●■◆关于极限的数学题（弱智级）◆●■

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)=25。x→0lim(2arcsinx)/3x(用罗比达法则)=x→0lim(2/√(1-x^2))/3=2/36。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