Question

这道题怎么算啊

Answer 1

lim(x->0) {∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) } ^(1/x^2)
x->0
e^[(1/2)t^2] ~ 1+ (1/2)t^2

∫(0->x^(2/3)) e^[(1/2)t^2] dt
~∫(0->x^(2/3)) (1+ (1/2)t^2) dt
= [ t + t^3]| (0->x^(2/3))
= x^(2/3) + x^2
∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) ~ 1+x^2

lim(x->0) {∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) } ^(1/x^2)
=lim(x->0) (1+x^2) ^(1/x^2)
=e

追问

答案为e∧(1/6)

追答

lim(x->0) {∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) } ^(1/x^2)
x->0
e^[(1/2)t^2] ~ 1+ (1/2)t^2

∫(0->x^(2/3)) e^[(1/2)t^2] dt
~∫(0->x^(2/3)) (1+ (1/2)t^2) dt
= [ t + (1/6)t^3]| (0->x^(2/3))
= x^(2/3) + (1/6)x^2

∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) ~ 1+x^2
lim(x->0) {∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) } ^(1/x^2)
=lim(x->0) (1+(1/6)x^2) ^(1/x^2)
=e^(1/6)

案为e∧(1/6)
lim(x->0) {∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) } ^(1/x^2)
x->0
e^[(1/2)t^2] ~ 1+ (1/2)t^2

∫(0->x^(2/3)) e^[(1/2)t^2] dt
~∫(0->x^(2/3)) (1+ (1/2)t^2) dt
= [ t + (1/6)t^3]| (0->x^(2/3))
= x^(2/3) +(1/6) x^2

∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) ~ 1+(1/6)x^2

lim(x->0) {∫(0->x^(2/3)) e^[(1/2)t^2] dt + 1 - x^(2/3) } ^(1/x^2)
=lim(x->0) (1+(1/6)x^2) ^(1/x^2)

=e^(1/6)