2019年推薦 電大高等數(shù)學(xué)基礎(chǔ)復(fù)習(xí)小抄

1、..一、單項(xiàng)選擇題一、單項(xiàng)選擇題11下列各函數(shù)對中，（C）中的兩個(gè)函數(shù)相等C.，3ln)(xxf?xxgln3)(?1⒉設(shè)函數(shù)的定義域?yàn)?，則函數(shù)的圖形關(guān)于（C）對稱C.軸)(xf)(????)()(xfxf??y設(shè)函數(shù)的定義域?yàn)?，則函數(shù)的圖形關(guān)于（D）對稱D.坐標(biāo)原點(diǎn))(xf)(????)()(xfxf??.函數(shù)的圖形關(guān)于（A）對稱(A)坐標(biāo)原點(diǎn)2eexxy???1⒊下列函數(shù)中為奇函數(shù)是（B）B.xxycos?下列函數(shù)中為奇函數(shù)是（A）

2、A.xxy??3下列函數(shù)中為偶函數(shù)的是（D）D)1ln(2xy??21下列極限存計(jì)算不正確的是（D）D.01sinlim???xxx22當(dāng)時(shí)，變量（C）是無窮小量C.0?xxx1sin當(dāng)時(shí)，變量（C）是無窮小量C0?x1e?x.當(dāng)時(shí)，變量（D）是無窮小量D0?x)1ln(?x下列變量中，是無窮小量的為（B）B????ln10xx??31設(shè)在點(diǎn)x=1處可導(dǎo)，則（D）D.)(xf????hfhfh)1()21(lim0)1(2f??設(shè)在可導(dǎo)

3、，則（D）D)(xf0x????hxfhxfh)()2(lim000)(20xf??設(shè)在可導(dǎo)，則（D）D.)(xf0x????hxfhxfh2)()2(lim000)(0xf??設(shè)，則（A）Axxfe)(????????xfxfx)1()1(lim0e32.下列等式不成立的是（D）D.)1(lnxdxdx?下列等式中正確的是（B）B.2)1(xdxxd??41函數(shù)的單調(diào)增加區(qū)間是（D）D.14)(2???xxxf)2(???函數(shù)在區(qū)間內(nèi)

4、滿足（A）A.先單調(diào)下降再單調(diào)上升542???xxy)66(?.函數(shù)在區(qū)間（－5，5）內(nèi)滿足（A）A先單調(diào)下降再單調(diào)上升62???xxy.函數(shù)在區(qū)間內(nèi)滿足（D）D.單調(diào)上升622???xxy)52(51若的一個(gè)原函數(shù)是，則（D）D.)(xfx1??)(xf32x.若是的一個(gè)原函數(shù)，則下列等式成立的是（A）。A)(xF)(xf)()()(aFxFdxxfxa???52若，則（B）B.xxfcos)(????xxfd)(cx?cos下列等式

5、成立的是（D）D.)(d)(ddxfxxfx??（B）B.??xxfxxd)(dd32)(32xfx（D）D??xxxfxd)(dd2xxxfd)(2⒌3若，則（B）B.???cxFxxf)(d)(??xxfxd)(1cxF?)(2補(bǔ)充：無窮積分收斂的是函數(shù)的圖形關(guān)于????xefexxd)(ceFx???)(dxx???121xxxf???1010)(y軸對稱。..（2）利用連續(xù)函數(shù)性質(zhì)：有定義，則極限)(0xf)()(lim00xf

6、xfxx??類型類型1：利用重要極限，，計(jì)算1sinlim0??xxxkxkxx??sinlim0kxkxx??tanlim011求解：xxx5sin6sinlim0?565sin6sinlim5sin6sinlim00?????xxxxxxxx12求解：0tanlim3xxx???xxx3tanlim031131tanlim310????xxx13求解：=xxx3tanlim0?xxx3tanlim0?3313.33tanlim0??

7、??xxx類型類型2：因式分解并利用重要極限，化簡計(jì)算。1)()sin(lim????axaxax1)sin(lim????axaxax21求解：=)1sin(1lim21????xxx)1sin(1lim21????xxx2)11(1)1.()1sin()1(lim1???????????xxxx22解：??21sin1lim1xxx???211111)1(1.)1()1sin(lim1)1sin(lim121????????????

8、xxxxxxx23解：)3sin(34lim23????xxxx2)1(lim)3sin()1)(3(lim)3sin(34lim3323?????????????xxxxxxxxxx類型類型3：因式分解并消去零因子，再計(jì)算極限31解：=4586lim224?????xxxxx4586lim224?????xxxxx??????)1)(4()2)(4(lim4xxxxx3212lim4????xxx322236lim12xxxxx???

9、???????????2233332625limlimlim123447xxxxxxxxxxxxx???????????????????33解423lim222????xxxx4121lim)2)(2()1)(2(lim423lim22222???????????????xxxxxxxxxxxx其他：，0sin21limsin11lim2020??????xxxxxx221sinlim11sinlim00??????xxxxx，????

10、???5456lim22xxxxx1lim22???xxx??????54362lim22xxxxx3232lim22???xxx（0807考題）計(jì)算解：=xxx4sin8tanlim0?xxx4sin8tanlim0?248.4sin8tanlim0???xxxxx（0801考題.）計(jì)算解xxx2sinlim0???xxx2sinlim021sinlim210??xxx（0707考題.）=)1sin(32lim21?????xxxx4