=lim =lim方法2：
?2cosx?2cos2x12x?2sin2x24x?22x?0?1316?1213?lim124cosxsinx?4sin2x24x23x?0?13?12
13x?0?12?????
2?sin2x?x2cos2x??1cosx??1?2 lim?2?cotx?=lim?????2222?=lim? x?0?xx?0?x?0xsinxsinxx?????? =lim??x?0?1?(x?1)cos?x422x????1??21?(x?1)(cos2x?1)??2? =lim?4x?0x??????24?1(2x)(2x)24??1?(x?1)(1?1???0(x)?22!4!? =lim?4x?0??x????1164?2424x?0(x))?1?(2x?2x?2?2x?224 =lim?4x?0x?????? ???22 =lim34?
x?0x33. 求下列极限
x4(1) limnlnnn??(nn?1)
n解. limnlnnn??(n?1)?limnn?1nn??lnn 令nn?1?x limxln(1?x)x?0?1
(2) lim1?e1?e?nx?nxn??
x?0?1???0 x?0 ??1x?0?b??, 其中a > 0, b > 0 ??n解. lim1?e1?e?nx?nxn???(3) lim?n?????解. lim?n????na?2nna?2n?1?cb?? x?1/n,c?b/a alim????x?0?2?n1xlim?x?x?0??ae??ln(1?c)?ln2xx
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=aelimx?0ln(1?c)?ln2?xxlimclnc?x?aex?01?cx?ac?aba?ab
?2?x2(1?cosx)?4. 设f(x)??1?1x??cost2dt?x0x?0x?0 x?0试讨论f(x)在x?0处的连续性与可导性.
1解. f?'(0)?limf(x)?f(0)?x?0x?lim?x?0?xxx0costdt?1x?lim?x?02?x0costdt?xx22
?limx?0cosx?1?2??lim?x?0122x22x2x2?0
(1?cosx)?1?lim?x?0 f?'(0)?limf(x)?f(0)?x?0 ?limx?0x2sinx?2x??lim?x?02(1?cosx)?xx323x2?lim?x?0x2(cosx?1)6x
?0
所以 f'(0)?0, f(x)在x?0处连续可导. 5. 求下列函数的间断点并判别类型
1(1) f(x)?2x?11
2x?111解. f(0)?limx?0?2?x1?1?1, f(0)?lim?x?0?2x?11??1
2x?12x?1所以x = 0为第一类间断点. ?x(2x??)x?0??2cosx(2) f(x)??
1?sinx?02?x?1?解. f(+0) =－sin1, f(－0) = 0. 所以x = 0为第一类跳跃间断点; limf(x)?limsinx?1x?11x?12不存在. 所以x = 1为第二类间断点;
f(??2)不存在, 而limx??x(2x??)?22cosx??2,所以x = 0为第一类可去间断点;
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limx??k??x(2x??)?22cosx??, (k = 1, 2, …) 所以x =?k???2为第二类无穷间断点.
1??x?0?xsin6. 讨论函数f(x)?? 在x = 0处的连续性. x
x?0?ex???解. 当??0时lim(x?sinx?0?1x1x)不存在, 所以x = 0为第二类间断点;
当??0, lim(x?sinx?0?)?0, 所以
???1时,在 x = 0连续, ???1时, x = 0为第一类跳跃间断点.
7. 设f(x)在[a, b]上连续, 且a < x1 < x2 < … < xn < b, ci (I = 1, 2, 3, …, n)为任意正数, 则在(a, b)内至少存在一个?, 使 f(?)?c1f(x1)?c2f(x2)???cnc1?c2???cn.
证明: 令M =max{f(xi)}, m =min{f(xi)}
1?i?n1?i?n所以 m ?
c1f(x1)?c2f(x2)???cnc1?c2???cn? M
所以存在?( a < x1 ? ? ? xn < b), 使得f(?)?c1f(x1)?c2f(x2)???cnc1?c2???cn
8. 设f(x)在[a, b]上连续, 且f(a) < a, f(b) > b, 试证在(a, b)内至少存在一个?, 使f(?) = ?. 证明: 假设F(x) = f(x)－x, 则F(a) = f(a)－a < 0, F(b) = f(b)－b > 0
于是由介值定理在(a, b)内至少存在一个?, 使f(?) = ?.
9. 设f(x)在[0, 1]上连续, 且0 ? f(x) ? 1, 试证在[0, 1]内至少存在一个?, 使f(?) = ?. 证明: (反证法) 反设?x?[0,1],?(x)?f(x)?x?0. 所以?(x)?f(x)?x恒大于0或恒小于0. 不妨设?x?[0,1],?(x)?f(x)?x?0. 令m?min?(x), 则m?0.
0?x?1因此?x?[0,1],?(x)?f(x)?x?m. 于是f(1)?1?m?0, 矛盾. 所以在[0, 1]内至少存在一个?, 使f(?) = ?.
10. 设f(x), g(x)在[a, b]上连续, 且f(a) < g(a), f(b) > g(b), 试证在(a, b)内至少存在一个?, 使 f(?) = g(?).
证明: 假设F(x) = f(x)－g(x), 则F(a) = f(a)－g(a) < 0, F(b) = f(b)－g(b) > 0 于是由介值定理在(a, b)内至少存在一个?, 使f(?) = ?. 11. 证明方程x5－3x－2 = 0在(1, 2)内至少有一个实根. 证明: 令F(x) = x5－3x－2, 则F(1) =－4 < 0, F(2) = 24 > 0 所以 在(1, 2)内至少有一个?, 满足F(?) = 0.
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f(x)??sin3x12. 设f(x)在x = 0的某领域内二阶可导, 且lim????0, 求32x?0x??xf(0),f'(0),f''(0)及limf(x)?3x2x?0.
sin3xf(x)?sin3x?xf(x)?sin3x解. lim???lim?lim?323x?0x?0x?0x?x?xxx?f(x)2?0. 所以
?sin3x? lim??f(x)??0. f(x)在x = 0的某领域内二阶可导, 所以f(x),f'(x)在x = 0
x?0?x?连续. 所以f(0) = －3. 因为
sin3xxx?f(x)2sin3x limx?0?0, 所以lim3??limx?0x?3?f(x)?3x2x?0?0, 所以
sin3xxx2 limf(x)?3x2x?0?lim3x?sin3xx3x?0?lim3?3cos3x3x2x?0
=lim3sin3x2xx?0?92
?limf(x)?3x?limx?x?0 f'(0)?lim由limf(x)?3x2f(x)?f(0)x?0f(x)?3x2x?0x?0?0?92?0
x?0?92, 将f(x)台劳展开, 得
12!xf''(0)x?0(x)?3222f(0)?f'(0)x? limx?0?92, 所以
12f''(0)?92, 于是
f''(0)?9.
(本题为2005年教材中的习题, 2008年教材中没有选入. 笔者认为该题很好, 故在题解中加
入此题)
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