Question

高数极限题求详解，谢谢

求x趋正无穷时 ， x^(1/2) [x^(1/x)-1] 极限。

谢谢～

Answer 1

lim x^(1/2) [x^(1/x)-1]=lim [x^(1/x)-1]/[x^(-1/2)]= lim [x^(1/x)-1]'/[x^(-1/2)]'
=lim x^(1/x)*(1-logx)/x^2/[-0.5*x^(-1.5)]
=lim x^(1/x)* lim (1-logx)/x^2/[-0.5*x^(-1.5)]
=1*0=0

追问

请问lim (1-logx)/x^2/[-0.5*x^(-1.5)] =0 是怎么得出的呢 ？

抱歉没太看懂……

追答

(1-logx)/x^2/[-0.5*x^(-1.5)](化简)
=2(logx-1)/x^0.5
而lim(logx-1)/x^0.5
=limlog(x)/x^0.5-lim1/x^0.5
=lim[log(x)]'/[x^0.5]'-0
=lim[1/x]/[0.5x^(-0.5)]
=lim2*1/x^0.5
=0

Answer 2

不好意思，那个二分之一后的式子是怎么个关系，与二分之一？