lim(x→0+)[(x^x-1)/xlnx] 这个极限怎么求啊?
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lim(x→0+)[(x^x-1)/xlnx] 这个极限怎么求啊?
lim(x→0+)[(x^x-1)/xlnx]
=lim(x→0+)[(e^(xlnx)-1)/xlnx]
=lim(x→0+)[xlnx/xlnx]
=1
e^x-1和x是等价无穷小
lim(x→0+)(xlnx)
=lim(x→0+)(lnx/(1/x))
=lim(x→0+)((1/x)/(-1/x^2))
=lim(x→0+)(-x)
=0
可令t=xlnx,
lim(x→0+)[(x^x-1)/xlnx]
=lim(x→0+)[(e^(xlnx)-1)/xlnx]
=lim(x→0+)[(x^x-1)/...
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lim(x→0+)(xlnx)
=lim(x→0+)(lnx/(1/x))
=lim(x→0+)((1/x)/(-1/x^2))
=lim(x→0+)(-x)
=0
可令t=xlnx,
lim(x→0+)[(x^x-1)/xlnx]
=lim(x→0+)[(e^(xlnx)-1)/xlnx]
=lim(x→0+)[(x^x-1)/xlnx]
=lim(x→0+)[(e^(xlnx)-1)/xlnx]
=lim(t→0+)[(e^t-1)/t)
=lim(t→0+)e^t)
=1
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令t=xlnx,
lim(x→0+)[(x^x-1)/xlnx]
=lim(x→0+)[(e^(xlnx)-1)/xlnx]
=lim(x→0+)[(x^x-1)/xlnx]
=lim(x→0+)[(e^(xlnx)-1)/xlnx]
=lim(t→0+)[(e^t-1)/t)
=lim(t→0+)e^t)
=1
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