用洛必达法则求下列极限

用洛必达法则求下列极限 习题3,2 1, 用洛必达法则求下列极限: xln(1，) (1)lim, x,0x ,xxee, (2), lim,0xsinx x,asinsin (3)lim, x,ax,a sin3x (4)lim, x,,tan5x lnsinxlim (5), 2,(,,2x)x,2 mmx,a (6), limnnx,ax,a lntan7x (7)lim, x,，0lntan2xtanx (8), lim,tan3xx,2 1ln(1，)xlim (9), x,，,arccotx 2ln(1，x) (10) lim, x,0secx,cosxlimxcot2x (11), x,0 122x (12), limxe,x0 21,,lim (13), ,,,2x,1x1,x1,,, axlim(1，) (14), x,,x sinx (15), limx,，x0 1tanxlim() (16), ,，x0x 1 ln(1，x)1x1，lim,lim,lim,1 解 (1), x,0x,0x,0x11，x x,xx,xeeee,， (2), lim,lim,200x,x,xxsincos sinx,sinacosxlim,lim,cosa (3), x,ax,ax,a1 xxsin33cos33 (4)limlim, ,,,2x,,x,,tan55x5sec5x 2lnsincot1csc1,xxxlimlimlim (5), ,,,,,2,,,2(,2)(2)428,,,,x(2),,xx,x,x,222 11mmm,m,x,amxmxmm,n (6), lim,lim,,a11nnn,n,x,ax,anx,anxna 12x,sec7,72xxxlntan77tan27sec2,2xtan7 (7), lim,lim,lim,lim,12x,，x,，x,，x,，00001xxlntan22tan722xsec7,7x,sec2,2xtan2 22tanxsecx1cos3x12cos3x(,sin3x),3lim,lim,lim,lim (8) 22,,,,tan3x332cosx(,sinx)sec3x,3cosxx,x,x,x,2222 xxcos3,3sin3 , ,,lim,,lim,3,,xxcos,sinx,x,22 11,(,)211xln(1，)1，2xx1，22xx (9), lim,lim,lim,lim,lim,12x,，,x,，,x,，,x,，,x,，,1xxarccot1，22xx，2x1， 222ln(1，x)cosxln(1，x)x22lim,lim,lim (10)(注: cosx,ln(1，x)~x) 22x,0x,0x,0secx,cosx1,cosx1,cosx xx2 , ,lim,lim,1x,0x,0xxx,2cos(,sin)sin x11limxcot2xlimlim (11), ,,,2x,x,x,000tan22xsec2x2, 112ttxeee122x,,,,，,limxelimlimlim (12)(注: 当x,0时～ ，, ,,，,t2,0,0,，,,，,xxtt11tx2x 211x11,,,,limlimlim (13), ,,,,,,,22x,1x,1x,1x12x2,x1x1,,,, axln(1，)axx (14)因为～ lim(1，),limexx,,,,x 1a(),,2aaxln(1)1，，aaxaxx而 ～ lim(ln(1)limlimlimlimx，,,,,,ax,,x,,x,,x,,x,,111xx，a,2xx axln(1，)axax所以 , lim(1，),lime,exx,,,,x , sinxsinxlnx (15)因为～ limx,lime,，,，x0x0 12xxlnsinxlimsinln,lim,lim,,lim,0xx而 ～ x,，x,，x,，x,，0000cscx,cscx,cotxxcosx sinxsinxlnx0所以 , limx,lime,e,1,，,，x0x0 1tanx,tanxlnxlim(),e (16)因为～ x,，0x 12xxlnsinx而 ～ xxlimtanln,lim,lim,,lim,02x,，x,，x,，x,，0000xxcotx,csc 1tanx,tanxlnx0lim(),lime,e,1x,，0x,，0x所以 , xx，sinlim 2, 验证极限存在～ 但不能用洛必达法则得出, x,,x xxx，sinsinx，sinxlim,lim(1，),1lim 解 ～ 极限是存在的, x,,x,,x,,xxx,(x，sinx)1，cosx但不存在～ 不能用洛必达法则, lim,lim,lim(1，cosx)x,,x,,x,,,(x)1 12xsinxlim 3, 验证极限存在～ 但不能用洛必达法则得出, ,0xsinx 1122xsinxsinx1xxlim,lim,xsin,1,0,0lim 解 ～ 极限是存在的, ,0x,0x,0xsinxsinxxsinx 1112,(xsin)2xsincos,xxx但不存在～ 不能用洛必达法则, limlim,x,0x,0,(sinx)cosx 1,1x(1x),，x[] x,0,f(x), 4, 讨论函数在点x,0处的连续性, e,1,,,2e x,0, 111,,,222 解 ～ ～ f(0),elimf(x),lime,e,f(0)xx00,,,, 1111x[ln(1，x),1](1，x)xxx因为 limf(x),lim[],lime～ x,，0x,,0x,,0e 11,11ln(1x)x11，,,1x，而 ～ lim[ln(1x)1]limlimlim，,,,,,,2x,，0x,，0x,，0x,，0xx2x2(1x)2，x 11111x[ln(1)1]x，,,(1，x)xxx2所以 limf(x),lim[],lime,e,f(0), 000x,，x,,x,,e 因此f(x)在点x,0处连续,