求极限lim趋向0 积分【X,0】 tan^2 tdt/ln(1+x^3)
人气:424 ℃ 时间:2020-03-24 09:30:29
解答
lim(x→0)[(∫(0,x)tan²tdt)/ln(1+x³)] =lim(x→0) [d(∫(0,x)tan²tdt)/dx]/[dln(1+x³)/dx](洛比达法则)
=lim(x→0){(tan²x)/[3x²/(1+x³)]}(……d∫(0,x)f(t)dt / dx = f(x))
=lim(x→0)[(tan²x)(1+x²)/(3x²)]
=lim(x→0)[(tan²x)/x²]·lim(x→0)(1+x²)/3
=[(dtan²x/dx)/(dx²/dx)
=lim(x→0)[(2tanx)/(2x)]·1/3
=lim(x→0)[(dtanx/dx)/(dx/dx)]·lim(x→0)sec²x·1/3(洛比达法则)
=lim(x→0)sec²x·1/3
=1/3
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