高數(shù)極限經(jīng)典60題分步驟詳解

1、高數(shù)極限經(jīng)典高數(shù)極限經(jīng)典60題分步驟詳解題分步驟詳解1.求數(shù)列極限)sin1(sinlimnnn????本題求解極限的關(guān)鍵是運(yùn)用三角函數(shù)和差化積公式，將算式進(jìn)行轉(zhuǎn)化，進(jìn)而求出極限，過(guò)程如下：＝?nnsin1sin??21sin21cos2nnnn????＝)1121sin(21cos2nnnnnnnn?????????＝)121sin(21cos2nnnn????)(0???n＝0?)sin1(sinlimnnn????這是因?yàn)?，?dāng)時(shí)，

2、，而是有界函數(shù)，有??n0)1(21sin???nn21cosnn??界函數(shù)與無(wú)窮小的乘積仍然是無(wú)窮小，所以原式極限為0。2.令=，求數(shù)列極限Sn???nkkk1)!1(Snn??lim解：?)!1(1!1)!1(????nnnn＝????nkkk1)!1())!1(1!1()!1)!1(1()!41!31()!31!21()!21!11(????????????nnnn?＝1)(1)!1(1?????nn所以，=1]＝1Snn??li

3、m[lim??n)!1(1??n且求證＝0（且），從而證明＝0。具體過(guò)程如下：xxxq???lim1?q0?qnnnq??lim將函數(shù)進(jìn)行恒等變形，得到xxq＝xxxq???limxxqx)1(lim???且，1?q?0?q。11??q可知，當(dāng)時(shí)，.即為型未定式。???x??xq)1(xxqx)1(lim?????所以，應(yīng)用洛必達(dá)法則，得到＝＝＝0.xxxq??limxxqx)1(lim??)1ln()1(1limqqxx??因此可知（