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求极限lim(x→∞)(sin1/x-cos1/x)^x
人气:380 ℃ 时间:2020-03-12 06:49:22
解答
x→∞
1/x→0
cos1/x→1
sin1/x~1/x
lim(x→∞)(sin1/x-cos1/x)^x
=lim(x→∞)(1/x-1)^x
=-lim(x→∞)(1-1/x)^x
=-lim(x→∞)(1+(-1/x)^(-x)*(-1)
=-e^(-1)
=-1/e1/x→0cos1/x→1sin1/x~1/xlim(x→∞)(sin1/x-cos1/x)^x这个是根据等价无穷小的代换吗是啊,x→∞1/x→01/x就是无穷小可是等价无穷小好像不可以直接用在相加或相除的极限式里面 ,只可以用在乘除的运算中括号内只有一个其实是可以的lim(x→∞)(sin1/x-cos1/x)^x=lim(x→∞)(sin1/x-1)^x=-lim(x→∞)(1-sin1/x)^x=-lim(x→∞)(1+(-sin1/x)]^1/(-sin1/x)*(-sin1/x)*x=-lim(x→∞)e^(-sin1/x)/(1/x)=-lim(1/x→0)e^(-sin1/x)/(1/x)=-e^(-1)=-1/e
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