√x2+1−1 | ||
lim (przy x−> 0) | . Czegokolwiek nie zrobię, i tak wychodzi zero w | |
√x2+25−5 |
x | √x2 + 25 | |||
[ H ] = lim | * | = 5 | ||
√x2+1 | x |
√x2+1−1 | (√x2+1−1)(√x2+25+5) | ||
= | |||
√x2+25−5 | (√x2+25−5)(√x2+25+5) |
(√x2+1−1)(√x2+25+5) | ||
= | ||
x2 |
(√x2+1+1)(√x2+1−1)(√x2+25+5) | ||
= | ||
x2((√x2+1+1)) |
x2(√x2+25+5) | ||
= | ||
x2(√x2+1+1) |
√x2+25+5 | ||
= | ||
√x2+1+1 |
√x2+25+5 | √25+5 | |||
limx→0 | = | |||
√x2+1+1 | √1+1 |
5+5 | 10 | |||
= | = | |||
1+1 | 2 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
Kliknij po więcej przykładów | |
---|---|
Twój nick | |