專升本高數(shù)第一章練習(xí)題(帶答案)

1、第一部分：第一部分：1下面函數(shù)與為同一函數(shù)的是（）yx???2.Ayx?2.Byx?ln.xCye?.lnxDye?解：，且定義域，∴選Dlnlnxyexex??????????2已知是的反函數(shù)，則的反函數(shù)是（）?f??2fx??1.2Ayx????.2Byx????1.22Cyx????.22Dyx??解:令反解出：互換，位置得反函數(shù)，選A??2yfx?x??12xy??xy??12yx??3設(shè)在有定義，則下列函數(shù)為奇函數(shù)的是（）??

2、fx??????????.Ayfxfx???????.Byxfxfx?????????32.Cyxfx?????.Dyfxfx???解：的定義域且∴??32yxfx??????????????????3232yxxfxxfxyx???????選C4下列函數(shù)在內(nèi)無界的是（）??????21.1Ayx??.arctanByx?.sincosCyxx??.sinDyxx?解:排除法：A有界，B有界，C，21122xxxx???arctan2x

3、??sincos2xx??故選D5數(shù)列有界是存在的（）??nxlimnnx??A必要條件B充分條件C充分必要條件D無關(guān)條件解：收斂時，數(shù)列有界（即），反之不成立，（如有界，但不???nxnxnxM?????11n??收斂，選A.6當(dāng)時，與為等價無窮小，則=（）n??21sinn1knkAB1C2D212解：，選C?2211sinlimlim111nnkknnnn??????2k?解：.故22sin2cos21sin222xxxf????

4、????????????????221f???????????221fxx??.14設(shè)，的反函數(shù)，求??fxlnx???gx????1211xgxx????????fgx解:（1）求∴反解出：22():1xgxyx????x22xyyx???22xyy???互換位置得(2).xy()22gxxx???????lnln22fgxgxxx????????15設(shè)，求的值。32lim8nnnana???????????a解：故.3323liml

5、im1nnnnnaanana??????????????????????limnnaanaee?????8ae??ln83ln2a??16求??111lim12231nnnn?????????????????解：（1）拆項，11(1)(1)kkkkkk?????11121knkk???????11112231nn???????1111112231nn??????????????????????????111n???（2）原式=lim11