lim(x->∞)[(x+2)/(x-1)]^mx=e^(-4) 則m
lim(x->∞)[(x+2)/(x-1)]^mx=e^(-4) 則m
其他人氣：369 ℃時(shí)間：2020-05-14 05:08:32
優(yōu)質(zhì)解答
[(x+2)/(x-1)]^mx=[(x-1+3)/(x-1)]^mx=[1+3/(x-1)]^mx令a=(x-1)/3x=3a+1[1+3/(x-1)]^mx=(1+1/a)^m(3a+1)=(1+1/a)^3ma*(1+1/a)=[(1+1/a)^a]^(3m)*(1+1/a)x趨于0則a趨于0,所以1+1/a極限是1[(1+1/a)^a]^(3m)極限=e^(3m)...
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