作业帮计算极限lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)请给详细步骤！！！！！
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计算极限lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)请给详细步骤！！！！！
计算极限lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)
请给详细步骤！！！！！
计算极限lim→0+ [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)请给详细步骤！！！！！
1、分母用等价代换：1-cosx~(1/2)x²；
2、然后用罗比达法则,分子分母同时求导；
lim（x→0+) [∫(上限x,下限0)ln(t+e^t)dt] / (1-cosx)
=lim（x→0+) [∫(上限x,下限0)ln(t+e^t)dt] / [(1/2)x²]
=lim（x→0+) [ln(x+e^x)] / x
3、继续求导：
=lim（x→0+) [(1+e^x)/(x+e^x)] /1
=lim（x→0+) [(1+e^x)/(x+e^x)]
4、取极限：
=2
lim(x->0+) ∫(0到x) [ln(t + e^t) dt]/(1 - cosx)
= lim(x->0+) [d/dx ∫(0到x) (ln(t + e^t))]/[d/dx (1 - cosx)]
= lim(x->0+) [ln(x + e^x)]/sinx
= lim(x->0+) [d/dx ln(x + e^x)]/(d/dx sinx)
= l...
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lim(x->0+) ∫(0到x) [ln(t + e^t) dt]/(1 - cosx)
= lim(x->0+) [d/dx ∫(0到x) (ln(t + e^t))]/[d/dx (1 - cosx)]
= lim(x->0+) [ln(x + e^x)]/sinx
= lim(x->0+) [d/dx ln(x + e^x)]/(d/dx sinx)
= lim(x->0+) [(1 + e^x)/(x + e^x)]/cosx
= lim(x ->0+) (1 + e^x)/[(x + e^x)cosx]
= (1 + 1)/[(0 + 1)(1)]
= (2)/(1)
= 2
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