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1、第一章極限和連續(xù)第一節(jié)極限復(fù)習(xí)考試要求1.了解極限的概念(對(duì)極限定義等形式的描述不作要求)。會(huì)求函數(shù)在一點(diǎn)處的左極限與右極限,了解函數(shù)在一點(diǎn)處極限存在的充分必要條件。2.了解極限的有關(guān)性質(zhì),掌握極限的四則運(yùn)算法則。3.理解無(wú)窮小量、無(wú)窮大量的概念,掌握無(wú)窮小量的性質(zhì)、無(wú)窮小量與無(wú)窮大量的關(guān)系。會(huì)進(jìn)行無(wú)窮小量階的比較(高階、低階、同階和等價(jià))。會(huì)運(yùn)用等價(jià)無(wú)窮小量代換求極限。4.熟練掌握用兩個(gè)重要極限求極限的方法。 主要知識(shí)內(nèi)容(一)數(shù)列的極限1.數(shù)列定義按一定順序排列的無(wú)窮多個(gè)數(shù)稱為無(wú)窮數(shù)列,簡(jiǎn)稱數(shù)列,記作xn,數(shù)列中每一個(gè)數(shù)稱為數(shù)列的項(xiàng), 第n項(xiàng)xn為數(shù)列的一般項(xiàng)或通項(xiàng),例如(1)1,3,5,

2、(2n-1),(等差數(shù)列)(2)(等比數(shù)列)(3)(遞增數(shù)列)(4)1,0,1,0,(震蕩數(shù)列)都是數(shù)列。它們的一般項(xiàng)分別為(2n-1),。對(duì)于每一個(gè)正整數(shù)n,都有一個(gè)xn與之對(duì)應(yīng),所以說(shuō)數(shù)列xn可看作自變量n的函數(shù)xn=f(n),它的定義域是全體正整數(shù),當(dāng)自變量n依次取1,2,3一切正整數(shù)時(shí),對(duì)應(yīng)的函數(shù)值就排列成數(shù)列。在幾何上,數(shù)列xn可看作數(shù)軸上的一個(gè)動(dòng)點(diǎn),它依次取數(shù)軸上的點(diǎn)x1,x2,x3,.xn,。2.數(shù)列的極限定義對(duì)于數(shù)列xn,如果當(dāng)n時(shí),xn無(wú)限地趨于一個(gè)確定的常數(shù)a,則稱當(dāng)n趨于無(wú)窮大時(shí),數(shù)列xn以常數(shù)a為極限,或稱數(shù)列收斂于a,記作 比如:無(wú)限的趨向0,無(wú)限的趨向1否則,對(duì)于數(shù)

3、列xn,如果當(dāng)n時(shí),xn不是無(wú)限地趨于一個(gè)確定的常數(shù),稱數(shù)列xn沒(méi)有極限,如果數(shù)列沒(méi)有極限,就稱數(shù)列是發(fā)散的。比如:1,3,5,(2n-1),1,0,1,0,數(shù)列極限的幾何意義:將常數(shù)a及數(shù)列的項(xiàng)依次用數(shù)軸上的點(diǎn)表示,若數(shù)列xn以a為極限,就表示當(dāng)n趨于無(wú)窮大時(shí),點(diǎn)xn可以無(wú)限靠近點(diǎn)a,即點(diǎn)xn與點(diǎn)a之間的距離|xn-a|趨于0。比如:無(wú)限的趨向0無(wú)限的趨向1(二)數(shù)列極限的性質(zhì)與運(yùn)算法則1.數(shù)列極限的性質(zhì)定理1.1(惟一性)若數(shù)列xn收斂,則其極限值必定惟一。定理1.2(有界性)若數(shù)列xn收斂,則它必定有界。注意:這個(gè)定理反過(guò)來(lái)不成立,也就是說(shuō),有界數(shù)列不一定收斂。比如:1,0,1,0,有界

4、:0,12.數(shù)列極限的存在準(zhǔn)則定理1.3(兩面夾準(zhǔn)則)若數(shù)列xn,yn,zn滿足以下條件:(1),(2), 則定理1.4若數(shù)列xn單調(diào)有界,則它必有極限。3.數(shù)列極限的四則運(yùn)算定理。定理1.5(1)(2)(3)當(dāng)時(shí),(三)函數(shù)極限的概念1.當(dāng)xx0時(shí)函數(shù)f(x)的極限(1)當(dāng)xx0時(shí)f(x)的極限定義對(duì)于函數(shù)y=f(x),如果當(dāng)x無(wú)限地趨于x0時(shí),函數(shù)f(x)無(wú)限地趨于一個(gè)常數(shù)a,則稱當(dāng)xx0時(shí),函數(shù)f(x)的極限是a,記作或f(x)a(當(dāng)xx0時(shí))例y=f(x)=2x+1x1,f(x)?x1x1(2)左極限當(dāng)xx0時(shí)f(x)的左極限定義對(duì)于函數(shù)y=f(x),如果當(dāng)x從x0的左邊無(wú)限地趨于x0

5、時(shí),函數(shù)f(x)無(wú)限地趨于一個(gè)常數(shù)a,則稱當(dāng)xx0時(shí),函數(shù)f(x)的左極限是a,記作或f(x0-0)=a(3)右極限當(dāng)xx0時(shí),f(x)的右極限定義對(duì)于函數(shù)y=f(x),如果當(dāng)x從x0的右邊無(wú)限地趨于x0時(shí),函數(shù)f(x)無(wú)限地趨于一個(gè)常數(shù)a,則稱當(dāng)xx0時(shí),函數(shù)f(x)的右極限是a,記作或f(x0+0)=a例子:分段函數(shù),求,解:當(dāng)x從0的左邊無(wú)限地趨于0時(shí)f(x)無(wú)限地趨于一個(gè)常數(shù)1。我們稱當(dāng)x0時(shí),f(x)的左極限是1,即有當(dāng)x從0的右邊無(wú)限地趨于0時(shí),f(x)無(wú)限地趨于一個(gè)常數(shù)-1。我們稱當(dāng)x0時(shí),f(x)的右極限是-1,即有顯然,函數(shù)的左極限右極限與函數(shù)的極限之間有以下關(guān)系:定理1.6

6、當(dāng)xx0時(shí),函數(shù)f(x)的極限等于a的必要充分條件是反之,如果左、右極限都等于a,則必有。x1時(shí)f(x)?x1x1f(x)2對(duì)于函數(shù),當(dāng)x1時(shí),f(x)的左極限是2,右極限也是2。2.當(dāng)x時(shí),函數(shù)f(x)的極限(1)當(dāng)x時(shí),函數(shù)f(x)的極限y=f(x)xf(x)?y=f(x)=1+xf(x)=1+1定義對(duì)于函數(shù)y=f(x),如果當(dāng)x時(shí),f(x)無(wú)限地趨于一個(gè)常數(shù)a,則稱當(dāng)x時(shí),函數(shù)f(x)的極限是a,記作或f(x)a(當(dāng)x時(shí))(2)當(dāng)x+時(shí),函數(shù)f(x)的極限定義對(duì)于函數(shù)y=f(x),如果當(dāng)x+時(shí),f(x)無(wú)限地趨于一個(gè)常數(shù)a,則稱當(dāng)x+時(shí),函數(shù)f(x)的極限是a,記作這個(gè)定義與數(shù)列極限的定

7、義基本上一樣,數(shù)列極限的定義中n+的n是正整數(shù);而在這個(gè)定義中,則要明確寫出x+,且其中的x不一定是正整數(shù),而為任意實(shí)數(shù)。y=f(x)x+f(x)x? x+,f(x)=2+2例:函數(shù)f(x)=2+e-x,當(dāng)x+時(shí),f(x)?解:f(x)=2+e-x=2+,x+,f(x)=2+2所以(3)當(dāng)x-時(shí),函數(shù)f(x)的極限定義對(duì)于函數(shù)y=f(x),如果當(dāng)x-時(shí),f(x)無(wú)限地趨于一個(gè)常數(shù)a,則稱當(dāng)x-時(shí),f(x)的極限是a,記作x-f(x)?則f(x)=2+(x0)x-,-x+f(x)=2+2例:函數(shù),當(dāng)x-時(shí),f(x)?解:當(dāng)x-時(shí),-x+2,即有由上述x,x+,x-時(shí),函數(shù)f(x)極限的定義,不難

8、看出:x時(shí)f(x)的極限是a充分必要條件是當(dāng)x+以及x-時(shí),函數(shù)f(x)有相同的極限a。例如函數(shù),當(dāng)x-時(shí),f(x)無(wú)限地趨于常數(shù)1,當(dāng)x+時(shí),f(x)也無(wú)限地趨于同一個(gè)常數(shù)1,因此稱當(dāng)x時(shí)的極限是1,記作其幾何意義如圖3所示。f(x)=1+y=arctanx不存在。但是對(duì)函數(shù)y=arctanx來(lái)講,因?yàn)橛屑措m然當(dāng)x-時(shí),f(x)的極限存在,當(dāng)x+時(shí),f(x)的極限也存在,但這兩個(gè)極限不相同,我們只能說(shuō),當(dāng)x時(shí),y=arctanx的極限不存在。x)=1+y=arctanx不存在。但是對(duì)函數(shù)y=arctanx來(lái)講,因?yàn)橛?即雖然當(dāng)x-時(shí),f(x)的極限存在,當(dāng)x+時(shí),f(x)的極限也存在,但這兩

9、個(gè)極限不相同,我們只能說(shuō),當(dāng)x時(shí),y=arctanx的極限不存在。(四)函數(shù)極限的定理定理1.7(惟一性定理)如果存在,則極限值必定惟一。定理1.8(兩面夾定理)設(shè)函數(shù)在點(diǎn)的某個(gè)鄰域內(nèi)(可除外)滿足條件:(1),(2)則有。注意:上述定理1.7及定理1.8對(duì)也成立。下面我們給出函數(shù)極限的四則運(yùn)算定理定理1.9如果則(1)(2)(3)當(dāng)時(shí),時(shí),上述運(yùn)算法則可推廣到有限多個(gè)函數(shù)的代數(shù)和及乘積的情形,有以下推論:(1)(2)(3)用極限的運(yùn)算法則求極限時(shí),必須注意:這些法則要求每個(gè)參與運(yùn)算的函數(shù)的極限存在,且求商的極限時(shí),還要求分母的極限不能為零。另外,上述極限的運(yùn)算法則對(duì)于的情形也都成立。(五)無(wú)

10、窮小量和無(wú)窮大量1.無(wú)窮小量(簡(jiǎn)稱無(wú)窮小)定義對(duì)于函數(shù),如果自變量x在某個(gè)變化過(guò)程中,函數(shù)的極限為零,則稱在該變化過(guò)程中,為無(wú)窮小量,一般記作常用希臘字母,來(lái)表示無(wú)窮小量。定理1.10函數(shù)以a為極限的必要充分條件是:可表示為a與一個(gè)無(wú)窮小量之和。注意:(1)無(wú)窮小量是變量,它不是表示量的大小,而是表示變量的變化趨勢(shì)無(wú)限趨于為零。(2)要把無(wú)窮小量與很小的數(shù)嚴(yán)格區(qū)分開(kāi),一個(gè)很小的數(shù),無(wú)論它多么小也不是無(wú)窮小量。(3)一個(gè)變量是否為無(wú)窮小量是與自變量的變化趨勢(shì)緊密相關(guān)的。在不同的變化過(guò)程中,同一個(gè)變量可以有不同的變化趨勢(shì),因此結(jié)論也不盡相同。例如:振蕩型發(fā)散 (4)越變?cè)叫〉淖兞恳膊灰欢ㄊ菬o(wú)窮小量

11、,例如當(dāng)x越變?cè)酱髸r(shí),就越變?cè)叫。皇菬o(wú)窮小量。(5)無(wú)窮小量不是一個(gè)常數(shù),但數(shù)“0”是無(wú)窮小量中惟一的一個(gè)數(shù),這是因?yàn)椤?.無(wú)窮大量(簡(jiǎn)稱無(wú)窮大)定義;如果當(dāng)自變量(或)時(shí),的絕對(duì)值可以變得充分大(也即無(wú)限地增大),則稱在該變化過(guò)程中,為無(wú)窮大量。記作。注意:無(wú)窮大()不是一個(gè)數(shù)值,“”是一個(gè)記號(hào),絕不能寫成或。3.無(wú)窮小量與無(wú)窮大量的關(guān)系無(wú)窮小量與無(wú)窮大量之間有一種簡(jiǎn)單的關(guān)系,見(jiàn)以下的定理。定理1.11在同一變化過(guò)程中,如果為無(wú)窮大量,則為無(wú)窮小量;反之,如果為無(wú)窮小量,且,則為無(wú)窮大量。當(dāng)無(wú)窮大無(wú)窮小當(dāng)為無(wú)窮小無(wú)窮大4.無(wú)窮小量的基本性質(zhì)性質(zhì)1有限個(gè)無(wú)窮小量的代數(shù)和仍是無(wú)窮小量;性質(zhì)

12、2有界函數(shù)(變量)與無(wú)窮小量的乘積是無(wú)窮小量;特別地,常量與無(wú)窮小量的乘積是無(wú)窮小量。性質(zhì)3有限個(gè)無(wú)窮小量的乘積是無(wú)窮小量。性質(zhì)4無(wú)窮小量除以極限不為零的變量所得的商是無(wú)窮小量。5.無(wú)窮小量的比較定義設(shè)是同一變化過(guò)程中的無(wú)窮小量,即。(1)如果則稱是比較高階的無(wú)窮小量,記作;(2)如果則稱與為同階的無(wú)窮小量;(3)如果則稱與為等價(jià)無(wú)窮小量,記為;(4)如果則稱是比較低價(jià)的無(wú)窮小量。當(dāng)?shù)葍r(jià)無(wú)窮小量代換定理:如果當(dāng)時(shí),均為無(wú)窮小量,又有且存在,則。均為無(wú)窮小又有這個(gè)性質(zhì)常常使用在極限運(yùn)算中,它能起到簡(jiǎn)化運(yùn)算的作用。但是必須注意:等價(jià)無(wú)窮小量代換可以在極限的乘除運(yùn)算中使用。常用的等價(jià)無(wú)窮小量代換有:

13、當(dāng)時(shí),sinxx;tanx;arctanxx;arcsinxx;(六)兩個(gè)重要極限1.重要極限重要極限是指下面的求極限公式令這個(gè)公式很重要,應(yīng)用它可以計(jì)算三角函數(shù)的型的極限問(wèn)題。其結(jié)構(gòu)式為:2.重要極限重要極限是指下面的公式: 其中e是個(gè)常數(shù)(銀行家常數(shù)),叫自然對(duì)數(shù)的底,它的值為e=2.718281828495045其結(jié)構(gòu)式為:重要極限是屬于型的未定型式,重要極限是屬于“”型的未定式時(shí),這兩個(gè)重要極限在極限計(jì)算中起很重要的作用,熟練掌握它們是非常必要的。(七)求極限的方法:1.利用極限的四則運(yùn)算法則求極限;2.利用兩個(gè)重要極限求極限;3.利用無(wú)窮小量的性質(zhì)求極限;4.利用函數(shù)的連續(xù)性求極限;

14、5.利用洛必達(dá)法則求未定式的極限;6.利用等價(jià)無(wú)窮小代換定理求極限?;緲O限公式 (2)(3)(4)例1.無(wú)窮小量的有關(guān)概念(1)9601下列變量在給定變化過(guò)程中為無(wú)窮小量的是a.b.c.d. 答ca.發(fā)散d.(2)0202當(dāng)時(shí),與x比較是a.高階的無(wú)窮小量b.等價(jià)的無(wú)窮小量c.非等價(jià)的同階無(wú)窮小量d.低階的無(wú)窮小量答b解:當(dāng),與x是極限的運(yùn)算:0611解:答案-1例2.型因式分解約分求極限(1)0208 答解:(2)0621計(jì)算答解:例3.型有理化約分求極限(1)0316計(jì)算 答解:(2)9516 答解: 例4.當(dāng)時(shí)求型的極限 答 (1)0308一般地,有例5.用重要極限求極限(1)9603

15、下列極限中,成立的是a.b.c.d. 答b(2)0006 答解:例6.用重要極限求極限(1)0416計(jì)算 答解析解一:令解二:03060601(2)0118計(jì)算 答 解:例7.用函數(shù)的連續(xù)性求極限0407 答0解:,例8.用等價(jià)無(wú)窮小代換定理求極限0317 答0解:當(dāng)例9.求分段函數(shù)在分段點(diǎn)處的極限(1)0307設(shè)則在的左極限答1解析(2)0406設(shè),則 答1解析例10.求極限的反問(wèn)題(1)已知?jiǎng)t常數(shù)解析解法一:,即,得.解法二:令,得,解得.解法三:(洛必達(dá)法則)即,得.(2)若求a,b的值.解析型未定式.當(dāng)時(shí),.令于是,得.即,所以.04020017,則k=_.(答:ln2)待添加的隱藏文

16、字內(nèi)容3解析前面我們講的內(nèi)容: 極限的概念;極限的性質(zhì);極限的運(yùn)算法則;兩個(gè)重要極限;無(wú)窮小量、無(wú)窮大量的概念;無(wú)窮小量的性質(zhì)以及無(wú)窮小量階的比較。第二節(jié)函數(shù)的連續(xù)性復(fù)習(xí)考試要求1.理解函數(shù)在一點(diǎn)處連續(xù)與間斷的概念,理解函數(shù)在一點(diǎn)處連續(xù)與極限存在之間的關(guān)系,掌握判斷函數(shù)(含分段函數(shù))在一點(diǎn)處連續(xù)性的方法。2.會(huì)求函數(shù)的間斷點(diǎn)。3.掌握在閉區(qū)間上連續(xù)函數(shù)的性質(zhì)會(huì)用它們證明一些簡(jiǎn)單命題。4.理解初等函數(shù)在其定義區(qū)間上的連續(xù)性,會(huì)利用函數(shù)連續(xù)性求極限。主要知識(shí)內(nèi)容(一)函數(shù)連續(xù)的概念1.函數(shù)在點(diǎn)x0處連續(xù)定義1設(shè)函數(shù)y=f(x)在點(diǎn)x0的某個(gè)鄰域內(nèi)有定義,如果當(dāng)自變量的改變量x(初值為x0)趨近于0

17、時(shí),相應(yīng)的函數(shù)的改變量y也趨近于0,即則稱函數(shù)y=f(x)在點(diǎn)x0處連續(xù)。函數(shù)y=f(x)在點(diǎn)x0連續(xù)也可作如下定義:定義2設(shè)函數(shù)y=f(x)在點(diǎn)x0的某個(gè)鄰域內(nèi)有定義,如果當(dāng)xx0時(shí),函數(shù)y=f(x)的極限值存在,且等于x0處的函數(shù)值f(x0),即定義3設(shè)函數(shù)y=f(x),如果,則稱函數(shù)f(x)在點(diǎn)x0處左連續(xù);如果,則稱函數(shù)f(x)在點(diǎn)x0處右連續(xù)。由上述定義2可知如果函數(shù)y=f(x)在點(diǎn)x0處連續(xù),則f(x)在點(diǎn)x0處左連續(xù)也右連續(xù)。2.函數(shù)在區(qū)間a,b上連續(xù)定義如果函數(shù)f(x)在閉區(qū)間a,b上的每一點(diǎn)x處都連續(xù),則稱f(x)在閉區(qū)間a,b上連續(xù),并稱f(x)為a,b上的連續(xù)函數(shù)。這里,

18、f(x)在左端點(diǎn)a連續(xù),是指滿足關(guān)系:,在右端點(diǎn)b連續(xù),是指滿足關(guān)系:,即f(x)在左端點(diǎn)a處是右連續(xù),在右端點(diǎn)b處是左連續(xù)??梢宰C明:初等函數(shù)在其定義的區(qū)間內(nèi)都連續(xù)。3.函數(shù)的間斷點(diǎn)定義如果函數(shù)f(x)在點(diǎn)x0處不連續(xù)則稱點(diǎn)x0為f(x)一個(gè)間斷點(diǎn)。由函數(shù)在某點(diǎn)連續(xù)的定義可知,若f(x)在點(diǎn)x0處有下列三種情況之一:(1)在點(diǎn)x0處,f(x)沒(méi)有定義;(2)在點(diǎn)x0處,f(x)的極限不存在;(3)雖然在點(diǎn)x0處f(x)有定義,且存在,但,則點(diǎn)x0是f(x)一個(gè)間斷點(diǎn)。,則f(x)在a.x=0,x=1處都間斷b.x=0,x=1處都連續(xù)c.x=0處間斷,x=1處連續(xù)d.x=0處連續(xù),x=1處間斷

19、解:x=0處,f(0)=0f(0-0)f(0+0)x=0為f(x)的間斷點(diǎn)x=1處,f(1)=1f(1-0)=f(1+0)=f(1)f(x)在x=1處連續(xù) 答案c9703設(shè),在x=0處連續(xù),則k等于a.0 b. c. d.2分析:f(0)=k答案b例30209設(shè)在x=0處連續(xù),則a=解:f(0)=e0=1f(0)=f(0-0)=f(0+0)a=1 答案1(二)函數(shù)在一點(diǎn)處連續(xù)的性質(zhì)由于函數(shù)的連續(xù)性是通過(guò)極限來(lái)定義的,因而由極限的運(yùn)算法則,可以得到下列連續(xù)函數(shù)的性質(zhì)。 定理1.12(四則運(yùn)算)設(shè)函數(shù)f(x),g(x)在x0處均連續(xù),則(1)f(x)g(x)在x0處連續(xù)(2)f(x)g(x)在x0

20、處連續(xù)(3)若g(x0)0,則在x0處連續(xù)。定理1.13(復(fù)合函數(shù)的連續(xù)性)設(shè)函數(shù)u=g(x)在x=x0處連續(xù),y=f(u)在u0=g(x0)處連續(xù),則復(fù)合函數(shù)y=fg(x)在x=x0處連續(xù)。在求復(fù)合函數(shù)的極限時(shí),如果u=g(x),在x0處極限存在,又y=f(u)在對(duì)應(yīng)的處連續(xù),則極限符號(hào)可以與函數(shù)符號(hào)交換。即定理1.14(反函數(shù)的連續(xù)性)設(shè)函數(shù)y=f(x)在某區(qū)間上連續(xù),且嚴(yán)格單調(diào)增加(或嚴(yán)格單調(diào)減少),則它的反函數(shù)x=f-1(y)也在對(duì)應(yīng)區(qū)間上連續(xù),且嚴(yán)格單調(diào)增加(或嚴(yán)格單調(diào)減少)。(三)閉區(qū)間上連續(xù)函數(shù)的性質(zhì)在閉區(qū)間a,b上連續(xù)的函數(shù)f(x),有以下幾個(gè)基本性質(zhì),這些性質(zhì)以后都要用到。定

21、理1.15(有界性定理)如果函數(shù)f(x)在閉區(qū)間a,b上連續(xù),則f(x)必在a,b上有界。定理1.16(最大值和最小值定理)如果函數(shù)f(x)在閉區(qū)間a,b上連續(xù),則在這個(gè)區(qū)間上一定存在最大值和最小值。定理1.17(介值定理)如果函數(shù)f(x)在閉區(qū)間a,b上連續(xù),且其最大值和最小值分別為m和m,則對(duì)于介于m和m之間的任何實(shí)數(shù)c,在a,b上至少存在一個(gè),使得推論(零點(diǎn)定理)如果函數(shù)f(x)在閉區(qū)間a,b上連續(xù),且f(a)與f(b)異號(hào),則在a,b內(nèi)至少存在一個(gè)點(diǎn),使得f()=0(四)初等函數(shù)的連續(xù)性由函數(shù)在一點(diǎn)處連續(xù)的定理知,連續(xù)函數(shù)經(jīng)過(guò)有限次四則運(yùn)算或復(fù)合運(yùn)算而得的函數(shù)在其定義的區(qū)間內(nèi)是連續(xù)函數(shù)

22、。又由于基本初等函數(shù)在其定義區(qū)間內(nèi)是連續(xù)的,可以得到下列重要結(jié)論。定理1.18初等函數(shù)在其定義的區(qū)間內(nèi)連續(xù)。利用初等函數(shù)連續(xù)性的結(jié)論可知:如果f(x)是初等函數(shù),且x0是定義區(qū)間內(nèi)的點(diǎn),則f(x)在x0處連續(xù)也就是說(shuō),求初等函數(shù)在定義區(qū)間內(nèi)某點(diǎn)處的極限值,只要算出函數(shù)在該點(diǎn)的函數(shù)值即可。04070611 例1.證明三次代數(shù)方程x3-5x+1=0在區(qū)間(0,1)內(nèi)至少有一個(gè)實(shí)根.證:設(shè)f(x)=x3-5x+1f(x)在0,1上連續(xù)f(0)=1 f(1)=-3由零點(diǎn)定理可知,至少存在一點(diǎn)(0,1)使得f()=0,3-5+1=0即方程在(0,1)內(nèi)至少有一個(gè)實(shí)根。本章小結(jié)函數(shù)、極限與連續(xù)是微積分中最

23、基本、最重要的概念之一,而極限運(yùn)算又是微積分的三大運(yùn)算中最基本的運(yùn)算之一,必須熟練掌握,這會(huì)為以后的學(xué)習(xí)打下良好的基礎(chǔ)。這一章的內(nèi)容在考試中約占15%,約為22分左右?,F(xiàn)將本章的主要內(nèi)容總結(jié)歸納如下:一、概念部分重點(diǎn):極限概念,無(wú)窮小量與等價(jià)無(wú)窮小量的概念,連續(xù)的概念。極限概念應(yīng)該明確極限是描述在給定變化過(guò)程中函數(shù)變化的性態(tài),極限值是一個(gè)確定的常數(shù)。函數(shù)在一點(diǎn)連續(xù)性的三個(gè)基本要素:(1)f(x)在點(diǎn)x0有定義。(2)存在。(3)。常用的是f(x0-0)=f(x0+0)=f(x0)。二、運(yùn)算部分重點(diǎn):求極限,函數(shù)的點(diǎn)連續(xù)性的判定。1.求函數(shù)極限的常用方法主要有:(1)利用極限的四則運(yùn)算法則求極限

24、;對(duì)于“”型不定式,可考慮用因式分解或有理化消去零因子法。(2)利用兩個(gè)重要極限求極限;(3)利用無(wú)窮小量的性質(zhì)求極限; (4)利用函數(shù)的連續(xù)性求極限;若f(x)在x0處連續(xù),則。(5)利用等價(jià)無(wú)窮小代換定理求極限;(6)會(huì)求分段函數(shù)在分段點(diǎn)處的極限;(7)利用洛必達(dá)法則求未定式的極限。2.判定函數(shù)的連續(xù)性,利用閉區(qū)間上連續(xù)函數(shù)的零點(diǎn)定理證明方程的根的存在性。emloyment tribunals sort out disagreements between employers and employees. you may need to make a claim to an employme

25、nt tribunal if: you dont agree with the disciplinary action your employer has taken against you your employer dismisses you and you think that you have been dismissed unfairly. for more informu, take advice from one of the organisations listed underfur ther help. employment tribunals are less formal

26、 than some other courts, but it is still a legal process and you will need to give evidence under an oath or affirmation. most people find making a claim to an employment tribunal challenging. if you are thinking about making a claim to an employment tribunal, you should get help straight away from

27、one of the organisations listed underfurther help. ation about dismissal and unfair dismissal, seedismissal. you can make a claim to an employment tribunal, even if you haventappealedagainst the disciplinary action your employer has taken against you. however, if you win your case, the tribunal may

28、reduce any compensation awarded to you as a result of your failure to appeal. remember that in most cases you must make an application to an employment tribunal within three months of the date when the event you are complaining about happened. if your application is received after this time limit, t

29、he tribunal will not usually accept i. if you are worried about how the time limits apply to you if you are being represented by a solicitor at the tribunal, they may ask you to sign an agreement where you pay their fee out of your compensation if you win the case. this is known as adamages-based ag

30、reement. in england and wales, your solicitor cant charge you more than 35% of your compensationif you win the case. you are clear about the terms of the agreement. it might be best to get advice from an experienced adviser, for example, at a citizens advice bureau. to find your nearest cab, includi

31、ng those that give advice by e-mail, click onnearest cab. for more information about making a claim to an employment tribunal, seeemployment tribunals. the (lack of) air up there watch m cay man islands-based webb, the head of fifas anti-racism taskforce, is in london for the football associations 1

32、50th anniversary celebrations and will attend citys premier league match at chelsea on sunday. i am going to be at the match tomorrow and i have asked to meet ya ya toure, he told bbc sport. for me its about how he felt and i would like to speak to him first to find out what his experience was. uefa

33、 hasopened disciplinary proceedings against cskafor the racist behaviour of their fans duringcitys 2-1 win. michel platini, president of european footballs governing body, has also ordered an immediate investigation into the referees actions. cska said they were surprised and disappointed by toures

34、complaint. in a statement the russian side added: we found no racist insults from fans of cska. age has reached the end of the beginning of a word. may be guilty in his seems to passing a lot of different life became the appearance of the same day; may be back in the past, to oneself the paranoid we

35、ird belief disillusionment, these days, my mind has been very messy, in my mind constantly. always feel oneself should go to do something, or write something. twenty years of life trajectory deeply shallow, suddenly feel something, do it.一字開(kāi)頭的年齡已經(jīng)到了尾聲。或許是愧疚于自己似乎把轉(zhuǎn)瞬即逝的很多個(gè)不同的日子過(guò)成了同一天的樣子;或許是追溯過(guò)去,對(duì)自己那些近

36、乎偏執(zhí)的怪異信念的醒悟,這些天以來(lái),思緒一直很凌亂,在腦海中不斷糾纏??傆X(jué)得自己似乎應(yīng)該去做點(diǎn)什么,或者寫點(diǎn)什么。二十年的人生軌跡深深淺淺,突然就感覺(jué)到有些事情,非做不可了。the end of our life, and can meet many things really do?而窮盡我們的一生,又能遇到多少事情是真正地非做不可? during my childhood, think lucky money and new clothes are necessary for new year, but as the advance of the age, will be more and

37、 more found that those things are optional; junior high school, thought to have a crush on just means that the real growth, but over the past three years later, his writing of alumni in peace, suddenly found that isnt really grow up, it seems is not so important; then in high school, think dont want

38、 to give vent to out your inner voice can be in the high school children of the feelings in a period, but was eventually infarction when graduation party in the throat, later again stood on the pitch he has sweat profusely, looked at his thrown a basketball hoops, suddenly found himself has already

39、cant remember his appearance. baumgartner the disappointing news: mission aborted. r plays an important role in this mission. starting at the ground, conditions have to be very calm - winds less than 2 mph, with no precipitation or humidity and limited cloud cover. the balloon, with capsule attached

40、, will move through the lower level of the atmosphere (the troposphere) where our day-to-day weather lives. it will climb higher than the tip of mount everest (5.5 miles/8.85 kilometers), drifting even higher than the cruising altitude of commercial airliners (5.6 miles/9.17 kilometers) and into the stratosphere. as he crosses the boundary layer (called the tropopause),e can expect a lot of turbulence. we often close ourselves off when traumatic events happen in our lives; instead of letting the world soften us, we let it

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