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limx→0 5x%sin3x/tAn2x

用洛必达法则处理之,如上图,望采纳/

等价无穷小替换:lim(x→0)(tan2x-sin3x)/x=lim(x→0) tan2x/x - lim(x→0) sin3x/x=lim(x→0) 2x/x - lim(x→0) 3x/x=2-3=-1

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

解: X→ 0 tan2x/sin3x =2x/3x =2/3 极限为: 2/3

这是一个 0/0 型的极限,可以使用罗必塔法则来计算: =lim[1+sin(3x)]' /[tan(2x)]' =lim{3*cos(3x)/[1+sin(3x)]} / {2*[sec(2x)]^2} =lim[3*cos0/(1+sin0)] /[2*(sec0)^2] =lim(3/1) /(2 * 1) =3/2

用洛必达法则求 =(tan2x+x^2)'/(sin3x)'=(2(sec2x)^2+2x)/(3cos3x) =(2+0)/3=2/3 x→0

tan∧2x/2是什么意思?

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

=lim tan²x / sin3x =lim x² / (3x) =lim x/3 =0

罗比达法则,求导 1:[1+tan^2_(3x)]*3-2cos2x=1 2:(a-bcos(bx))/k(1+tan^2_kx)=(a-b)/k

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