Question

洛必达法则求极限:x趋向于0,lim(sin x/x)的1/x²次幂

Answer 1

J(x) = [(sinx)/x]^(1/x^2)

lim(x->0) lnJ(x) = lim(x->0) ln[(sinx)/x] / x^2

= lim(x->0)(xcosx - sinx)/(2x^2sinx)

= lim(x->0) (cosx-xsinx-cosx) / 2(2xsinx+x^2cosx)

= 0.5 lim(x->0) -sinx / (2sinx+xcosx)

= -0.5 lim(x->0) cosx /(2cosx+cosx-xsinx)

= -0.5 (1/3)

= - 1/6 //:lnJ=-1/6

因此：lim(x->) [(sinx)/x]^(1/x^2) = e^(-1/6)