【正文】
n2lnlim????1ar c t anln2lnlim??????xxx? 原極限 .2??? e.2111a r c t a n1lim22??????????xxxx)1(12l i m22xxxe???????.2??? e 注:數(shù)列極限利用函數(shù)極限來求 210)1l n (1)1(limxxxxx xx???????)1()1l n ()1(lim0 xxxxxex ??????2)1(2)1l n (lim0exxex??? ????xex xx???10)1(l i m例 )00(例 6. 設(shè) f(x)在 x0二階可導(dǎo) , 求 20000)(2)()(limhxfhxfhxfh?????解 : 00但不可再用洛必達(dá)法則 , hhxfhxfh 2)()(lim 000???????原式下一步應(yīng)利用二階導(dǎo)數(shù)定義 : 00hxfhxfxfhxfh 2)]()([)]()([lim 00000???????????))](()([21 00 xfxf ??????? )( 0xf ???.,)21(ar c c os3217baxaxx b無窮小,求為等價(jià)與時(shí),當(dāng)例 ?????122121)21(113lim)21(ar c c os3lim????????????bxbxxabxxax?解1)21(36lim121??????bx xab132,01 ????? bab有一階連續(xù)導(dǎo)數(shù)在證明有二階連續(xù)導(dǎo)數(shù),且在設(shè)例0)(,0)0(0)()(,0)0(),()(8??????????????xxFxfxxxfxFfxf)0()0()(lim)(lim)(lim000fx fxfx xfxFxxx????????解:.0)( 連續(xù)在 ?xxF))0()(lim)0()(lim)0(00 xfxxfxFxFFxx????????2)0(2)0()(lim)0()(lim020fxfxfxfxxfxx????????????,02)0(0)()()(2?????????????xfxxxfxfxxF連續(xù)在 0)(2)0()()(lim)(lim200????????????xxFfxxfxfxxFxx