Question

求解三道高等数学极限计算。

佛了，这仨题都不会，大佬给个过程，谢谢😭

Answer 1

(5)=1

追答

追问

谢谢，感谢老师解答😊😊😊

Answer 2

(5)
L =lim(x->+∞) ( π/2- arctanx ) ^(1/lnx)

lnL
=lim(x->+∞) ln( π/2- arctanx ) /lnx (∞/∞分子分母分别求导）
=lim(x->+∞) { -1/[( π/2- arctanx ).(1+x^2)] }/ (1/x)
=lim(x->+∞) -x/[( π/2- arctanx ).(1+x^2)]
=lim(x->+∞) -x^2/(1+x^2) .lim(x->+∞) 1/ { x[( π/2- arctanx )] }
=-lim(x->+∞) 1/ { x[( π/2- arctanx )] }
=-lim(x->+∞) (1/x) / [( π/2- arctanx )] (0/0分子分母分别求导）
=-lim(x->+∞) (-1/x^2) / [ -1/(1+x^2)]
=-lim(x->+∞) (1+x^2)/x^2
= -1
L = e^(-1)
lim(x->+∞) ( π/2- arctanx ) ^(1/lnx)= e^(-1)
(6)
let
y=1/x
y->0
cosy ~ 1 -(1/2)y^2
lim(x->+∞) [ cos(1/x) ]^(x^2)
=lim(y->0) [ cosy ]^(1/y^2)
=lim(y->0) [ 1-(1/2)y^2 ]^(1/y^2)
=e^(-1/2)
(7)
x->0
分母
sinx = x-(1/6)x^3 +o(x^3)
x-sinx = (1/6)x^3 +o(x^3)
分子
sinx = x-(1/6)x^3 +o(x^3)
e^(sinx)
=e^[x-(1/6)x^3 +o(x^3)]
= 1+ [x-(1/6)x^3 ] +(1/2) [x-(1/6)x^3] ^2 +(1/6)[x-(1/6)x^3] ^3 +o(x^3)
= 1+ [x-(1/6)x^3] +(1/2)[x^2+o(x^3)]^2 +(1/6)[x^3 +o(x^3)] +o(x^3)
=1 +x +(1/2)x^2 + o(x^3)
e^x = 1 +x +(1/2)x^2 +(1/6)x^3+ o(x^3)
e^x - e^(sinx)
= 1 +x +(1/2)x^2 +(1/6)x^3+ o(x^3) -[1 +x +(1/2)x^2 + o(x^3)]
=(1/6)x^3+ o(x^3)
lim(x->0) [ e^x - e^(sinx) ]/(x-sinx)
=lim(x->0) (1/6)x^3/[(1/6)x^3]
=1