Question

数学问题如图

如图，求解

Answer 1

x->0

分子

sinx = x-(1/6)x^3 +o(x^3)

e^(sinx) 

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

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

= 1+x +(1/2)x^2  +o(x^3)

e^x = 1+ x +(1/2)x^2 + (1/6)x^3 +o(x^3)

e^x - e^(sinx)

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

= (1/6)x^3 +o(x^3)

分母

ln(1+x+x^2)

=(x+x^2)  -(1/2)(x+x^2)^2  +o(x^2)

=(x+x^2)  -(1/2)x^2 +o(x^2)

=x +(1/2)x^2  +o(x^2)

ln(1-x+x^2)

=(-x+x^2)  -(1/2)(-x+x^2)^2  +o(x^2)

=(-x+x^2)  -(1/2)x^2  +o(x^2)

=-x +(1/2)x^2 +o(x^2)

ln(1+x+x^2) +ln(1-x+x^2)

=[x +(1/2)x^2  +o(x^2)] +[-x +(1/2)x^2 +o(x^2)]

=x^2 +o(x^2)

x[ln(1+x+x^2) +ln(1-x+x^2)] = x^3 +o(x^3)

lim(x->0) [e^x - e^(sinx) ]/{ x[ ln(1+x+x^2) +ln(1-x+x^2) ] }

=lim(x->0) (1/6)x^3/ x^3

=1/6