高數(shù)六大定理(下)

1、200000()( )()()()()2!fxf xf xfxxxxx( )00()()( ),!nnnfxxxR xn1.帶拉格朗日型余項(xiàng)2.帶皮亞諾型余項(xiàng)200000()( )()()()()2!fxf xf xfxxxxx( )000()()() !nnnfxxxo xxn2(0)( )(0)(0)2!ff xffxx( )(0)()!nnnfxo xn1,sin ,cos ,ln(1),1xexxxx泰勒公式的主要應(yīng)用：建立函數(shù)與高階導(dǎo)數(shù)的關(guān)系.1.帶“皮式”余項(xiàng)的泰勒公式：用于計(jì)算：求極限找等價(jià)無窮小；確定無窮小的階；展開原則：AB上下同階原則 AB階數(shù)最低原則sin xx331()

2、3! xxxo x331()6 xo x無窮小的運(yùn)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已知極限其中, c k為常數(shù)，且0,c 則( )= , =(B)= , =3, = (D) =3, =11( )222211( )33A kckcC kckc 0arctanlimkxxxx3013lim0kxxcx所以13,.3kc豬0，40sinsinsin sinlim.xxxxx求極限=30sinsi

4、nsinlim.xxxx=33301sin(sin)6limxxoxx=1.60sinlim1xxx331sin()3o豬豬豬豬sinxx:2.帶“拉式”余項(xiàng)的泰勒公式：用于證明等式和不等式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)、函數(shù)零點(diǎn)或方程根唯一(二)、函數(shù)零點(diǎn)或方程根個(gè)數(shù)的討論( )( )()( )( )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】證明：欲證即證構(gòu)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在單調(diào)增加，( )fxxa故有( )( )0,.f xf axa又( )0, f a ( )0 fx在單調(diào)增加，( )f xax2()ln (0).bbaababa所以( )0,f b ()(lnln