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习题1.4
1.直接用?-?说法证明下列各极限等式:(1)limx?a(a?0);(2)limx2?a2;(3)limex?ea;(4)limcosx?cosa.x?ax?ax?ax?a证(1)???0,要使|x?a|?只需|x-a||x-a||x-a|??,由于?,x?ax?aa|x?a|??,|x?a|?a?.取??a?,则当|x?a|??时,|x?a|??,故limx?a. x?aa(2)???0,不妨设|x?a|?1.要使|x2?a2|?|x?a||x?a|??,由于.取??min{,1},则当|x?a|??时,1?|2a|1?|2a||x?a|?|x?a|?|2a|?1?|2a|,只需(1?|2a|)|x?a|??,|x?a|?|x2?a2|??,故limx2?a2.x?a??(3) ???0,设x?a.要使|ex?ea|?ea(ex?a?1)??,即0?(ex?a?1)??ex?a,1?e?1?a?ea,????0?x?a?ln?1?a?,取??min{,1},则当0?x?a??时,|ex?ea|??,1?|2a|?e?故limex?ea. 类似证limex?ea.故limex?ea.x?a?x?a?x?a(4)???0,要使|cosx?cosa|?2sinx?ax?ax?ax?asin?2sinsin?|x?a|,2222x?a取???,则当|x?a|??时,|cosx?cosa|??,故limcosx?cosa.2.设limf(x)?l,证明存在a的一个空心邻域(a??,a)?(a,a??),使得函数u?f(x)在x?a该邻域内使有界函数.证对于??1,存在??0,使得当 0?|x-a|??时,|f(x)?l|?1,从而|f(x)|?|f(x)?l?l|?|f(x)?l|?|l|?1?|l|?M. 3.求下列极限:(1?x)2?12x?x2x(1)lim?lim?lim(1?)?1.x?0x?0x?02x2x2??x???x?2sin2???sin?2??11?cosx2?1?????12?1.?(2)lim?lim?limx?0x?0x2x22x?0?x?22?2???x?a?ax1(3)lim?lim?(a?0).x?0x?0xx(x?a?a)2ax2?x?2?2(4)lim2?.x?12x?2x?3?3x2?x?2?2(5)lim2?.x?02x?2x?3?32
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(2x?3)20(2x?2)10230(6)lim?30?1.x??(2x?1)302(7)limx?01?x?1?x2x?lim?1.x?0x(1?x?1?x)x3?x2?x?1?3x2?x?2?1(8)lim??3??lim?limx??1x?1x??1(x?1)(x2?x?1)x??1(x?1)(x2?x?1)x?1??(x?1)(x?2)(x?2)?3?lim?lim???1.x??1(x?1)(x2?x?1)x??1(x2?x?1)3(9)limx?41?2x?3(1?2x?3)(x?2)(1?2x?3)?limx?4x?2(x?2)(x?2)(1?2x?3)?lim(2x?8)(x?2)244??.x?4(x?4)(1?2x?3)63ny??yn?n.n(n?1)2y?xn?1(1?y)n?12(10)lim?lim?limx?1x?1y?0y?0yy2(11)limx2?1?x2?1?lim?0.22x??x??x?1?x?1a0xm?a1xm?1??amam(12)lim(b?0)?.nx?0bxn?bxn?1??bnbn01??
(13)lima0x?a1x?x??bxn?bxn?1?01mm?1?a0/b0,m?n?am?(a0b0?0)??0, n?m?bn??, m?n.?x4?81?8/x4(14)lim2?lim?1.x??x?1x??1?1/x21?3x?31?2x(15)limx?0x?x2?limx?03(1?3x?1?2x)(1?3x?1?3x1?2x?1?2x)(x?x2)(31?3x?31?3x31?2x?31?2x)5x222233323332?limx?0x(1?x)(31?3x?31?3x31?2x?31?2x)55?lim?.22x?033333(1?x)(1?3x?1?3x1?2x?1?2x)(16)a?0,lim??lim?x?a?0???x?a1??lim????2222x?a?0x?a?0?x?ax?a?x?a?(x?a)(x?a)1???x?ax?a(x?a)x?a??x?a?x?a
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?(x?a)1??lim???x?a?0x?ax?a(x?a)x?a???x?a1?1?lim???.??x?a?0?x?a(x?a)x?a2a??x
sinx?1?4.利用lim?1及lim?1???e求下列极限:x?x??x?x?sin?xsin?x?(1)lim?limlimcos?x?.x?0tan?xx?0sin?xx?0?sin(2x2)sin(2x2)2x2(2)lim?limlim?10?02x?x?0x?03x2x3xtan3x?sin2xtan3xsin2x321(3)lim?lim?lim???.x?0x?0x?0sin5xsin5xsin5x555xx(4)lim?lim?2.x?0?1?cosxx?0?x2sin2x?ax?acossinsinx?sina22?cosa.(5)lim?limx?ax?ax?ax?a2?k?(6)lim?1??x???x?(7)lim(1?5y)y?0?x?k??lim?1??x???x?x(?k)kx??kk????lim?1????x???x?????5?k?e?k.1/y??lim(1?5y)1/(5y)??e?5.???y?0?x100?1??1???1??(8)lim?1???lim?1???lim?1????e.x??x???x??x??x???x??5.给出limf(x)???及limf(x)???的严格定义.x?ax???x?100同
limf(x)???:对于任意给定的A?0,存在??0,使得当0?|x-a|??时f(x)?A.x?ax???limf(x)???:对于任意给定的A?0,存在??0,使得当x???时f(x)??A.义句转换 方法一
1）用同义词或同义短语替换。
①She got to China in 1950．→She___ _China in 1950．
②Be careful with your handwriting．→___ __to your handwriting．
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2）用反义词或反义短语替换。
①She hardly speaks at the meeting．→She__ __ ___ _ __ __at the meeting．
②My watch doesn’t work well．→Something_ __ __ _ _ __my watch．
3）用短语替换从句或用从句替换短语，例：
①After we had breakfast，we went to school．→__ __，we went to school．
②We can’t finish the work without your help．→We can’t finish the work__ __you____us．
方法二 ---- 转换法
这种方法是用不同句型、句式、语态、引语等方法改写句子，使其意思相同。
1）句型转换（这种转换通常是用另一种句型替换原来的句型）。如： The Great Pyramid is the biggest of all the Pyramid（金字塔）．→ The Great Pyramid is__ __any other pyramid．
2）句式转换（这种转换通常是感叹句的转换或状语从句的转换等）。如： He went to bed after he had finished his homework．→
He_ __ ___ to bed___ _he had finished his homework．
3）语态转换（这种转换通常是主动语态变被动语态或被动语态变主动语态）。例如：
They made her work fourteen hours a day．→
She was___ _ __ __ __ __fourteen hours a day．
4）引语转换（这种转换是指直接引语变间接引语或间接引语变直接引语）。如： ①“Don’t make faces in class！”the teacher said to the student．→ The teacher__ __the student__ __make faces in class． ②Tom asked Jack if he had ever been to China．→
“___ _you ever__ __to China？”Tom asked Jack．
方法三 ---- 合并法
（这是指用连词将两个简单句合并成一个简单句或复合句，使其意思不变） 1）用并列连词 both…and…，neither…nor…，either…or…，not…but…，not only…but also…等将两个简单句合并成一个新的简单句，如： ①Tom is good at English．Tom is also good at French．→
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Tom is good at____ ____English____ ____French．
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最新高等数学( 北大版)答案一习题1.4
