用泰勒公式求极限limx→0tan(tanx)-sin(sinx)/tanx-sinx 详细过程? 考研高等数学 泰勒公式的应用lim(x趋近于0)(tan(t...
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sinx=x-x^3/6+o(x^3)
sin(sinx)=sinx-(sinx)^3/6+o(sinx)
=x-x^3/6+o(x^3)-(x-x^3/6+o(x^3))^3/6 +o(sinx)
=x-x^3/3+o(x^3)
tanx=x+x^3/3+o(x^3)
tan(tanx)=x+x^3/3+(x+x^3+o(x^3))^3/3+o(x^3)
=x+2/3 x^3+o(x^3)
lim(x-->0)(tan(tanx)-sin(sinx))/(x-sinx)
=lim(x-->0)(x+2/3 x^3+o(x^3)-x+1/3 x^3-o(x^3))/(1/6 x^3-o(x^3))
=lim(x-->0)(x^3+o(x^3))/(1/6 x^3-o(x^3))
=6
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具体回答如下:
分母 = sinx/cosx-sinx =sinx(1/cosx-1)=sinx(1-cosx)/cosx
分母是等价于 x/2的
对分子我们做等价变形
分子 = (tan(tanx)-tanx) +(tanx -sinx) +(sinx -sin(sinx))
令 p1 = lim (tan(tanx)-tanx)/(tanx -sinx)
lim (tan(tanx)-tanx)/(x³/2)
再令 f(x)=tanx
则p1的分子是 f(tanx)-f(x)=f'(c)(tanx -x)(这里用了中值定理,c在x与tanx之间)
当 x→0时,c→0,f'(c)=sec²c→1
p1 = lim (tanx-x)/ (x³/2)=2/3
p2 = lim (tanx -sinx)/(tanx - sinx)=1
p3 = (sinx -sin(sinx))/(tanx-sinx)=(sinx -sin(sinx))/(x³/2)
所以原式=p1+p2+p3 =2
极限的意义:
和实数运算的相容性,譬如:如果两个数列{xn} ,{yn} 都收敛,那么数列{xn+yn}也收敛,而且它的极限等于{xn} 的极限和{yn} 的极限的和。
与子列的关系,数列{xn} 与它的任一平凡子列同为收敛或发散,且在收敛时有相同的极限;数列{xn} 收敛的充要条件是:数列{xn} 的任何非平凡子列都收敛。
简单计算一下即可,答案如图所示
sinx=x-x^3/6+o(x^3)
sin(sinx)=sinx-(sinx)^3/6+o(sinx)
=x-x^3/6+o(x^3)-(x-x^3/6+o(x^3))^3/6 +o(sinx)
=x-x^3/3+o(x^3)
tanx=x+x^3/3+o(x^3)
tan(tanx)=x+x^3/3+(x+x^3+o(x^3))^3/3+o(x^3)
=x+2/3 x^3+o(x^3)
tanx-sinx=x^3/2+o(x^3)
所以求极限
=lim(x-->0)(x^3+o(x^3))/(x^3/2 x^3-o(x^3))
=2.
大概过程就是如此,满意请采纳,
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