | ||||||||||||||||||
W(x) → | = − | ||||||||||||||||||
|
|
| 2 | |||||||||||
W(x) → − | = 2ex | ||||||||||
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