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【正文】 m)()](1l i m [ xfxgxg exf ??? )]()[(l i m xfxge ??.)()(l i m xfxge ?? ))(~)](1l n [( xfxf ???310)]1s i n1 t a n1(1[l i m xx xx ??????原式310]si n1 si nt a n1[lim xx xxx?????301si n1si nt a nl i mxxxxx?? ??? 30 1c o s)s i n1( )c o s1(s i nl i m xxx xxx ?? ?? ?xxxxxxx c o s)si n1(1c o s1si nl i m20 ?????? ??21.21e?? 原式例 5 ).(,1)(l i m,2)(l i m,)(023xpxxpxxxpxpxx求且是多項(xiàng)式設(shè)??????解 ,2)(l i m23???? xxxpx?),(2)( 23 為待定系數(shù)其中可設(shè) babaxxxxp ?????,1)(l i m0?? xxpx?又)0(~2)( 23 ?????? xxbaxxxxp.1,0 ?? ab從而得 xxxxp ??? 23 2)(故例 6 .1,2c o s1,1)( 的連續(xù)性討論?????????xxxxxf ?解 改寫成將 )( xf?????????????????1,111,2co s1,1)(xxxxxxxf.),1(),1,1(),1,()( 內(nèi)連續(xù)在顯然 ??????xf,1時當(dāng) ??x???? )(l i m1 xfx ????? )1(lim 1 xx .2???? )(l i m1 xfx ????? 2coslim 1 xx .0)(lim)(lim 11 xfxf xx ?? ???? ??.1)( 間斷在故 ??xxf,1時當(dāng) ?x??? )(lim 1 xfx ???? 2cosl
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