求【1+x)/(1-x)】的x次方,当x趋于无穷时的极限
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(x->∞)lim
[(1+x)/(1-x)]^x
=(x->∞)lim
[(-1)(x+1)/(x-1)]^x
=(x->∞)lim
(-1)^x
[((x-1+2)/(x-1)]^x
=(x->∞)lim
(-1)^x
[1+2/(x-1)]^x
令(x-1)/2=t,则x=2t+1
=(x->∞)lim
(-1)^x
(t->∞)
lim
(1+1/t)^(2t+1)
=(x->∞)lim
(-1)^x
(t->∞)
lim
(1+1/t)^t(2+1/t)
利用重要极限(t->∞)lim(1+1/t)^t=e
=(x->∞)lim
(-1)^x
(t->∞)
lim
e^(2+1/t)
=(x->∞)lim
(-1)^x
e^2
当x为奇数时,原式=-e^2
当x为偶数时,原式=e^2
两种情况下极限不相等,所以原式不存在极限
[(1+x)/(1-x)]^x
=(x->∞)lim
[(-1)(x+1)/(x-1)]^x
=(x->∞)lim
(-1)^x
[((x-1+2)/(x-1)]^x
=(x->∞)lim
(-1)^x
[1+2/(x-1)]^x
令(x-1)/2=t,则x=2t+1
=(x->∞)lim
(-1)^x
(t->∞)
lim
(1+1/t)^(2t+1)
=(x->∞)lim
(-1)^x
(t->∞)
lim
(1+1/t)^t(2+1/t)
利用重要极限(t->∞)lim(1+1/t)^t=e
=(x->∞)lim
(-1)^x
(t->∞)
lim
e^(2+1/t)
=(x->∞)lim
(-1)^x
e^2
当x为奇数时,原式=-e^2
当x为偶数时,原式=e^2
两种情况下极限不相等,所以原式不存在极限
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