www.whkt.net > 求极限lim(0—∞)sin^3x/x^3

求极限lim(0—∞)sin^3x/x^3

lim(x→0)(sinx-xcosx)/(sin^3x)=lim(x→0)[(sinx-xcosx)]'/(sin^3x)'=lim(x→0)(cosx-cosx+xsinx)/[3(sin^2x)cosx]=lim(x-0) x/[3(sinxcosx)]=lim(x-0) x/[3sin(2x)/2]=lim(x-0) x'/[3sin(2x)/2]'=lim(x-0) 1/[3cos(2x)]=1/3

lim(x→0)sin^3x/tanx^3用等价无穷小代换=lim(x→0)3x/x^3用洛必达法则=lim(x→0)3/3x^2=无穷

洛必达法则,lim(3x-sin3x)/x=lim(3-3cos3x)/3x=lim(9sin3x)/6x=lim27cos3x/6=27/6=9/2

x→0则x→0所以sinx~x所以原式=lim(x→0)x/x=1

原式=(5x+1/x)/(3-1/x)*sin(1/x)=(5+1/x)/(3-1/x)*x*sin(1/x)=(5+1/x)/(3-1/x)*sin(1/x)/(1/x) x→∞1/x→0 所以sin(1/x)/(1/x)极限是1 所以原来极限=5/3

lim sin(3x-3)/sin(x-1) x->1 =lim sin(3t)/sin(t) t->0=lim(3t/t)t->0=3

lim sin(x^3)/(sinx)^3=lim[ sin(x^3)*x^3/x^3]/(sinx*x/x)^3=limx^3/x^3=1

lim(x ->0)sin x /x ^3+3x=lim(x ->0)sin x /[x(x ^2+3)]=lim(x ->0)x /[x(x ^2+3)]=lim(x ->0)1 /(x ^2+3)=1/3

limx→0(1/sin^3x-1/x^3)=limx→0[(x^3-sin^3x)/(xsinx)^3]=limx→0{[x^3-(x-x^3/6)^3]/x^6]}=limx→0{(x^5/2-x^7/12+x^9/216)/x^6]}=limx→0{(1-x^2/12+x^4/216)/x]}→无穷,极限不存在,是否看错了题目?把平方看成了立方,把立方改成平方:lim(x→0)(1/

因为sinx~x是不对的,此处不能忽略高次项,应该是sinx~(x-x^3/6)

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