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求极限(sin x+3x)/(tAn x+2x)趋近于0

lim(x->0) (sinx +3x)/(tanx +2x) =lim(x->0) (x +3x)/(x +2x) =4/3

分子分母同时除以x 原式=lim(x~0)【1-(sin2x/x)】/【1+tan3x/x】 =(1-2)/(1+3) =-1/4

lim (sin2x)/(tan2x) x→0 =lim (sin2x)/(sin2x/cos2x) x→0 =lim cos2x x→0 = cos0 = 1

解:lim(x->0)[tan(3x)/(2x)] =lim(x->0)[((3/2)/cos(3x))*(sin(3x)/(3x))] ={lim(x->0)[(3/2)/cos(3x)]}*{lim(x->0)[sin(3x)/(3x)]} =(3/2)*1 (应用重要极限lim(z->0)(sinz/z)=1) =3/2。

 limsin3x/tan5x =lim3cos3x/[5(sec5x)^2] =(3/5)limcos3x(cos5x)^2 =(3/5)cos3π(cos5π)^2 =-3/5 limtanx/x=limx/x=1

x→0则2x→0,3x→0 所以sin2x和2x是等价无穷小 tan3x和3x是等价无穷小 所以原式=lim(x→0)(2x/3x)=2/3

limx趋近于0 (tan3x-sin2x)/x=limx趋近于0 ( 3tan3x sec^2 x-2cos2x)=0-2=-2 用罗必塔法则

lim arctan3x / sin2x 上下同时除以x =lim arctan3x/x / sin2x/x =(3/2)*lim arctan3x/3x / sin2x/2x 因为, lim sin2x/2x=1(重要的极限) lim arctan3x/3x 换元3x=t, =lim arctant/t 再换元t=tanu =lim u/tanu =lim u/sinu * lim cosu =1*1 =...

=lim(sin3x+2xcos3x)/cos3x(sin2x+3x) =lim(3sin3x/3x+2cos3x)/cos3x(2sin2x/2x+3) =(3+2)/1(2+3)=1

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