哪错了，求助

都是自学吗? 字那么丑

x->0
√(1+x^2) = 1+(1/2)x^2+o(x^3)
x+√(1+x^2) = 1+x+(1/2)x^2+o(x^3)
ln[x+√(1+x^2)]
=[x+(1/2)x^2] -(1/2)[x+(1/2)x^2]^2 +(1/3)[x+(1/2)x^2]^3+o(x^3)
=[x+(1/2)x^2] -(1/2)[x^2+x^3+o(x^2)]+(1/3)[x^3+o(x^3)]+o(x^3)
=x -(1/2+1/3)x^3 +o(x^3)
=x -(1/6)x^3 +o(x^3)
[x+√(1+x^2)]^(1/x)
=e^{ln[x+√(1+x^2)]/x}
=e^{[x -(1/6)x^3 +o(x^3)]/x }
=e^[1-(1/6)x^2 +o(x^2)]
[x+√(1+x^2)]^(1/x)/e
=e^[-(1/6)x^2 +o(x^2)]
=1 -(1/6)x^2 +o(x^2)
lim(x->0) [x+√(1+x^2)]^(1/x^2)/ e^(1/x)
=lim(x->0) { [x+√(1+x^2)]^(1/x)/e }^(1/x)
=lim(x->0) [1 -(1/6)x^2]^(1/x)
=1