原式=lim【x→0】[1+(arctanx-x)/x]^(1/x²)
=e^lim【x→0】(arctanx-x)/x³
=e^lim【x→0】{[1/(1+x²)]-1}/(3x²)
=e^lim【x→0】-1/[3(1+x²)]
=e^(-1/3)
不明白可以追问,如果有帮助,请选为满意回答!
先取自然对数
lim(x→0) ln[(arctanx/x)^(1/x^2)]
=lim(x→0) ln(arctanx/x)/x^2
=lim(x→0) [ln(arctanx)-lnx]/x^2 (0/0)
=lim(x→0) {1/[arctanx*(1+x^2)]-1/x}/(2x)
=lim(x→0) {[x-arctanx*(1+x^2)/[x*arctanx*(1+x^2)]}/(2x)
=lim(x→0) [x-arctanx*(1+x^2)]/(2x^3) (0/0)
=lim(x→0) [1-1-2x*arctanx]/(6x^2)
=lim(x→0) [-2x^2]/(6x^2)
=-1/3
所以
lim(x→0) [(arctanx/x)^(1/x^2)]
=lim(x→0) e^ln[(arctanx/x)^(1/x^2)]
=e^(-1/3)