求极限limx→0(∫(x–0)sin(xt)^2dt)/x^5

Question

求极限limx→0 (∫(x–0) sin(xt)^2dt)/x^5

Answer 1

先对分子做变换
∫[0→x] sin(xt)^2 dt
令xt=u,则dt=du/x,u：0→x^2
=∫[0→x^2] sin(u^2) du /x
则原极限化为：
lim[x→0] ∫[0→x^2] sin(u^2) du / x^6
洛必达法则
=lim[x→0] 2xsin(x^4) / (6x^5)
=lim[x→0] 2x^5 / (6x^5)
=1/3
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