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lim当x趋近于零时,tAn3x/2x

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。

解:lim(x->0)[tan(3x)/(2x)] =lim(x->0)[(tan(3x))'/(2x)'] (0/0型极限,应用洛必达法则) =lim(x->0)[3(secx)^2/2] =3/2。

这种做法是错的 等价替换适用于乘、除,不适用加减。当你学到泰勒公式之后,你就明白为什么。 (x→0)lim(tan3x-sin2x)/sin5x=(x→0)lim(tan3x-sin2x)/5x =(x→0)lim(3sex3x-2cos2x)/5=(x→0)lim(3-2)/5=1/5

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(x→π/4) tan2x*tan(π/4-x) =lim(x→π/4) 2tanx/(1-tan^2x)*[tan(π/4)-tan(x)]/[1+tan(π/4)*tan(x)] =lim(x→π/4) 2tanx/(1-tan^2x)*[1-tan(x)]/[1+tan(x)] =lim(x→π/4) 2tanx/(1+tanx)^2 =2tanπ/4/(1+tanπ/4)^2 =2/(1+1)^2 =2/4 =1/2

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

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

答: lim(x→0) sin(2x) / tan(3x) =lim(x→0) sin(2x) cos(3x) / sin(3x) =lim(x→0) sin(2x) / sin(3x) =lim(x→0) 2x /(3x) =2/3

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

泰勒公式。。。把上面和下面展开系数一除,结果出来了

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