x(5)limx[ln(2?x)?lnx]?lim2?x?ln2?x?lim2ln??1?2??2x??x??2xx???x?x?2limx??ln???1?2?2?x??2x?2???2?ln???lim????1?x?? x????2lne?2.(6)令x?1?t，则当x?1时，t?0.
lim1?x1x?1lnx??limtt?0ln(1?t)??1??1limln?1t???1lne??1.t?0ln(1?t)??lim(1t?0?t)t??34. 利用取对数的方法求下列幂指函数的极限：
1(1)lim?1xxxexx?x?x?a?b?c?x?0; (2)limx?0??3??;
(3)lim?x????sin1x?cos1?x?1?xx??; (4)limx????1?x2??.1解：(1)令y?(ex?x)x，则lny?1xln(ex?x) 于是：
lnex?ln?x?limx?0lny?lim1x?0xln?ex?x??lim1x?0xlnex??x???1?ex???1?ex???limx?0x xxe?lim?e11xxx?0??1?x?e??1?x?ex??????limx?0ex?limx?0ln???1??xln??1ex?? ?1?1?lne?2即ln?limy??2 即limy?e21 即lim?exx?0?x?xx?0x?0?e2. 1xxx(2)令y??x?a?b?c?1ax?bx?cx?3??，则lny?xln3 于是
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1ax?bx?cxlim(lny)?limlnx?0x?0x331??ax?bx?cx?3?ax?bx?cx?3??limln?1????x?0x??3????xxxxxax?bx?cx?33a?b?c?3?a?b?c?3?a?b?cx?3 ?lim?limln?1??x?0x?03x3??31?ax?1bx?1cx?1???ax?bx?cx?3?ax?bx?cx?3??lim??????ln?lim?1??3x?0?xx?0xx???3????1?(lna?lnb?lnc)?lne?ln3abc3xxx3y?ln3abc, 故limy?即lim(lny)?ln3abc, 即lnlimx?0x?0x?0??3abc
即 lim?x?0?a?b?c?3?abc. ?3??xxxx1x11?1???1(3)令y??sin?cos?，则lny?xln?sin?cos? xx??xx??于是
??11????11?sin?cos?1?limlny?limxln?1??sin?cos?1?xx?x??x???xx?????????11??1?sin?cos?1?x??x111???11??sin1?cos1?1??limx?sin?cos?1??ln?1??sin?cos?1??xxx???xx???xx??11??1sin1?cos????11??xx?11?sin?cos?1???lim????ln?lim?1??sin?cos?1??xx?x??11xx???????x?????x?x?2?1?1??1sin????2x??lim?lim?x???lne?(1?0)?lne?1 x??1??x??1??xx??
y?1 故limy?e 即limlny?1 从而lnlimx??x??x????1??1即 lim?sin?cos??e.
x???xx?x 22
x(4)令y????1?1??1?x2??，则lny?xln??1?x2?? 于是：
1x2lim(lnx2x??y)?limx??xln???1?1???1??x2???limx??xln????1?x2????x2x2?lim1x??xln???1?1?1?1?x2???limx??x?limx??ln??1?x2?? ?0?lne?0即 lim(lnyx??y)?0, ln?limx????0 x?limx??y?1 即lim?1?x????1?x2???1. 35. 求下列函数在指定点处的左、右极限，并说明在该点处函数的极限是否存在？?(1)f(x)?x?,x?0, 在x?0处; ?x?1x?0,?x?2,(2)f(x)?x?2? 在x?2处. ?1?x?2x?0解：(1)limf(x)?limx?x?0?x?0?x?limxx?0?x?1, lx?i0m?fx(?)x?l0?ixxm??x?0lxxim?? 因为 xlim?0?f(x)?xlim?0?f(x) 所以limx?0f(x)不存在.
(2)f(1xlim?2?x)?xlim?2?x?2???, xlim?2?f(x)?xlim(?2?x?2)?4
因为xlim?2?f(x)不存在，所以limx?2f(x)不存在. 36. 研究下列函数的连续性，并画出图形：
(1)f(x)???x2,0?x?1, (2)f(x)??x,x?1,?2?x,1?x?2;??1,x?1; (3)f(x)?limnx?n?x1?x2nn??nx?n?x; (4)f(x)?limn??1?x2nx.
解：(1)由初等函数的连续性知，f(x)在（0，1），（1，2）内连续，
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1f(x)?lim(2?x)?1, limf(x)?limx?1 又?lim????x?1x?1x?1x?12?limf(x)?1, 而f(1)?1，?f(x)在x?1处连续，
x?1f(x)?limx?0?f(0)，知f(x)在x?0处右连续， 又，由lim??x?0x?02综上所述，函数f(x)在[0，2)内连续. 函数图形如下：
图1-2
(2) 由初等函数的连续性知f(x)在(??,?1),(?1,1),(1,??)内连续，又由
x??1?limf(x)?lim?1?1, lim?f(x)?lim?x??1,
x??1x??1x??1知lim?f(x)不存在，于是f(x)在x??1处不连续.
x??1f(x)?limx?1, limf(x)?lim1?1, 又由lim????x?1x?1x?1x?1及f(1)?1知limf(x)?f(1)，从而f(x)在x=1处连续，
x?1综上所述，函数f(x)在(??,?1)及(?1,??)内连续，在x??1处间断.函数图形如下：
图1-3
nx?n?xn2x?1?lim2x??1, (3)∵当x<0时，f(x)?limxn??n?n?xn??n?1n0?n0?0, 当x=0时，f(x)?lim0n??n?n0nx?n?x当x>0时，f(x)?limxn??n?n?x1n2x?1n2x?1 ?lim2x?limn??n?1n??1?1n2x1?24
?f(x)?limn?nn??nx?n?xx?x??1,x?0,???0,x?0, ?1,x?0.?由初等函数的连续性知f(x)在(??,0),(0,??)内连续，
f(x)?lim1?1, limf(x)?lim(?1)??1 又由 lim????x?0x?0x?0x?0知limf(x)不存在，从而f(x)在x?0处间断.综上所述，函数f(x)在(??,0),(0,??)内
x?0连续，在x?0处间断.图形如下：
图1-4
1?x2nx?0, (4)当|x|=1时，f(x)?limn??1?x2n1?x2nx?x, 当|x|<1时，f(x)?limn??1?x2n1?x2n当|x|>1时，f(x)?limn??1?x2n?1??1?2?x?lim?x?n?x??x
n???1??1?2??x?n?x,?即 f(x)??0,??x,?x?1,x?1, x?1.由初等函数的连续性知f(x)在(－∞,－1),(－1,1),(1,+∞)内均连续，又由
x??1?limf(x)?lim?(?x)?1, lim?f(x)?lim?x??1
x??1x??1x??1知limf(x)不存在，从而f(x)在x??1处不连续.
x??1f(x)?lim(?x)??1, limf(x)?limx?1 又由 lim????x?1x?1x?1x?1知limf(x)不存在，从而f(x)在x?1处不连续.
x?1综上所述，f(x)在(－∞,－1),(－1,1),(1,+∞)内连续，在x??1处间断.
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