極限的概念及其運(yùn)算

極限的概念及其運(yùn)算高考要求極限的概念及其滲透的思想，在數(shù)學(xué)中占有重要的地位，它是人們研究許多問題的工具舊教材中原有的數(shù)列極限一直是歷年高考中重點(diǎn)考查的內(nèi)容之一本節(jié)內(nèi)容主要是指導(dǎo)考生深入地理解極限的概念，并在此基礎(chǔ)上能正確熟練地進(jìn)行有關(guān)極限的運(yùn)算問題重難點(diǎn)歸納1學(xué)好數(shù)列的極限的關(guān)鍵是真正從數(shù)列的項(xiàng)的變化趨勢理解數(shù)列極限學(xué)好函數(shù)的極限的關(guān)鍵是真正從函數(shù)值或圖象上點(diǎn)的變化趨勢理解函數(shù)極限2運(yùn)算法則中各個(gè)極限都應(yīng)存在都可推廣到任意有限個(gè)極限的情況，不能推廣到無限個(gè)在商的運(yùn)算法則中，要注意對式子的恒等變形，有些題目分母不能直接求極限3注意在平時(shí)學(xué)習(xí)中積累一些方法和技巧，如SKIPIF1<0SKIPIF1<0典型題例示范講解例1已知SKIPIF1<0(SKIPIF1<0－ax－b)=0,確定a與b的值例2設(shè)數(shù)列a1,a2,…,an,…的前n項(xiàng)的和Sn和an的關(guān)系是Sn=1－ban－SKIPIF1<0,其中b是與n無關(guān)的常數(shù)，且b≠－1(1)求an和an－1的關(guān)系式；(2)寫出用n和b表示an的表達(dá)式；(3)當(dāng)0＜b＜1時(shí)，求極限SKIPIF1<0Sn例3求SKIPIF1<0學(xué)生鞏固練習(xí)1an是(1+x)n展開式中含x2的項(xiàng)的系數(shù)，則SKIPIF1<0等于A2B0C1D－12若三數(shù)a,1,c成等差數(shù)列且a2,1,c2又成等比數(shù)列，則SKIPIF1<0的值是()A0B1C0或1D不存在3SKIPIF1<0=_________4若SKIPIF1<0=1,則ab的值是_________5在數(shù)列{an}中，已知a1=SKIPIF1<0,a2=SKIPIF1<0,且數(shù)列{an+1－SKIPIF1<0an}是公比為SKIPIF1<0的等比數(shù)列，數(shù)列{lg(an+1－SKIPIF1<0an}是公差為－1的等差數(shù)列(1)求數(shù)列{an}的通項(xiàng)公式；(2)Sn=a1+a2+…+an(n≥1),求SKIPIF1<0Sn6設(shè)f(x)是x的三次多項(xiàng)式，已知SKIPIF1<0=1,試求SKIPIF1<0的值(a為非零常數(shù))7已知數(shù)列{an},{bn}都是由正數(shù)組成的等比數(shù)列，公式分別為p、q,其中p＞q,且p≠1,q≠1,設(shè)cn=an+bn,Sn為數(shù)列{cn}的前n項(xiàng)和，求SKIPIF1<0的值8已知數(shù)列{an}是公差為d的等差數(shù)列，d≠0且a1=0,bn=2SKIPIF1<0(n∈N*),Sn是{bn}的前n項(xiàng)和，Tn=SKIPIF1<0(n∈N*)(1)求{Tn}的通項(xiàng)公式；(2)當(dāng)d＞0時(shí)，求SKIPIF1<0Tn參考答案1解析SKIPIF1<0，SKIPIF1<0答案A2解析SKIPIF1<0答案C二、3解析SKIPIF1<0SKIPIF1<0答案SKIPIF1<04解析原式=SKIPIF1<0SKIPIF1<0∴a·b=8SKIPIF1<0答案8SKIPIF1<05解(1)由{an+1－SKIPIF1<0an}是公比為SKIPIF1<0的等比數(shù)列，且a1=SKIPIF1<0,a2=SKIPIF1<0,∴an+1－SKIPIF1<0an=(a2－SKIPIF1<0a1)(SKIPIF1<0)n-1=(SKIPIF1<0－SKIPIF1<0×SKIPIF1<0)(SKIPIF1<0)n-1=SKIPIF1<0,∴an+1=SKIPIF1<0an+SKIPIF1<0①又由數(shù)列{lg(an+1－SKIPIF1<0an)}是公差為－1的等差數(shù)列，且首項(xiàng)lg(a2－SKIPIF1<0a1)=lg(SKIPIF1<0－SKIPIF1<0×SKIPIF1<0)=－2,∴其通項(xiàng)lg(an+1－SKIPIF1<0an)=－2+(n－1)(－1)=－(n+1),∴an+1－SKIPIF1<0an=10－(n+1),即an+1=SKIPIF1<0an+10－(n+1)②①②聯(lián)立解得an=SKIPIF1<0［(SKIPIF1<0)n+1－(SKIPIF1<0)n+1］(2)Sn=SKIPIF1<0SKIPIF1<06解由于SKIPIF1<0=1，可知，f(2a)=0 ①同理f(4a)=0②由①②可知f(x)必含有(x－2a)與(x－4a)的因式，由于f(x)是x的三次多項(xiàng)式，故可設(shè)f(x)=A(x－2a)(x－4a)(x－C),這里A、C均為待定的常數(shù)，SKIPIF1<0SKIPIF1<0,即4a2A－2aCA=－1③同理，由于SKIPIF1<0=1,得A(4a－2a)(4a－C)=1,即8a2A－2aCA=1④由③④得C=3a,A=SKIPIF1<0,因而f(x)=SKIPIF1<0(x－2a)(x－4a)(x－3a),SKIPIF1<0SKIPIF1<0由數(shù)列{an}、{bn}都是由正數(shù)組成的等比數(shù)列，知p＞0,q＞0SKIPIF1<0當(dāng)p＜1時(shí),q＜1,SKIPIF1<0SKIPIF1<08解(1)an=(n－1)d,bn=2SKIPIF1<0=2(n－1)dSn=b1+b2+b3+…+bn=20+2d+22d+…+2(n－1)d由d≠0,2d≠1,∴Sn=SKIPIF1<0∴Tn=SKIPIF1<0(2)當(dāng)d＞0時(shí)，2d＞1SKIPIF1<0課前后備注極限的概念及其運(yùn)算高考要求極限的概念及其滲透的思想，在數(shù)學(xué)中占有重要的地位，它是人們研究許多問題的工具舊教材中原有的數(shù)列極限一直是歷年高考中重點(diǎn)考查的內(nèi)容之一本節(jié)內(nèi)容主要是指導(dǎo)考生深入地理解極限的概念，并在此基礎(chǔ)上能正確熟練地進(jìn)行有關(guān)極限的運(yùn)算問題重難點(diǎn)歸納1學(xué)好數(shù)列的極限的關(guān)鍵是真正從數(shù)列的項(xiàng)的變化趨勢理解數(shù)列極限學(xué)好函數(shù)的極限的關(guān)鍵是真正從函數(shù)值或圖象上點(diǎn)的變化趨勢理解函數(shù)極限2運(yùn)算法則中各個(gè)極限都應(yīng)存在都可推廣到任意有限個(gè)極限的情況，不能推廣到無限個(gè)在商的運(yùn)算法則中，要注意對式子的恒等變形，有些題目分母不能直接求極限3注意在平時(shí)學(xué)習(xí)中積累一些方法和技巧，如SKIPIF1<0SKIPIF1<0典型題例示范講解例1已知SKIPIF1<0(SKIPIF1<0－ax－b)=0,確定a與b的值命題意圖在數(shù)列與函數(shù)極限的運(yùn)算法則中，都有應(yīng)遵循的規(guī)則，也有可利用的規(guī)律，既有章可循，有法可依因而本題重點(diǎn)考查考生的這種能力也就是本知識的系統(tǒng)掌握能力知識依托解決本題的閃光點(diǎn)是對式子進(jìn)行有理化處理，這是求極限中帶無理號的式子常用的一種方法錯(cuò)解分析本題難點(diǎn)是式子的整理過程繁瑣，稍不注意就有可能出錯(cuò)技巧與方法有理化處理解SKIPIF1<0SKIPIF1<0要使上式極限存在，則1－a2=0,當(dāng)1－a2=0時(shí)，SKIPIF1<0∴SKIPIF1<0解得SKIPIF1<0例2設(shè)數(shù)列a1,a2,…,an,…的前n項(xiàng)的和Sn和an的關(guān)系是Sn=1－ban－SKIPIF1<0,其中b是與n無關(guān)的常數(shù)，且b≠－1(1)求an和an－1的關(guān)系式；(2)寫出用n和b表示an的表達(dá)式；(3)當(dāng)0＜b＜1時(shí)，求極限SKIPIF1<0Sn命題意圖歷年高考中多出現(xiàn)的題目是與數(shù)列的通項(xiàng)公式，前n項(xiàng)和Sn等有緊密的聯(lián)系有時(shí)題目是先依條件確定數(shù)列的通項(xiàng)公式再求極限，或先求出前n項(xiàng)和Sn再求極限，本題考查學(xué)生的綜合能力知識依托解答本題的閃光點(diǎn)是分析透題目中的條件間的相互關(guān)系錯(cuò)解分析本題難點(diǎn)是第(2)中由(1)中的關(guān)系式猜想通項(xiàng)及n=1與n=2時(shí)的式子不統(tǒng)一性技巧與方法抓住第一步的遞推關(guān)系式，去尋找規(guī)律解(1)an=Sn－Sn－1=－b(an－an－1)－SKIPIF1<0=－b(an－an－1)+SKIPIF1<0(n≥2)解得an=SKIPIF1<0(n≥2)SKIPIF1<0SKIPIF1<0SKIPIF1<0例3求SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0學(xué)生鞏固練習(xí)1an是(1+x)n展開式中含x2的項(xiàng)的系數(shù)，則SKIPIF1<0等于A2B0C1D－12若三數(shù)a,1,c成等差數(shù)列且a2,1,c2又成等比數(shù)列，則SKIPIF1<0的值是()A0B1C0或1D不存在3SKIPIF1<0=_________4若SKIPIF1<0=1,則ab的值是_________5在數(shù)列{an}中，已知a1=SKIPIF1<0,a2=SKIPIF1<0,且數(shù)列{an+1－SKIPIF1<0an}是公比為SKIPIF1<0的等比數(shù)列，數(shù)列{lg(an+1－SKIPIF1<0an}是公差為－1的等差數(shù)列(1)求數(shù)列{an}的通項(xiàng)公式；(2)Sn=a1+a2+…+an(n≥1),求SKIPIF1<0Sn6設(shè)f(x)是x的三次多項(xiàng)式，已知SKIPIF1<0=1,試求SKIPIF1<0的值(a為非零常數(shù))7已知數(shù)列{an},{bn}都是由正數(shù)組成的等比數(shù)列，公式分別為p、q,其中p＞q,且p≠1,q≠1,設(shè)cn=an+bn,Sn為數(shù)列{cn}的前n項(xiàng)和，求SKIPIF1<0的值8已知數(shù)列{an}是公差為d的等差數(shù)列，d≠0且a1=0,bn=2SKIPIF1<0(n∈N*),Sn是{bn}的前n項(xiàng)和，Tn=SKIPIF1<0(n∈N*)(1)求{Tn}的通項(xiàng)公式；(2)當(dāng)d＞0時(shí)，求SKIPIF1<0Tn參考答案1解析SKIPIF1<0，SKIPIF1<0答案A2解析SKIPIF1<0答案C二、3解析SKIPIF1<0SKIPIF1<0答案SKIPIF1<04解析原式=SKIPIF1<0SKIPIF1<0∴a·b=8SKIPIF1<0答案8SKIPIF1<05解(1)由{an+1－SKIPIF1<0an}是公比為SKIPIF1<0的等比數(shù)列，且a1=SKIPIF1<0,a2=SKIPIF1<0,∴an+1－SKIPIF1<0an=(a2－SKIPIF1<0a1)(SKIPIF1<0)n-1=(SKIPIF1<0－SKIPIF1<0×SKIPIF1<0)(SKIPIF1<0)n-1=SKIPIF1<0,∴an+1=SKIPIF1<0an+SKIPIF1<0①又由數(shù)列{lg(an+1－SKIPIF1<0an)}是公差為－1的等差數(shù)列，且首項(xiàng)lg(a2－SKIPIF1<0a1)=lg(SKIPIF1<0－SKIPIF1<0×SKIPIF1<0)=－2,∴其通項(xiàng)lg(an+1－SKIPIF1<0an)=－2+(n－1)(－1)=－(n+1),∴an+1－SKIPIF1<0an=10－(n+1),即an+1=SKIPIF1<0an+10－(n+1)②①②聯(lián)立解得an=SKIPIF1<0［(SKIPIF1<0)n+1－(SKIPIF1<0)n+1］(2)Sn=SKIPIF1<0