Question

一道高数极限题求解！

Answer 1

lim(n->∞){ [ n^3 -n^2 + (1/2)n] .e^(1/n) - √(1+n^6) }

consider

let

y=1/x

y->0

e^y = 1+y+(1/2)y^2 +(1/6)y^3 +o(y^3)

[ 1 -y + (1/2)y^2] .e^y 

=[ 1 -y + (1/2)y^3] .[1+y+(1/2)y^2 +(1/6)y^3 +o(y^3) ]

=[ 1+y+(1/2)y^2 +(1/6)y^3 ] - y[1+y+(1/2)y^2 +(1/6)y^3 ]

+(1/2)y^2.[ 1+y+(1/2)y^2 +(1/6)y^3 ]  +o(y^3)

=[ 1+y+(1/2)y^2 +(1/6)y^3 ] + [-y-y^2-(1/2)y^3+o(y^3) ]

+[ (1/2)y^2+(1/2)y^3+o(y^3) ]  +o(y^3)

=1 + (1/6)y^3 +o(y^3)

√(1+y^6) = 1 +(1/2)y^6 +o(y^6) = 1+ o(y^3)

[ 1 -y + (1/2)y^2] .e^y - √(1+y^6)  =  (1/6)y^3 +o(y^3)

lim(x->∞){ [ x^3 -x^2 + (1/2)x] .e^(1/x) - √(1+x^6) }

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

=lim(y->0) (1/6)y^3 / y^3

=1/6

=>

lim(n->∞){ [ n^3 -n^2 + (1/2)n] .e^(1/n) - √(1+n^6) } =1/6

Answer 2