Question

第二题和第四题，求解？

Answer 1

(2)
lim(x->0) [e^(ax) - 1]/(2x) =5
lim(x->0) ax/(2x) =5
a/2 =5
a=10
(4)
x->0
sinx = x-(1/6)x^3 +o(x^3)
x-sinx = (1/6)x^3 +o(x^3)
lim(x->0) (x-sinx)/[x^2.sinx]
=lim(x->0) (x-sinx)/x^3
=lim(x->0) (1/6)x^3/x^3
=1/6

Answer 2

lim(x→0)[(e^ax-1)/2x=5
lim(x→0)[(a·e^ax)/2=5
a=10
lim(x→0)[(x-sinx)/x²sinx
=lim(x→0)[(1-cosx)/(2xsinx+x²cosx)
=lim(x→0)[(sinx)/(2sinx+2xcosx+2xcosx-x²sinx)
=lim(x→0)[(1)/(2+2cosx+2cosx-x²)
=1/6