資源描述：
《洛必達(dá)法則習(xí)題.pdf》由會員上傳分享，免費(fèi)在線閱讀，更多相關(guān)內(nèi)容在教育資源-天天文庫。
1、習(xí)題3?21.用洛必達(dá)法則求下列極限:ln(1+x)(1)lim;x→0xx?xe?e(2)lim;x→0sinxsinx?sina(3)lim;x→ax?asin3x(4)lim;x→πtan5xlnsinx(5)lim;π(2x)2x→π?2mmx?a(6)lim;x→axn?anlntan7x(7)lim;x→+0lntan2xtanx(8)lim;x→πtan3x21ln(1+)x(9)lim;x→+∞arccotx2ln(1+x)(10)lim;x→0secx?cosx(11)limxcot2x;x→012x2(12)limxe;x→0?21?(13)lim???;
2、x→1?x2?1x?1?ax(14)lim1(+);x→∞xsinx(15)limx;x→+01tanx(16)lim().x→+0x1ln(1+x)1+x1解(1)lim=lim=lim=1.x→0xx→01x→01+xx?xx?xe?ee+e(2)lim=lim=2.x→0sinxx→0cosxsinx?sinacosx(3)lim=lim=cosa.x→ax?ax→a1sin3x3cos3x3(4)lim=lim=?.x→πtan5xx→π5sec25x52lnsinxcotx1?cscx1(5)lim=lim=?lim=?.ππ?2π(2π?2x)?(?)24π?28
3、x→(2x)x→x→222mmm?1m?1x?amxmxmm?n(6)lim=lim==a.x→axn?anx→anxn?1nan?1n12?sec7x?72lntan7xtan7x7tan2x7sec2x?2(7)lim=lim=lim=lim=1.x→+0lntan2xx→+0122x→+0tan7x2x→+0sec27x?7?sec2x?2tan2x22tanxsecx1cos3x12cos3x(?sin3x)?3(8)lim=lim=lim=limπtan3xπsec23x?33πcos2x3π2cosx(?sinx)x→x→x→x→2222cos3x?3sin3x=
4、?lim=?lim=3.x→πcosxx→π?sinx2211?(?)11x2ln(1+)1+2xx1+x2x2(9)lim=lim=lim=lim=lim=1.x→+∞arccotxx→+∞1x→+∞x+x2x→+∞1+2xx→+∞221+x222ln(1+x)cosxln(1+x)x22(10)lim=lim=lim(注:cosx?ln(1+x)~x)x→0secx?cosxx→01?cos2xx→01?cos2x2xx=lim=lim=1.x→0?2cosx(?sinx)x→0sinxx11(11)limxcot2x=lim=lim=.x→0x→0tan2xx→0sec
5、22x?2211x2tt2x2eee1(12)limxe=lim=lim=lim=+∞(注:當(dāng)x→0時(shí),t=→+∞).x→0x→01t→+∞tt→+∞1x22x?21?1?x?11(13)lim???=lim=lim=?.x→1?x2?1x?1?x→1x2?1x→12x2aaxln(1+)(14)因?yàn)閘im1(+)x=limex,x→∞xx→∞1a?(?)aax2ln(1+)1+axxaxa而limx(ln(1+)=lim=lim=lim=lim=a,x→∞xx→∞1x→∞1x→∞x+ax→∞1?x2xaaxln(1+)xxa所以lim1(+)=lime=e.x→∞xx→∞.
6、sinxsinxlnx(15)因?yàn)閘imx=lime,x→+0x→+012lnxxsinx而limsinxlnx=lim=lim=?lim=0,x→+0x→+0cscxx→+0?cscx?cotxx→+0xcosxsinxsinxlnx0所以limx=lime=e=1.x→+0x→+01tanx?tanxlnx(16)因?yàn)閘im()=e,x→+0x12lnxxsinx而limtanxlnx=lim=lim=?lim=0,x→+0x→+0cotxx→+0?csc2xx→+0xlim(1tanx=e?tanxlnx=e0=所以)lim1.x→+0xx→+0x+sinx2.驗(yàn)證極限
7、lim存在,但不能用洛必達(dá)法則得出.x→∞xx+sinxsinxx+sinx解lim=lim1(+)=1,極限lim是存在的.x→∞xx→∞xx→∞x(x+sinx)′1+cosx但lim=lim=lim1(+cosx)不存在,不能用洛必達(dá)法則.x→∞(x)′x→∞1x→∞21xsinx3.驗(yàn)證極限lim存在,但不能用洛必達(dá)法則得出.x→0sinx2121xsinxsinxx1x解lim=lim?xsin=1?0=0,極限lim是存在的.x→0sinxx→0sinxxx→0sinx2111(xsin)′
