早教吧作业答案频道 -->其他-->
limx-->0(1/x^2-1/tanx^2)x趋近与0x平方分之一减去tanx平方分之一
题目详情
limx-->0 (1 /x^2 - 1/tanx^2) x趋近与0 x平方分之一减去tanx平方分之一
▼优质解答
答案和解析
lim(x-->0) (1 /x^2 - 1/tanx^2)
=lim(x-->0) (tan^2x-x^2) /(x^2 *tanx^2)
=lim(x-->0) (tan^2x-x^2) /(x^4)
=lim(x-->0) (sin^2x-cos^2x*x^2) /(cos^2x*x^4)
=lim(x-->0) (sin^2x-cos^2x*x^2) /(x^4) (0/0)
=lim(x-->0) (2sinxcosx+2cosxsinx*x^2-2cos^2x*x) /(4x^3)
=lim(x-->0) (sinx+sinx*x^2-cosx*x) /(2x^3) (0/0)
=lim(x-->0) (cosx+cosx*x^2+2xsinx+sinx*x-cosx) /(6x^2)
=lim(x-->0) (cosx*x^2+3xsinx) /(6x^2)
=lim(x-->0) (cosx*x+3sinx) /(6x) (0/0)
=lim(x-->0) (-sinx*x+cosx+3cosx) /(6)
=2/3
=lim(x-->0) (tan^2x-x^2) /(x^2 *tanx^2)
=lim(x-->0) (tan^2x-x^2) /(x^4)
=lim(x-->0) (sin^2x-cos^2x*x^2) /(cos^2x*x^4)
=lim(x-->0) (sin^2x-cos^2x*x^2) /(x^4) (0/0)
=lim(x-->0) (2sinxcosx+2cosxsinx*x^2-2cos^2x*x) /(4x^3)
=lim(x-->0) (sinx+sinx*x^2-cosx*x) /(2x^3) (0/0)
=lim(x-->0) (cosx+cosx*x^2+2xsinx+sinx*x-cosx) /(6x^2)
=lim(x-->0) (cosx*x^2+3xsinx) /(6x^2)
=lim(x-->0) (cosx*x+3sinx) /(6x) (0/0)
=lim(x-->0) (-sinx*x+cosx+3cosx) /(6)
=2/3
看了 limx-->0(1/x^2...的网友还看了以下：
设M={y∈R|y=x平方+1,x∈N},P={y∈R|y=-x平方+1,x∈N},求M∩P.写出 2020-04-12 …
已知X*m次方=6,X*n次方=3,则X*2m-n次方的值为多少 2020-05-14 …
设f(x)=(1+x)的m次方+(1+x)n次方展开式中x系数是10(m,n属于N+)1.求f(x 2020-05-14 …
方程x^n+a1x^(n-1)+……+an-1x+an=0的根设方程x^n+a1x^(n-1)+… 2020-06-22 …
列举法表示下列集合A={x∈N|（9-x）分之9∈N}C={y|y=负x方+6,x∈N,y∈N}D 2020-08-01 …
整式方程n(nx-1)=x+1解法如下：n的平方x-n=x+1(n方-1)x=n+1(n+1)(n 2020-08-02 …
数列的极限不知道的不要乱回答证明：方程x^n+x^(n-1)+……+x^2+x=1,在（0,1）上 2020-08-02 …
四元n次不定方程问题∶给出一个四元不定方程x^n+y^n+z^n=t^n我要问当实数n满足什么条件 2020-08-02 …
求证当整数n>2时,关于x,y,z的不定方程x^n+y^n=z^n.（(x,y)=(x,z)=(y 2020-08-02 …
证明方程x^n+x^(n-1)+……x^2+x=1(n>=2),在(0,1)内存在唯一实数根x0,并 2020-11-01 …