当x→0时{[e^2-(1+1/x)]^(2/x)}/x的极限是多少
2个回答
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不对吧?应该是
lim(x→0){[e^2-(1+x)^(2/x)]}/x,
才会有结果。此时,利用
e^x - 1 ~ x (x→0),
及罗比达法则,可得
lim(x→0){[e^2-(1+x)^(2/x)]}/x
= lim(x→0){[e^2 - e^(2/x)ln(1+x)]}/x
= lim(x→0)[e^(2/x)ln(1+x)]*lim(x→0){e^[2 - (2/x)ln(1+x)] - 1}/x
= lim(x→0)[e^(2/x)ln(1+x)]*lim(x→0){[2 - (2/x)ln(1+x)]/x}
= 2(e^2)*lim(x→0){[x - ln(1+x)]/x²} (0/0)
= 2(e^2)*lim(x→0){[1 - 1/(1+x)]/(2x)}
= 2(e^2)*(1/2)
= ……
lim(x→0){[e^2-(1+x)^(2/x)]}/x,
才会有结果。此时,利用
e^x - 1 ~ x (x→0),
及罗比达法则,可得
lim(x→0){[e^2-(1+x)^(2/x)]}/x
= lim(x→0){[e^2 - e^(2/x)ln(1+x)]}/x
= lim(x→0)[e^(2/x)ln(1+x)]*lim(x→0){e^[2 - (2/x)ln(1+x)] - 1}/x
= lim(x→0)[e^(2/x)ln(1+x)]*lim(x→0){[2 - (2/x)ln(1+x)]/x}
= 2(e^2)*lim(x→0){[x - ln(1+x)]/x²} (0/0)
= 2(e^2)*lim(x→0){[1 - 1/(1+x)]/(2x)}
= 2(e^2)*(1/2)
= ……
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