数学系2002级三、四班数学分析补充材料(三) (单变元函数极限、单变元连续函数习题,2002年11月) 部分难题解析 补充题(一) 17(2)and(3).Pr any given n E N+, because En+p= en+ m+1)+(n+2 n+p)! where pEN+ P23, and (n+) <(n+i-1): 3=3,, P, we may get n+2 (n+1)n+21(n+1)(n+3 (n+p)! (n+1)!n+2(m+1)!(n+2)!"(n+p-1)!(n+p) (n+1)!n+ Thus, there holds the following inequ (n+1)!n+2 En+pen Furthermore, (ii) implies that n+2 (n+1)!(n+1) Notice that en+p→e(asp→∞). Letting p→oin(i).Then, there holds the inequality e- en ≤ (n+1)(n+1) which is the second inequality in(2). The first inequality is easy Set 0n=n!n(e-en). From(2), it is easy to see that(3)is true 习题1.3的第3题(见《简明数学分析》) (1). First, to give the formula for S:Sn=2n·3·sin360° (2)By using continuously the formula sin(2 z)=2 sin r cos ar, to give the equality: sin 60 2n-]sIn (3). Notice that the facts sin 10o 60° -and1>c0s可≥ Co It may be obtained: sin 2n.3 2 2n-1
1 2002 2002 11 17(2) and (3). Proof Set en =1+1+ 1 2! + ... + 1 n! , n ∈ N+. For any given n ∈ N+, because en+p = en + 1 (n + 1)! + 1 (n + 2)! + ··· + 1 (n + p)!, (i) where p ∈ N+, p ≥ 3, and n + 2 (n + j)! < 1 (n + j − 1)!, j = 3, ··· , p, we may get en+p − en = 1 (n + 1)! + 1 n + 2 1 (n + 1)! + n + 2 (n + 3)! ··· + n + 2 (n + p)! < 1 (n + 1)! + 1 n + 2 1 (n + 1)! + 1 (n + 2)! ··· + 1 (n + p − 1)! + 1 (n + p)! = 1 (n + 1)! + 1 n + 2 (en+p − en). Thus, there holds the following inequality en+p − en < 1 (n + 1)! + 1 n + 2 (en+p − en). (ii) Furthermore, (ii) implies that en+p − en < n + 2 (n + 1)!(n + 1). (iii) Notice that en+p → e (as p → ∞). Letting p → ∞ in (iii). Then, there holds the inequality e − en ≤ n + 2 (n + 1)!(n + 1) < 1 n!n, (v) which is the second inequality in (2). The first inequality is easy. Set θn = n!n(e − en). From (2), it is easy to see that (3) is true. 1.3 3 ( Ideas of Proof: (1). First, to give the formula for Sn: Sn = 2n · 3 · sin 360◦ 2n · 3 ; (2). By using continuously the formula sin (2x) = 2 sin x cos x, to give the equality: sin 60◦ = 2n−1 cos 60◦ 2 · cos 60◦ 22 ··· cos 60◦ 2n−1 sin 60◦ 2n−1 ; (3). Notice that the facts sin 360◦ 2n · 3 = sin 60◦ 2n−1 , and 1 > cos 60◦ 2j ≥ cos 60◦ 2 = √3 2 , j = 1, 2, ..., n − 1. It may be obtained: sin 360◦ 2n · 3 < √3 2 · 1 2n−1 · 2 √3 n−1 = √3 2 · 1 √3 n−1 .
2 (4). By (3), it is easy to see that 360° 0<S n+1.3 In ont :2n+1.3.sin -0600 360 cos on+l 360° 3 (5). To prove that (Sn) is a Cauchy sequence 0<S+2-Sn<∑|Sk+1-58|<∑ (6). The conclusion is obvious from (5) 补充习题 、证明:任何点集的内点全体是开集 2、设limf(x)=a,且f(x)在xo有定义。问在x→xo的过程中,x可否取到xo?是否必有 a=f(zo)? 试用ε-δ语言来叙述当工→xo时f(x)不收敛于a 4、用分析定义证明 31 1 (4)lim (x-2)(x-1) (5 (7)lim x→∞x+1=∞ 5、下列命题是否正确?若正确,请给岀证明;若不正确,请举出反例。 (1)limf(x)=a充分必要条件是imf(x)=|l (2)若mnfx)=a则inUf()2=a2 (3)若limf()=a,则limf(x)=a (4)若Hn。f(x)与imn(()+9(a)都存在,则1in,9()存在 (5)若imf(x)与lim(f(x)g(x)都存在,则limg(x)存在 (6)若在xo的某邻域内f(x)>0,并且limf(x)=a,那么必有a>0
2 (4). By (3), it is easy to see that 0 < Sn+1 − Sn =2n+1 · 3 · sin 360◦ 2n+1 · 3 − 2n · 3 · sin 360◦ 2n · 3 =2n+1 · 3 · sin 360◦ 2n+1 · 3 1 − cos 360◦ 2n+1 · 3 =2n+1 · 3 · sin 360◦ 2n+1 · 3 · 2 · sin2 360◦ 2n+2 · 3 < 9 4 · 2 3 √3 n+1 (5). To prove that {Sn} is a Cauchy sequence: 0 < Sn+p − Sn < n +p k=n |Sk+1 − Sk| < 9 4 · n +p k=n 2 3 √ 3 k+1. (6). The conclusion is obvious from (5). 1 ! !"!# 2 " limx→x0 f(x) = a, Æ f(x) # x0 $$%#&# x → x0 %&' x '(() x0 Æ!(*$ a = f(x0)? 3 )* ε − δ +,+-., x → x0 / f(x) -0.1 a. 4 * $% (1) limx→1 x − 3 x2 − 9 = 1 2 ; (2) limx→3 x − 3 x2 − 9 = 1 6 ; (3) limx→1 x − 1 √x − 1 = 2; (4) limx→1 (x − 2)(x − 1) x − 3 = 0; (5) limt→1 t(t − 1) t2 − 1 = 1 2 ; (6) limx→∞ x − 1 x + 2 = 1; (7) limx→3 x x2 − 9 = ∞; (8) limx→∞ x2 + x x + 1 = ∞. 5 2/3 !(45Æ645701 26-4573145# (1) limx→x0 f(x) = a *896! limx→x0 |f(x)| = |a|; (2) 6 limx→x0 f(x) = a, : limx→x0 [f(x)]2 = a2; (3) 6 limn→∞ f( 1 n) = a, : lim x→0+ f(x) = a; (4) 6 limx→x0 f(x) ; limx→x0 (f(x) + g(x)) 78#: limx→x0 g(x) 8#2 (5) 6 limx→x0 f(x) ; limx→x0 (f(x)g(x)) 78#: limx→x0 g(x) 8#2 (6) 6# x0 <9= f(x) > 0 :Æ lim x→x− 0 f(x) = a >?*$ a > 0.
3 6、证明: (1)f(x)=a台→,imf()=,im、f(x)=a (2)imf(x)=a←÷limf(x)=limf(x)=a 7、证明:函数极限的惟一性、局部保号性与局部保序性。 8、下列运算有无错误?错在何处? 0 sIn (1) sIn lim r lim 9、设m,n∈N+,求下列极限 (1)Iim (1+ma)"-(1+na) (x+t) (x∈R); (4)lim sin 2r- sin 3x cos(a +h)-cos (7)lim √+x2+x)-(①1+x2-x)n (9)lim(1-r)tan- (10) lin e cos T (11) lim sin 5a-sin 3r 12)lim cos- cos 3.x SIn 2a (13)lim1-2x; (14)lim(1+ (15)lir (16)imn(cos=)(a≠0) cos(n arccos (n为奇数); (18)lim (cos vn+I-cos V (19) lim cos =cos (20) lim sin(x√n2+1) (21)lim(sin z) (22)lim sin -+cos +In 10、设{a,b是一个有限闭区间,如果vro∈[a,b,limf(x)存在且有限,证明:f(x)在[,b 上有界 11、设∫:(a,b)→R是无界函数,证明:存在数列{xn}c(a,b),使得limf(xn)=∞ 12、设f:园,+∞)→B,证明:、f(如)存在且有限台→v>0,3M>0使得Vx,z2>M 恒有|f(x1)-f(x2川<E
3 6 (1) limx→∞ f(x) = a ⇐⇒ lim x→+∞ f(x) = lim x→−∞ f(x) = a ; (2) limx→x0 f(x) = a ⇐⇒ lim x→x+ 0 f(x) = lim x→x− 0 f(x) = a. 7 @A;<=>A;;<=BA# 8 2/CD$E?FÆ?# @Æ (1) limx→0 sin x x = limx→0 sin x limx→0 x = 0 0 = 1; (2) limx→∞ sin x x = limx→∞ sin x limx→∞ x = 0; (3) limx→0 x sin 1 x = limx→0 x limx→0 sin 1 x = 0. 9 " m, n ∈ N+ G2/ (1) limx→0 (1 + mx)n − (1 + nx)m x2 ; (2) limx→1 m 1 − xm − n 1 − xn ; (3) limt→0 (x + t)n − xn t (x ∈ R); (4) limx→0 √ n 1 + x − 1 x ; (5) limx→0 sin 2x − sin 3x x ; (6) limh→0 cos (x + h) − cos x h ; (7) limx→0 ( √1 + x2 + x)n − ( √1 + x2 − x)n x ; (8) limx→0+ x 1 x ; (9) limx→1 (1 − x) tan πx 2 ; (10) limx→0 x2 1 − cos x ; (11) limx→0 sin 5x − sin 3x sin 2x ; (12) limx→0 cos x − cos 3x x . (13) limx→0 √x 1 − 2x; (14) limx→∞ 1 + 2 x −x ; (15) limx→∞ x2 − 1 x2 + 1 x2 ; (16) lim x→+∞ cos a x x2 (a = 0); (17) limx→0 cos (n arccos x) x (nHI); (18) lim n→+∞ cos √ n + 1 − cos √n ; (19) lim n→+∞ cos x 2 · cos x 22 ··· cos x 2n ; (20) lim n→+∞ sin (π n2 + 1); (21) lim x→π 2 (sin x) tan x; (22) limx→∞ sin 1 x + cos 1 x x ; (23) lim n→+∞ n + x n − 1 n ; (24) lim n→+∞ n + ln n n − ln n n ln n . 10 " [a, b] !A$ BJCKD ∀x0 ∈ [a, b], limx→x0 f(x) 8#Æ$ f(x) # [a, b] L$E# 11 " f : (a, b) → R !EE 8#/ {xn} ⊂ (a, b), MF limn→∞ f(xn) = ∞. 12 " f : [a, +∞) → R, lim x→+∞ f(x) 8#Æ$ ⇐⇒ ∀ε > 0, ∃M > 0 MF ∀x1, x2 > M, G$ |f(x1) − f(x2)| < ε
13、设∫:{a,+∞)→R是单调增函数,证明下面三个命题是等价的 (1)limf(xn)存在且有限;(2){f(m)}是收敛数列;(③3)∫在[a,+∞)上有上界 14、利用无穷小的等价代换求下列极限: Cos 1+sin22-l ar tan r (4)l-mo sin(2r) (5)lim (i+tan E-1(V1+1-1 (1 os a)tan I (1 15、设fg:ACR→R是两个函数,且vx∈A,f(x)>0,则称形如f(x)(x)的函数为幂指函 数。若limf(x)=1,img(x)=∞,则极限limf(x)()属于1型不定式。对于这类不定式,一般利 用等式f(x)9(x)=e()mf()转化为讨论0·∞型不定式img(x)lnf(x)的极限问题 (1)设g1(x)~g2(x)证明:若imf(x)%()存在,则imf(x)91(x)=limf(x)92(x) (2)假定 lim In a=lnxo,即对数函数y=mnx是连续的(见有关章节),证明:若limf(x) a>0, lim g(r)=b, Du lim f(r) 9i(a)=ab (3)求下列极限:im(1+sinr)x 16、证明:函数f在x0点连续←→在x既左连续又右连续。 17、两个在xo处不连续函数之和在xo是否一定不连续?若其中一个在xo处连续,一个在 处不连续,则它们的和在x0处是否一定不连续? 18、证明:若∫连续,则|升也连续。逆命题成立吗 19、证明:若连续函数在有理点上函数值为零,则此函数恒为零。 20、若函数∫在叵连续,恒正,按照定义证明在叵可连续。 21、若∫和g都在叵a可连续。试证明:max{f(x),g(x)}以及min{f(x),9()}都在[a,b连续。 22、若∫是连续的。证明:对于任何c>0,函数 若f(x) (x)={f(x,若|f(x)≤ C,若f(x)>c 23、若∫在[a,b连续,a<x1<x2<…<xn<b,则在{x1,xn]中必有ξ使 f()=f(x)+f(x2)+…+f(xn) 24、设函数∫在(-∞,+∞)满足 Lipschitz条件:彐M>0,使得Ⅵx,y∈(-∞,+∞)恒有 f(x)-f(y)≤M|x-列证明:∫在(-∞,+∞)上一致连续
4 13 " f : [a, +∞) → R ! HN 2OPA3 !IJ (1) lim x→+∞ f(xn) 8#Æ$ 2 (2) {f(n)} !0./2 (3)f # [a, +∞) L$LE# 14 K*EQRIJLMG2/ (1) limx→0 1 − cos x sin2 x ; (2) limx→1 5x2 − 2(1 − cos2 x) 3x2 + 4 tan2 x ; (3) limx→0 1 + sin2 x − 1 x tan x ; (4) limx→0 tan(tan x) sin (2x) ; (5) limx→0 ( √3 1 + tan x − 1)(√1 + x2 − 1) tan x − sin x ; (6) lim x→0− (1 − √ n cos x) tan x (1 − cos x)3/2 . 15 " f,g : A ⊂ R → R !NAÆ ∀x ∈ A, f(x) > 0, :OSK f(x)g(x) H TU V #6 lim f(x)=1, lim g(x) = ∞, : lim f(x)g(x) W1 1∞XYZ #P1[Q-$\RK *I\ f(x)g(x) = eg(x) ln f(x) ]SH^T 0 · ∞ _-$\ lim g(x) ln f(x) & U 1 " g1(x) ∼ g2(x). 6 lim f(x)g1(x) 8#: lim f(x)g1(x) = lim f(x)g2(x) ; 2 V$ limx→x0 ln x = ln x0 WP y = ln x ! $X`Y 6 limx→x0 f(x) = a > 0, limx→x0 g(x) = b, : lim f(x)g1(x) = ab; 3 G2/ limx→0 (1 + sin x) 1 x ; limx→∞(cos 1 x ) x2 . 16 f # x0 ⇐⇒ # x0 Za bc # 17 NA# x0 @- d[# x0 !($- Æ6e'A# x0 @ A# x0 @- :fg[# x0 @!($- Æ 18 6 f : |f| h #i3 \]^Æ 19 6 #$_ LjH`:aGH`# 20 6 f # [a, b] G4bk$% 1 f # [a, b] # 21 6 f [ g 7# [a, b] #) max{f(x), g(x)} lc min{f(x), g(x)} 7# [a, b] # 22 6 f ! # P1 c > 0 F(x) = −c, 6 f(x) < −c f(x), 6 |f(x)| ≤ c c, 6 f(x) > c ! # 23 6 f # [a, b] a<x1 < x2 < ··· < xn < b, :# [x1, xn] '*$ ξ M f(ξ) = f(x1) + f(x2) + ··· + f(xn) n . 24 " f # (−∞, +∞) dm Lipschitz 96 ∃M > 0, MF ∀x, y ∈ (−∞, +∞) G$ |f(x) − f(y)| ≤ M|x − y|. f # (−∞, +∞) Ln #
5 25、证明:函数f:I→R在0∈I处连续←→Vxn∈I,xn→mo(n→0),有limf(xn)=f(xo) 26、设函数∫:I→R在x∈Ⅰ连续,且f(xo)>0。证明:存在xo的一个邻域,在该邻域内 f(x)≥q>0 27、讨论下列函数的连续性,并画出图形 (1)f(x)= (2)f(x) 4 sIn (3)f()=,x≠0 (4)f(x)={] 28、讨论下列函数在指定点处的连续性。若是间断点,说明它的类型 sIn x<0, (1)f(x) (2)f(x) 2x-3,x≠0,x=3 1,x≥0 x=0, 29、讨论下列函数的连续性。若有间断点,说明间断点的类型 (1)f(x) (2)f(x) ,x≠0 1,x≥0; P (q>0,q,p为互质的整数 <1 (3)f(x)= (4)f(x)= 0,x为无理数; x-1x>1 n丌x,x为有理数 x,|x|≤1, (5)f(x) (6)f(x) 0,x为无理数 x|≥ 其中(3)所表示的函数叫黎曼( Riemann)函数 30、试确定常数,使下列函数在x=0处连续: arctan 0, (1)f(x) (2)f(x) 0; 个+x-1,x≠0 0 (3)f(x)= (4)f(x) 1 (5)f()=(1+x),x≠0 ≠0, (6)f(x)
5 25 f : I → R # x0 ∈ I @ ⇐⇒ ∀xn ∈ I,xn → x0(n → ∞), $ limn→∞ f(xn) = f(x0). 26 " f : I → R # x0 ∈ I Æ f(x0) > 0 # 8# x0 A9=#e9= f(x) ≥ q > 0 # 27 ^T2/ A:f1oS# (1) f(x) = x2 − 4 x − 2 , x = 2, 4, x = 2; (2) f(x) = | sin x x |, x = 0, 1, x = 0; (3) f(x) = sin x |x| , x = 0, 1, x = 0; (4) f(x)=[x]. 28 ^T2/#p$ @ A#6!Cg qfQ_ (1) f(x) = x sin 1 x , x< 0, 1, x ≥ 0, x = 0; (2) f(x) = 2 − 1 x − 3 , x = 0, 2, x = 0, x = 3; 29 ^T2/ A#6$Cg qCg Q_ (1) f(x) = sin x x , x< 0, x2 − 1, x ≥ 0; (2) f(x) = e − 1 x2 , x = 0, 2, x = 0; (3) f(x) = 1 q , x = p q (q > 0, q, pHhrs), 0, xHE_; (4) f(x) = cos πx 2 , x ≤ 1, |x − 1|, x> 1; (5) f(x) = sin πx, xH$_, 0, xHE_; (6) f(x) = x, |x| ≤ 1, 1, |x| ≥ 1. e' 3 tiujkl Riemann U 30 )5$mM2/# x = 0 @ (1) f(x) = a + x, x ≤ 0, sin x, x > 0; (2) f(x) = arctan 1 x , x< 0, a + √x, x ≥ 0; (3) f(x) = √1 + x − 1 √3 1 + x − 1 , x = 0, c, x = 0; (4) f(x) = sin x · sin 1 x , x = 0, c, x = 0; (5) f(x) = (1 + x) 1 x , x = 0, c, x = 0; (6) f(x) = tan x x , x = 0, c, x = 0