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lim(x→0)[x^4∕∫(0,sinx)ln(1+t)dt]的值
人气:487 ℃ 时间:2020-02-01 00:58:05
解答
当x→0时,ln(1 + sinx) sinx x
lim(x→0) x^4/∫(0→sinx) ln(1 + t) dt
= lim(x→0) 4x^3/[cosxln(1 + sinx)]
= lim(x→0) 4x^3/(cosxsinx)
= lim(x→0) 4x^3/x
= lim(x→0) 4x^2
= 0
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