●■◆关于极限的数学题(弱智级)◆●■

1个回答

  • 1.x→0lim(tan2x-sinx)/x

    =x→0lim(2cosx/cos2x-1)*sinx/x =1

    (因为x→0时,2cosx/cos2x=2,sinx/x=1)

    2.x →0 lim(cosx-cos3x)/x的平方

    =x→0lim(cosx-cos3x)/x^2(罗比达法则,上下同时求导)

    =x→0lim(-sinx+3sin3x)/2x(再用一次罗比达法则)

    =x→0lim(-cosx+3cos3x)/2=1

    3.x→0lim[tan(2x+x的3次方)]/sin(x-x的平方)

    =x→0lim[tan(2x+x^3)]/sin(x-x^2)(用罗比达法则)

    =x→0lim[sec^2(2x+x^3)(2+3x^2)]/cos(x-x^2)(1-2x)

    =2 (x→0,sec^2(2x+x^3)=1,cos(x-x^2)=1)

    4.x→正无穷大lim x*sin(2/x)

    =x→正无穷大lim sin(2/x)/(1/x)

    =t(=1/x)→0lim2*sin(2t)/(2t)=2

    5.x→0lim(2arcsinx)/3x(用罗比达法则)

    =x→0lim(2/√(1-x^2))/3

    =2/3

    6.x→0lim(tanx-sinx)/sinx的3次方

    =x→0lim(tanx-sinx)/sin^3x(用罗比达法则)

    =x→0lim(sec^2x-cosx)/3sin^2xcosx

    =x→0lim(sec^2x-cosx)/3sin^2x(用罗比达法则)

    =x→0lim(2secxtanxsecx+sinx)/6sinxcosx(用罗比达法则)

    =x→0lim(2tanx+sinx)/6sinx(用罗比达法则)

    =x→0lim(2sec^2x+cosx)/6cosx

    =1/2