用洛必达法则求下列极限
习
题
快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题
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,