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设f(0)=0,f'(0)=6,求lim(x趋近于0)=(f(x-sinx))/x^3=?
人气:199 ℃ 时间:2020-04-10 06:48:27
解答
lim(x趋近于0)=(f(x-sinx))/x^3 (分子趋于f(0)=0分母趋于0,罗比达法则)=lim(x->0) f'(x-sinx)*(1-cosx)/3x^2=lim(x->0) f'(x-sinx)*lim(x->0) (1-cosx)/3x^2=f'(0)*lim(x->0) {1-[1-2(sinx/2)^2}/3x^2=6*lim (x->0...
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