Question

当x趋于0 时x分之一的tanx次方的极限是

Answer 1

lim(x->0+) (1/x)^tanx
=lim(x->0+) e^{ln[(1/x)^tanx]}
=lim(x->0+) e^{ -tanxlnx }
= e^{ - lim(x->0+) (tanx/x) [(lnx)/(1/x)] }
= e^{ - lim(x->0+) (tanx/x) [(1/x)/(-1/x²)] }
= e^{ - 1*0 }
= 1
【解二：由 lim(x->0+) x^x = 1 】
lim(x->0+) (1/x)^tanx
=lim(x->0+) { 1/(x^x)}^(tanx/x)
= {1/1}^1
= 1

Answer 2

x-0,(1/x)^tanx
x-0,1/x-无穷
tanx-tan0=0
无穷^(0)
令u=(1/x)^tanx

lnu=tanxln(1/x)
limlnu=lnlimu=limtanxln(1/x)=limtanx/(1/ln(1/x)
x-0,分子tanx-tan0=0,1/-lnx=-(lnx)^(-1),-(-1)(lnx）^(-2)x1/x=1/(xln^2x)
分母-1/ln无穷-0
0/0型
sec^2x/1/(xln^2x)
=sec^2x*xln^2x。
x-0