高数六大定理(下)

1、200000()( )()()()()2!fxf xf xfxxxxx( )00()()( ),!nnnfxxxR xn1.带拉格朗日型余项2.带皮亚诺型余项200000()( )()()()()2!fxf xf xfxxxxx( )000()()() !nnnfxxxo xxn2(0)( )(0)(0)2!ff xffxx( )(0)()!nnnfxo xn1,sin ,cos ,ln(1),1xexxxx泰勒公式的主要应用：建立函数与高阶导数的关系.1.带“皮式”余项的泰勒公式：用于计算：求极限找等价无穷小；确定无穷小的阶；展开原则：AB上下同阶原则 AB阶数最低原则sin xx331()

2、3! xxxo x331()6 xo x无穷小的运算：232()()()o xo xo x235()()()o xo xo x235()()(0)kxo xo xk316x2220coslimln(1) 求xxxexxx【解析】22ln(1)()2 xxxo x22ln(1)()2 xxxo x+22ln(1)()2 xxxo x 424+ln(1)()2 xxxxo x2211()2! xexxo x244cos1()2!4! 又xxxo x2244221()22 2! xxxeo x=444041()12lim()2 原式xxo xxo x24421cos()12 xxexo x=4444

3、041()12lim()2 xo xxxo xx1.6 331sin6()xxo xx331arcsi()n6xo xxx331tan3()xxo xx331arc()tan3xo xxx316x:316x:313x:313x:sintanxxx 02x 0arctanlim,kxxxcx已知极限其中, c k为常数，且0,c 则( )= , =(B)= , =3, = (D) =3, =11( )222211( )33A kckcC kckc 0arctanlimkxxxx3013lim0kxxcx所以13,.3kc猪0，40sinsinsin sinlim.xxxxx求极限=30sinsi

4、nsinlim.xxxx=33301sin(sin)6limxxoxx=1.60sinlim1xxx331sin()3o猪猪猪猪sinxx:2.带“拉式”余项的泰勒公式：用于证明等式和不等式2()( )2 ()( )( )24abbaf bff af( ),( ).f af b( )()()()222abababf xffx21( )()2!2abfx( )()()222abab abf aff21()( )8abf( )()()222abab baf bff22()()8abf12aba22abb212()( )()( )( )2 ()242ababfff af bf12( )()( )2f

5、ff2()( )( )2 ()( )24ababf af bff2()( )2 ()( )( )24abbaf bff af(一)、函数零点或方程根唯一(二)、函数零点或方程根个数的讨论( )( )()( )( )f af af af afkk( )( )0.f af akk【解析】0lim(ln)xxxke ( )lnxf xxke( )ln0ef eekkelim(ln)xxxke ( )(1),xxxfxexex e( )xf xxae( )1xfxae ( )( ),g xh x xI( )( )( )0,.f xg xh xxI( )( )0,.f xf axa221ln1xxx【解

6、析】令( )1ln 1,0.f xxxx x11( )ln 112 11fxxxx1ln 122 12 1xxx令( )ln 122 1,0.g xxx x11( )0.11g xxx( )(0)0,g xg0( )()0.f xf x令( )( )( )(0)(1)(1)( )F xg xf xfxfxf x( )( )f bf aba( )fabMm【分析】证明：欲证即证构造2()ln (0).bbaababa()(lnln )2()bababa()(lnln )2()0bababa( )()(lnln )2() f xxaxaxaxa1lnln()baba令( )()(lnln )2() f xxaxaxa xa其中，( )(lnln )2; xafxxax221( )0axafxxxx在单调增加，( )fxxa故有( )( )0,.f xf axa又( )0, f a ( )0 fx在单调增加，( )f xax2()ln (0).bbaababa所以( )0,f b ()(lnln