lim | x3+4x2+3x | ||
x−>1− | |x−1| |
lim | x | −0,1 | |||
to |x| było by |−0,1|=0,1 czyli | =−1 ![]() | ||||
x−>0− | |x| | 0,1 |
x | x | |||
limx→0− | = limx→0− | = −1 (bo x<0 gdy x→0−) | ||
|x| | −x |
x3+4x2+3x | x3+4x2+3x | 8 | ||||
limx→1− | = limx→1− | = [ | ] = ∞ | |||
|x−1| | −x+1 | 0+ |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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