Question

求极限，要步骤，在线等

Answer 1

(1)
lim(x->0) (tanx - sinx)/ln(1+x^2) (0/0)
=lim(x->0) [(secx)^2 - cosx]/[2x/(1+x^2)]
=lim(x->0) (1+x^2) [(secx)^2 - cosx]/(2x) (0/0)
=lim(x->0) {(1+x^2) [2(secx)^2.tanx + sinx] +2x[(secx)^2 - cosx] }/2
=0

(2)
lim(x->0) sin(x^n)/(sinx)^m

case 1: n=m
lim(x->0) sin(x^n)/(sinx)^m
=lim(x->0)x^n/x^m
=1

case 2: n>m
lim(x->0) sin(x^n)/(sinx)^m
=lim(x->0)x^n/x^m
=lim(x->0)x^(n-m)
=0

case 3: n<m
lim(x->0) sin(x^n)/(sinx)^m
=lim(x->0)x^n/x^m
->∞

追问

亲，你第一题答案错了，是二分之一，还有，你好像有一道题没做

追答

(1)
lim(x->0) (tanx - sinx)/ln(1+x^2) (0/0)
=lim(x->0) [(secx)^2 - cosx]/[2x/(1+x^2)]
=lim(x->0) (1+x^2) [(secx)^2 - cosx]/(2x) (0/0)
=lim(x->0) {(1+x^2) [2(secx)^2.tanx + sinx] +2x[(secx)^2 - cosx] }/2
=0
没有错

lim(x->π/2) (1+cotx)^(2tanx)

let
1/y = cotx
x->π/2 , y->∞

lim(x->π/2) (1+cotx)^(2tanx)
=lim(y->∞) (1+1/y)^(2y)
=e^2