考研數(shù)學(xué)容易混淆的概念辨析歸納

高等數(shù)學(xué)部分易混淆概念第一章：函數(shù)與極限一、數(shù)列極限大小的判斷例1：判斷命題是否正確．若SKIPIF1<0，且序列SKIPIF1<0的極限存在，SKIPIF1<0解答：不正確．在題設(shè)下只能保證SKIPIF1<0，不能保證SKIPIF1<0．例如：SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0．例2．選擇題設(shè)SKIPIF1<0，且SKIPIF1<0（）A．存在且等于零B.存在但不一定等于零C．不一定存在D.一定不存在答：選項(xiàng)C正確分析：若SKIPIF1<0，由夾逼定理可得SKIPIF1<0，故不選A與D.取SKIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，但SKIPIF1<0不存在，所以B選項(xiàng)不正確，因此選C．例3．設(shè)SKIPIF1<0SKIPIF1<0（）A．都收斂于SKIPIF1<0B.都收斂，但不一定收斂于SKIPIF1<0C．可能收斂，也可能發(fā)散D.都發(fā)散答：選項(xiàng)A正確．分析：由于SKIPIF1<0，得SKIPIF1<0，又由SKIPIF1<0及夾逼定理得SKIPIF1<0因此，SKIPIF1<0，再利用SKIPIF1<0得SKIPIF1<0．所以選項(xiàng)A．二、無(wú)界與無(wú)窮大無(wú)界：設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，如果存在正數(shù)SKIPIF1<0，使得SKIPIF1<0則稱(chēng)函數(shù)SKIPIF1<0在SKIPIF1<0上有界，如果這樣的SKIPIF1<0不存在，就成函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)界；也就是說(shuō)如果對(duì)于任何正數(shù)SKIPIF1<0，總存在SKIPIF1<0，使SKIPIF1<0，那么函數(shù)SKIPIF1<0在SKIPIF1<0上無(wú)界．無(wú)窮大：設(shè)函數(shù)SKIPIF1<0在SKIPIF1<0的某一去心鄰域內(nèi)有定義（或SKIPIF1<0大于某一正數(shù)時(shí)有定義）．如果對(duì)于任意給定的正數(shù)SKIPIF1<0（不論它多么大），總存在正數(shù)SKIPIF1<0（或正數(shù)SKIPIF1<0），只要SKIPIF1<0適合不等式SKIPIF1<0（或SKIPIF1<0），對(duì)應(yīng)的函數(shù)值SKIPIF1<0總滿足不等式SKIPIF1<0則稱(chēng)函數(shù)SKIPIF1<0為當(dāng)SKIPIF1<0（或SKIPIF1<0）時(shí)的無(wú)窮大．例4：下列敘述正確的是：②如果SKIPIF1<0在SKIPIF1<0某鄰域內(nèi)無(wú)界，則SKIPIF1<0如果SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0某鄰域內(nèi)無(wú)界解析：舉反例說(shuō)明．設(shè)SKIPIF1<0，令SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，而SKIPIF1<0SKIPIF1<0故SKIPIF1<0在SKIPIF1<0鄰域無(wú)界，但SKIPIF1<0時(shí)SKIPIF1<0不是無(wú)窮大量，則①不正確．由定義，無(wú)窮大必?zé)o界，故②正確．結(jié)論：無(wú)窮大必?zé)o界，而無(wú)界未必?zé)o窮大．三、函數(shù)極限不存在SKIPIF1<0極限是無(wú)窮大當(dāng)SKIPIF1<0（或SKIPIF1<0）時(shí)的無(wú)窮大的函數(shù)SKIPIF1<0，按函數(shù)極限定義來(lái)說(shuō)，極限是不存在的，但是為了便于敘述函數(shù)的性態(tài)，我們也說(shuō)“函數(shù)的極限是無(wú)窮大”．但極限不存在并不代表其極限是無(wú)窮大．例5：函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0的極限不存在．四、如果SKIPIF1<0不能退出SKIPIF1<0例6：SKIPIF1<0，則SKIPIF1<0，但由于SKIPIF1<0在SKIPIF1<0的任一鄰域的無(wú)理點(diǎn)均沒(méi)有定義，故無(wú)法討論SKIPIF1<0在SKIPIF1<0的極限．結(jié)論：如果SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0的某一去心鄰域內(nèi)滿足SKIPIF1<0，則SKIPIF1<0．反之，SKIPIF1<0為無(wú)窮大，則SKIPIF1<0為無(wú)窮小。五、求函數(shù)在某點(diǎn)處極限時(shí)要注意其左右極限是否相等，求無(wú)窮大處極限要注意自變量取正無(wú)窮大和負(fù)無(wú)窮大時(shí)極限是否相等。例7．求極限SKIPIF1<0解：SKIPIF1<0，因而SKIPIF1<0時(shí)SKIPIF1<0極限不存在。SKIPIF1<0，因而SKIPIF1<0時(shí)SKIPIF1<0極限不存在。六、使用等價(jià)無(wú)窮小求極限時(shí)要注意：（1）乘除運(yùn)算中可以使用等價(jià)無(wú)窮小因子替換，加減運(yùn)算中由于用等價(jià)無(wú)窮小替換是有條件的，故統(tǒng)一不用。這時(shí)，一般可以用泰勒公式來(lái)求極限。（2）注意等價(jià)無(wú)窮小的條件，即在哪一點(diǎn)可以用等價(jià)無(wú)窮小因子替換例8：求極限SKIPIF1<0分析一：若將SKIPIF1<0寫(xiě)成SKIPIF1<0，再用等價(jià)無(wú)窮小替換就會(huì)導(dǎo)致錯(cuò)誤。分析二：用泰勒公式SKIPIF1<0原式SKIPIF1<0。例9：求極限SKIPIF1<0解：本題切忌將SKIPIF1<0用SKIPIF1<0等價(jià)代換，導(dǎo)致結(jié)果為1。SKIPIF1<0七、函數(shù)連續(xù)性的判斷（1）設(shè)SKIPIF1<0在SKIPIF1<0間斷，SKIPIF1<0在SKIPIF1<0連續(xù)，則SKIPIF1<0在SKIPIF1<0間斷。而SKIPIF1<0在SKIPIF1<0可能連續(xù)。例10．設(shè)SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0間斷，SKIPIF1<0在SKIPIF1<0連續(xù)，SKIPIF1<0在SKIPIF1<0連續(xù)。若設(shè)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0間斷，但SKIPIF1<0在SKIPIF1<0均連續(xù)。（2）“SKIPIF1<0在SKIPIF1<0點(diǎn)連續(xù)”是“SKIPIF1<0在SKIPIF1<0點(diǎn)連續(xù)”的充分不必要條件。分析：由“若SKIPIF1<0，則SKIPIF1<0”可得“如果SKIPIF1<0，則SKIPIF1<0”，因此，SKIPIF1<0在SKIPIF1<0點(diǎn)連續(xù)，則SKIPIF1<0在SKIPIF1<0點(diǎn)連續(xù)。再由例10可得，SKIPIF1<0在SKIPIF1<0點(diǎn)連續(xù)并不能推出SKIPIF1<0在SKIPIF1<0點(diǎn)連續(xù)。（3）SKIPIF1<0在SKIPIF1<0連續(xù)，SKIPIF1<0在SKIPIF1<0連續(xù)，則SKIPIF1<0在SKIPIF1<0連續(xù)。其余結(jié)論均不一定成立。第二章導(dǎo)數(shù)與微分一、函數(shù)可導(dǎo)性與連續(xù)性的關(guān)系可導(dǎo)必連續(xù)，連續(xù)不一定可導(dǎo)。例11．SKIPIF1<0在SKIPIF1<0連讀，在SKIPIF1<0處不可導(dǎo)。二、SKIPIF1<0與SKIPIF1<0可導(dǎo)性的關(guān)系（1）設(shè)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0連續(xù)，則SKIPIF1<0在SKIPIF1<0可導(dǎo)是SKIPIF1<0在SKIPIF1<0可導(dǎo)的充要條件。（2）設(shè)SKIPIF1<0，則SKIPIF1<0是SKIPIF1<0在SKIPIF1<0可導(dǎo)的充要條件。三、一元函數(shù)可導(dǎo)函數(shù)與不可導(dǎo)函數(shù)乘積可導(dǎo)性的討論設(shè)SKIPIF1<0，SKIPIF1<0在SKIPIF1<0連續(xù)，但不可導(dǎo)，又SKIPIF1<0存在，則SKIPIF1<0是SKIPIF1<0在SKIPIF1<0可導(dǎo)的充要條件。分析：若SKIPIF1<0，由定義SKIPIF1<0反之，若SKIPIF1<0存在，則必有SKIPIF1<0。用反證法，假設(shè)SKIPIF1<0，則由商的求導(dǎo)法則知SKIPIF1<0在SKIPIF1<0可導(dǎo)，與假設(shè)矛盾。利用上述結(jié)論，我們可以判斷函數(shù)中帶有絕對(duì)值函數(shù)的可導(dǎo)性。四、在某點(diǎn)存在左右導(dǎo)數(shù)時(shí)原函數(shù)的性質(zhì)（1）設(shè)SKIPIF1<0在SKIPIF1<0處存在左、右導(dǎo)數(shù)，若相等則SKIPIF1<0在SKIPIF1<0處可導(dǎo)；若不等，則SKIPIF1<0在SKIPIF1<0連續(xù)。（2）如果SKIPIF1<0在SKIPIF1<0內(nèi)連續(xù)，SKIPIF1<0，且設(shè)SKIPIF1<0則SKIPIF1<0在SKIPIF1<0處必可導(dǎo)且SKIPIF1<0。若沒(méi)有如果SKIPIF1<0在SKIPIF1<0內(nèi)連續(xù)的條件，即設(shè)SKIPIF1<0，則得不到任何結(jié)論。例11．SKIPIF1<0，顯然設(shè)SKIPIF1<0，但SKIPIF1<0，SKIPIF1<0，因此極限SKIPIF1<0不存在，從而SKIPIF1<0在SKIPIF1<0處不連續(xù)不可導(dǎo)。第三章微分中值定理與導(dǎo)數(shù)的應(yīng)用一、若SKIPIF1<0若SKIPIF1<0，不妨設(shè)SKIPIF1<0，則SKIPIF1<0，再由微分中值定理SKIPIF1<0SKIPIF1<0同理，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0若SKIPIF1<0，再由微分中值定理SKIPIF1<0SKIPIF1<0同理可證SKIPIF1<0時(shí)，必有SKIPIF1<0第八章多元函數(shù)微分法及其應(yīng)用8.1多元函數(shù)的基本概念1.SKIPIF1<0,SKIPIF1<0,使得當(dāng)SKIPIF1<0,SKIPIF1<0且SKIPIF1<0時(shí),有SKIPIF1<0,那么SKIPIF1<0成立了嗎?成立,與原來(lái)的極限差異只是描述動(dòng)點(diǎn)SKIPIF1<0與定點(diǎn)SKIPIF1<0的接近程度的方法不一樣,這里采用的是點(diǎn)的矩形鄰域,,而不是常用的圓鄰域,事實(shí)上這兩種定義是等價(jià)的.2.若上題條件中SKIPIF1<0的條件略去,函數(shù)SKIPIF1<0就在SKIPIF1<0連續(xù)嗎?為什么?如果SKIPIF1<0條件沒(méi)有,說(shuō)明SKIPIF1<0有定義,并且SKIPIF1<0包含在該點(diǎn)的任何鄰域內(nèi),由此對(duì)SKIPIF1<0,都有SKIPIF1<0,從而SKIPIF1<0,因此我們得到SKIPIF1<0SKIPIF1<0,即函數(shù)在SKIPIF1<0點(diǎn)連續(xù).3.多元函數(shù)的極限計(jì)算可以用洛必塔法則嗎?為什么?不可以,因?yàn)槁灞厮▌t的理論基礎(chǔ)是柯西中值定理.8.2偏導(dǎo)數(shù)1.已知SKIPIF1<0,求SKIPIF1<0令SKIPIF1<0,SKIPIF1<0那么解出SKIPIF1<0,SKIPIF1<0得SKIPIF1<0,所以SKIPIF1<0或者SKIPIF1<08.3全微分極其應(yīng)用1.寫(xiě)出多元函數(shù)連續(xù),偏導(dǎo)存在,可微之間的關(guān)系偏導(dǎo)數(shù)SKIPIF1<0,SKIPIF1<0連續(xù)SKIPIF1<0Z可微SKIPIF1<0SKIPIF1<0連續(xù)SKIPIF1<0SKIPIF1<0極限存在偏導(dǎo)數(shù)SKIPIF1<0,SKIPIF1<0連續(xù)SKIPIF1<0偏導(dǎo)數(shù)SKIPIF1<0,SKIPIF1<0存在2.判斷二元函數(shù)SKIPIF1<0SKIPIF1<0在原點(diǎn)處是否可微.對(duì)于函數(shù)SKIPIF1<0,先計(jì)算兩個(gè)偏導(dǎo)數(shù):SKIPIF1<0SKIPIF1<0又SKIPIF1<0令SKIPIF1<0,則上式為SKIPIF1<0因而SKIPIF1<0在原點(diǎn)處可微.8.4多元復(fù)合函數(shù)的求導(dǎo)法則1.設(shè)SKIPIF1<0,SKIPIF1<0可微,求SKIPIF1<0.SKIPIF1<08.5隱函數(shù)的求導(dǎo)設(shè)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0都是由方程SKIPIF1<0所確定的具有連續(xù)偏導(dǎo)數(shù)的函數(shù),證明SKIPIF1<0.對(duì)于方程SKIPIF1<0,如果他滿足隱函數(shù)條件.例如,具有連續(xù)偏導(dǎo)數(shù)且SKIPIF1<0,則由方程SKIPIF1<0可以確定函數(shù)SKIPIF1<0,即SKIPIF1<0是SKIPIF1<0,SKIPIF1<0的函數(shù),而SKIPIF1<0,SKIPIF1<0是自變量,此時(shí)具有偏導(dǎo)數(shù)SKIPIF1<0,SKIPIF1<0同理,SKIPIF1<0,所以SKIPIF1<0.8.6多元函數(shù)的極值及其求法1.設(shè)SKIPIF1<0在點(diǎn)SKIPIF1<0處具有偏導(dǎo)數(shù),若SKIPIF1<0,SKIPIF1<0則函數(shù)SKIPIF1<0在該點(diǎn)取得極值,命題是否正確?不正確,見(jiàn)多元函數(shù)極值存在的充分必要條件.2.如果二元連續(xù)函數(shù)在有界閉區(qū)域內(nèi)有惟一的極小值點(diǎn)，且無(wú)極大值，那么該函數(shù)是否在該點(diǎn)取得最小值？不一定，對(duì)于一元函數(shù)來(lái)說(shuō)上述結(jié)論是成立的，但對(duì)于多元函數(shù)，情況較為復(fù)雜，一般來(lái)說(shuō)結(jié)論不能簡(jiǎn)單的推廣。例如，二元函數(shù)SKIPIF1<0SKIPIF1<0，SKIPIF1<0由二元函數(shù)極值判別法：SKI