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limx→0+[∫(0→x^2)t^(3/2)dt]/[∫(0→x)t(t-sint)dt]
人气:277 ℃ 时间:2020-01-28 22:18:09
解答
答:
lim ( x→0+) [ ∫(0→x^2) t^(3/2)dt] / [ ∫(0→x)t(t-sint)dt ] (0---0型可导用洛必达法则)
=lim (x→0+) [ (x^2)^(3/2)*(x^2)' ] / [ x(x-sinx) ]
=lim(x→0+) (2x^4) / [x(x-sinx)]
=lim(x→0+) 2(x^3) /(x-sinx) 再次应用洛必达法则
=lim(x→0+) 6(x^2) /(1-cosx) 再次应用
=lim(x→0+) 12x/(sinx)
=12
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