求函數(shù)極限的方法三

求函數(shù)極限的方法三PAGEPAGE1一、求函數(shù)極限的方法1、運(yùn)用極限的定義例:用極限定義證明:SKIPIF1<0SKIPIF1<0SKIPIF1<0證:由SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0 取SKIPIF1<0則當(dāng)SKIPIF1<0時(shí),就有SKIPIF1<0由函數(shù)極限SKIPIF1<0定義有:SKIPIF1<0

2、利用極限的四則運(yùn)算性質(zhì)若SKIPIF1<0 SKIPIF1<0(I)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0(II)SKIPIF1<0(III)若B≠0則：SKIPIF1<0（IV）SKIPIF1<0（c為常數(shù)）上述性質(zhì)對(duì)于SKIPIF1<0例：求SKIPIF1<0解:SKIPIF1<0=SKIPIF1<03、約去零因式（此法適用于SKIPIF1<0）例:求解:原式=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0SKIPIF1<04、通分法（適用于SKIPIF1<0型）例:求SKIPIF1<0解:原式=SKIPIF1<0=SKIPIF1<0 =SKIPIF1<0 5、利用無(wú)窮小量性質(zhì)法（特別是利用無(wú)窮小量與有界量之乘積仍為無(wú)窮小量的性質(zhì)）設(shè)函數(shù)f(x)、g(x)滿(mǎn)足：SKIPIF1<0m、n、k、l為正整數(shù)。例：求下列函數(shù)極限①SKIPIF1<0、nSKIPIF1<0②SKIPIF1<0解:①令t=SKIPIF1<0則當(dāng)SKIPIF1<0時(shí)SKIPIF1<0,于是原式=SKIPIF1<0②由于SKIPIF1<0=SKIPIF1<0令：SKIPIF1<0則SKIPIF1<0SKIPIF1<0SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<011、利用函數(shù)極限的存在性定理定理:設(shè)在SKIPIF1<0的某空心鄰域內(nèi)恒有g(shù)(x)≤f(x)≤h(x)且有:SKIPIF1<0則極限SKIPIF1<0存在,且有SKIPIF1<0例:求SKIPIF1<0(a>1,n>0)解:當(dāng)x≥1時(shí),存在唯一的正整數(shù)k,使k≤x≤k+1于是當(dāng)n>0時(shí)有:SKIPIF1<0及SKIPIF1<0又SKIPIF1<0當(dāng)xSKIPIF1<0時(shí),kSKIPIF1<0有SKIPIF1<0SKIPIF1<0及SKIPIF1<0SKIPIF1<0SKIPIF1<0 SKIPIF1<0=012、用左右極限與極限關(guān)系(適用于分段函數(shù)求分段點(diǎn)處的極限，以及用定義求極限等情形)。定理：函數(shù)極限SKIPIF1<0存在且等于A的充分必要條件是左極限SKIPIF1<0及右極限SKIPIF1<0都存在且都等于A。即有：SKIPIF1<0SKIPIF1<0=SKIPIF1<0=A例：設(shè)SKIPIF1<0=SKIPIF1<0求SKIPIF1<0及SKIPIF1<0SKIPIF1<0由SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<013、羅比塔法則（適用于未定式極限）定理：若SKIPIF1<0此定理是對(duì)SKIPIF1<0型而言，對(duì)于函數(shù)極限的其它類(lèi)型，均有類(lèi)似的法則。注：運(yùn)用羅比塔法則求極限應(yīng)注意以下幾點(diǎn)：要注意條件，也就是說(shuō)，在沒(méi)有化為SKIPIF1<0時(shí)不可求導(dǎo)。應(yīng)用羅比塔法則，要分別的求分子、分母的導(dǎo)數(shù)，而不是求整個(gè)分式的導(dǎo)數(shù)。要及時(shí)化簡(jiǎn)極限符號(hào)后面的分式，在化簡(jiǎn)以后檢查是否仍是未定式，若遇到不是未定式，應(yīng)立即停止使用羅比塔法則，否則會(huì)引起錯(cuò)誤。4、當(dāng)SKIPIF1<0不存在時(shí)，本法則失效，但并不是說(shuō)極限不存在，此時(shí)求極限須用另外方法。例：求下列函數(shù)的極限①SKIPIF1<0②SKIPIF1<0解：①令f(x)=SKIPIF1<0,g(x)=lSKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0由于SKIPIF1<0但SKIPIF1<0從而運(yùn)用羅比塔法則兩次后得到SKIPIF1<0②由SKIPIF1<0故此例屬于SKIPIF1<0型，由羅比塔法則有：SKIPIF1<014、利用泰勒公式對(duì)于求某些不定式的極限來(lái)說(shuō)，應(yīng)用泰勒公式比使用羅比塔法則更為方便，下列為常用的展開(kāi)式：1、SKIPIF1<02、SKIPIF1<03、SKIPIF1<04、SKIPIF1<05、SKIPIF1<06、SKIPIF1<0上述展開(kāi)式中的符號(hào)SKIPIF1<0都有:SKIPIF1<0例:求SKIPIF1<0解:利用泰勒公式，當(dāng)SKIPIF1<0有SKIPIF1<0于是SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<015、利用拉格朗日中值定理定理:若函數(shù)f滿(mǎn)足如下條件：(I)f在閉區(qū)間上連續(xù)(II)f在(a,b)內(nèi)可導(dǎo)則在(a,b)內(nèi)至少存在一點(diǎn)SKIPIF1<0,使得SKIPIF1<0此式變形可為:SKIPIF1<0例:求SKIPIF1<0解:令SKIPIF1<0對(duì)它應(yīng)用中值定理得SKIPIF1<0即:SKIPIF1<0SKIPIF1<0連續(xù)SKIPIF1<0從而有:SKIPIF1<016、求代數(shù)函數(shù)的極限方法(1)有理式的情況，即若:SKIPIF1<0(I)當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0(II)當(dāng)SKIPIF1<0時(shí)有:①若SKIPIF1<0則SKIPIF1<0②若SKIPIF1<0而SKIPIF1<0則SKIPIF1<0③若SKIPIF1<0,SKIPIF1<0,則分別考慮若SKIPIF1<0為SKIPIF1<0的s重根,即:SKIPIF1<0也為SKIPIF1<0的r重根,即:SKIPIF1<0可得結(jié)論如下：SKIPIF1<0例：求下列函數(shù)的極限①SKIPIF1<0②SKIPIF1<0解:①分子，分母的最高次方相同，故SKIPIF1<0=SKIPIF1<0②SKIPIF1<0SKIPIF1<0SKIPIF1<0必含有（x-1）之因子，即有1的重根故有:SKIPIF1<0(2)無(wú)理式的情況。雖然無(wú)理式情況不同于有理式，但求極限方法完全類(lèi)同，這里就不再一一詳述.在這里我主要舉例說(shuō)明有理化的方法求極限。例：求SKIPIF1<0解:SKIPIF1<0SKIPIF1<0二、多種方法的綜合運(yùn)用上述介紹了求解極限的基本方法，然而，每一道題目并非只有一種方法。因此我們?cè)诮忸}中要注意各種方法的綜合運(yùn)用的技巧，使得計(jì)算大為簡(jiǎn)化。例:求SKIPIF1<0[解法一]:SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0=SKIPIF1<0注:此法采用羅比塔法則配合使用兩個(gè)重要極限法。[解法二]:SKIPIF1<0=SKIPIF1<0注：此解法利用“三角和差化積法”配合使用兩個(gè)重要極限法。[解法三]:SKIPIF1