洛必达法则习题.pdf

《洛必达法则习题.pdf》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、习题3−21.用洛必达法则求下列极限: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:当x→0时,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)因为lim1(+)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)因为limx=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)因为lim()=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.验证极限

7、lim存在,但不能用洛必达法则得出.x→∞xx+sinxsinxx+sinx解lim=lim1(+)=1,极限lim是存在的.x→∞xx→∞xx→∞x(x+sinx)′1+cosx但lim=lim=lim1(+cosx)不存在,不能用洛必达法则.x→∞(x)′x→∞1x→∞21xsinx3.验证极限lim存在,但不能用洛必达法则得出.x→0sinx2121xsinxsinxx1x解lim=lim⋅xsin=1⋅0=0,极限lim是存在的.x→0sinxx→0sinxxx→0sinx2111(xsin)′