请问，如图所示，limx→0（cosx）［sin（sin²x)/3x²+4x³ 极限怎么算？

limx→0 2sin²x/x²?

lim(x→0) sinx^2=0 lim(x→0) (sinx)^2=0 lim(x→0) sinx^2/(sinx)^2=lim(x→0) (sinx^2)'/[(sinx)^2]'=lim(x→0) (2xcosx^2)/(2sinxcosx)=lim(x→0) 2xcosx^2/(sin2x)=lim(x→0) cosx^2/(sin2x/2x) lim(x→0) cosinx^2=1 limx→0 sin2x/(2x)=1=1

limx趋近于0,x²+3x/sin²x

解:(x→0)lim[(x²+3x)/sin²x]=(x→0)lim{(1+3/x)/[(sinx)/x]²}=(x→0)lim(1+3/x)/(x→0)lim[(sinx)/x]²=(1+0)/1=1

解limx→0[cos(sinx) - cosx]/x^4

^等价无穷小替换:cosx~来1-x²/2,自当x→0时,sinx→0,则cos(sinx)~1-sin²x/2,limx→0[cos(sinx)-cosx]/x^zhidao4=limx→0[1-sin²x/2-cosx]/x^4=limx→0[cos²x/2-cosx+1/2]/x^4=limx→0[1-cosx]²/2x^4=limx→0[x²/2]²/2x^4=1/8.

求极限lim(x→0)(1/sin²x - 1/x²cos²x)

lim(x→0) 2x-2sinxcosx / 4x³ = lim(x→0) 1-cosx / 6x² 这一步,罗比达法则,计算错误 应该是 lim(x→0) 2x-2sinxcosx / 4x³=lim(x→0) 2x-sin2x / 4x³=lim(x→0) 1-cos2x / 6x²=lim(x→0) 2sin²x / 6x²=1/3

求极限lim(x,y)→(0,0) sin(x³+y³)/x²+y²

极限不存在假设以y=kx趋近(0,0)原极限=lim{x,y→0}kx^2/[(k^2+1)x^2]=k/(k^2+1)显然k不同极限不同,所以极限不存在

lim(x→0)(sin²x/x²)

sin(sinx)/x=sin(sinx)/sinx*sinx/x 当xx→0=1

求极限lim(x,y)→(0,0) sin(x³+y³)/x²+y²

设y=kx即可,不明白的话我可以做了拍照给你

limx - >0 (sinx - xcosx)/x^3 极限 0.0谢谢

罗比达法则解法.解:原式=lim(x->0)[(sinx-xcosx)/(sinx)^3] =lim(x->0)[(cosx-cosx+xsinx)/(3sin²x)] (0/0型极限,应用罗比达法则) =lim(x->0)[x/(3sinx)] (化简) =(1/3)lim(x->0)(x/sinx) =(1/3)*1 (应用重要极限lim(x->0)(sinx/x)=1) =1/3.

当x趋于0时,lim{x²/(sin²x/5)}

x/5趋于0 所以sinx/5~x/5 所以原式=limx²/(x/5)²=25

limx→0[sinx - sin(sinx)]sinx/x^4

limx→0[sinx-sin(sinx)]sinx/x^4=limx→0[sinx-sin(sinx)]/x^3=limx→0[cosx-cos(sinx)*cosx]/3x^2(洛必达)=limx→0[1-cos(sinx)]cosx/3x^2=limx→0[1-cos(sinx)]/3x^2(洛必达)=limx→0sin(sinx)cosx/6x=limx→0sin(sinx)/6x=limx→0sinx/6=1/6