(6)对ve>0,由于2-1_,_|3 x+2 因x→∞,不妨设1>2,则x+21>1-2,于是 3,要使_3 1<总成立,故 小同-2<e即可,即叫>3。取X=2+2,则当>X时,就 只要 75|+《,/“)国|=因x一3,不妨设-3<1,则2<x<4,从 (7)对vG>0,由 3/·要使x 1 只要二 >G即 颗0=m0则增0b=<时,二引,故=x (8)对vG>0,由 x2+=2,因x→∞,取X=G>0,则当>X时,發x2+>G总成 →∞x+1 (1)12x (3)lin (4)im①+x)(1+21+3x)-1 6=1 (1+x)5-(1+5x) x2+x5 (m,n为自然数) 5+6 (10) x+15 x2+3 (11) (12)lim 1 增正=+1+=2+ (3)1n2x2-x-1=2 (1+x)(1+2x)(1+32)-1=im(6+11+6x)=6 1) t21 (5)iB-1= t(vt-1)(t+vt+1
26 (6) È∀ε > 0ßdu x − 1 x + 2 − 1 = 3 x + 2 ßœx → ∞ßÿî|x| > 2ßK|x + 2| > |x| − 2ßu¥ 3 x + 2 < 3 |x| − 2 ßᶠ3 x + 2 < εßêá 3 |x| − 2 < ε=åß=|x| > 3 ε "X = 3 ε + 2ßK|x| > Xûß“ k x − 1 x + 2 − 1 < εo§·ß limx→∞ x − 1 x + 2 = 1 (7) È∀G > 0ßdu x x2 − 9 = x x + 3 1 x − 3 ßœx → 3ßÿî|x − 3| < 1ßK2 < x < 4ßl 2 7 < x x + 3 < 4 5 ßu¥ x x + 3 1 x − 3 > 2 7 1 x − 3 ßᶠx x2 − 9 > Gßêá 2 7 1 x − 3 > G= å"δ = min 2 7G , 1 > 0ßK0 < |x − 3| < δûß“k x x2 − 9 > Go§·ßlimx→3 x x2 − 9 = ∞ (8) È∀G > 0ßdu x 2 + x x + 1 = |x|ßœx → ∞ßX = G > 0ßK|x| > Xûß“k x 2 + x x + 1 > Go§ ·ß limx→∞ x 2 + x x + 1 = ∞ 2. ¶4ŵ (1) limx→0 x 2 − 1 2x2 − x − 1 (2) limx→1 x 2 − 1 2x2 − x − 1 (3) limx→∞ x 2 − 1 2x2 − x − 1 (4) limx→0 (1 + x)(1 + 2x)(1 + 3x) − 1 x (5) limt→1 t 2 (t − 1) t 2 − 1 (6) limt→1 t 2 − √ t √ t − 1 (7) limx→3 √ 1 + x − 2 x − 3 (8) limx→0 (1 + x) 5 − (1 + 5x) x2 + x5 (9) limx→0 (1 + mx) n − (1 + nx) m x2 £m, nèg,ͧ (10) limx→3 x 2 − 5 + 6 x2 − 8x + 15 (11) limx→∞ x 2 + 3x x2 (12) limx→∞ 5x − 7 2x + √ x )µ (1) limx→0 x 2 − 1 2x2 − x − 1 = 1 (2) limx→1 x 2 − 1 2x2 − x − 1 = limx→1 (x − 1)(x + 1) (2x + 1)(x − 1) = limx→1 x + 1 2x + 1 = 2 3 (3) limx→∞ x 2 − 1 2x2 − x − 1 = 1 2 (4) limx→0 (1 + x)(1 + 2x)(1 + 3x) − 1 x = limx→0 (6 + 11x + 6x 2 ) = 6 (5) limt→1 t 2 (t − 1) t 2 − 1 = limt→1 t 2 t + 1 = 1 2 (6) limt→1 t 2 − √ t √ t − 1 = limt→1 √ t( √ t − 1)(t + √ t + 1 √ t − 1 = limt→1 √ t(t + √ t + 1) = 3
(8)lin(1+x)5-(1+3≠mx2+1D23+ 1+mx)"-(1+nx)m 织n212⊥(3m3-Cmn 5+6 1 (01-8x+15=m(a-a(x-5)=mx-5=2 x2+3x (11)lim (12)lim 2x+√x2 3.设R(x) P(r) 式中P(x)和Q(x)为x的多项式,并且P(a)=Q(a)=0,问lim有哪些可能的值? 解:由于P(x)和Q(x)为的多项式且P(a)=Q(a)=0, 则P(x)=(x-a)P(x),Q(x)=(x-a)"Q1(x)(P1(a)≠0,Q1(x)≠0),于是imR(x)=li Q(r) (ar-a"Pi(ar) Er-a Q1(ar (1)当n=m时,limB(a)P(a) ()当n>m时,lm(x-0)m=∞且lmB.m2=B.(0≠0,故lmnF(x) (3)当nm时,(x-a)m-”=0且mnB()P(a),故m取 Q1(x)Q1(a) 4.求下列极限 cos(ar +h)-cos a h (4)lim(vz2+1-x) )lm,2 x→01-csx cos t-cos 3 r (8)lim sIn 5z-sin 32 (11) (12)1i
27 (7) limx→3 √ 1 + x − 2 x − 3 = limx→3 1 √ 1 + x + 2 = 1 4 (8) limx→0 (1 + x) 5 − (1 + 5x) x2 + x5 = limx→0 10x 2 + 10x 3 + 5x 4 + x 5 x2 + x5 = 10 (9) limx→0 (1 + mx) n − (1 + nx) m x2 = limx→0 (C 2 nm2 − C 2 mn 2 )x 2 + (C 3 nm3 − C 3 mn 3 )x 3 + · · · + mnx n − n mx m x2 = C 2 nm 2 − C 2 mn 2 = n 2m − m2n 2 (10) limx→3 x 2 − 5 + 6 x2 − 8x + 15 = limx→3 (x − 2)(x − 3) (x − 3)(x − 5) = limx→3 x − 2 x − 5 = − 1 2 (11) limx→∞ x 2 + 3x x2 = 1 (12) limx→∞ 5x − 7 2x + √ x = 5 2 3. R(x) = P(x) Q(x) ™•P(x)⁄Q(x)èxıë™ßøÖP(a) = Q(a) = 0ßØlimx→a k= åUäº )µduP(x)⁄Q(x)èxıë™ÖP(a) = Q(a) = 0ß KP(x) = (x − a) mP1(x), Q(x) = (x − a) nQ1(x)(P1(a) 6= 0, Q1(x) 6= 0)ßu¥limx→a R(x) = limx→a P(x) Q(x) = limx→a (x − a) mP1(x) (x − a) nQ1(x) ?ÿµ (1) n = mûßlimx→a R(x) = P1(a) Q1(a) (2) n > mûßlimx→a (x − a) m−n = ∞Ölimx→a P1(x) Q1(x) = P1(a) Q1(a) 6= 0ßlimx→a R(x) = ∞ (3) n < mûßlimx→a (x − a) m−n = 0Ölimx→a P1(x) Q1(x) = P1(a) Q1(a) ßlimx→a R(x) = 0 4. ¶e4ŵ (1) limx→0 sin 2x − sin 3x x (2) limh→0 cos(x + h) − cos x h (3) lim x→+∞ ( √ x2 + 1 − x) (4) lim x→−∞ ( √ x2 + 1 − x) (5) limx→0 x 2 1 − cos x (6) lim x→+∞ q x + p x + √ x x + 1 (7) limx→0 cos x − cos 3x x2 (8) limx→0 sin 5x − sin 3x sin 2x (9) limx→1 (1 − x) tan πx 2 (10) limx→a sin x − sin a x − a (11) limx→0 ( √ 1 + x2 + x) n − ( √ 1 + x2 − x) n x (12) limx→0 x 1 x )µ
sin 2r -sin 3 r sin 2 r sin 3x hh I+h)-cosr 2 lim 2 sin 2r+h=_sinz √x2+1+x 0 x→+ (4)lim(√x2+1-x)=+∞ +√x+√ cos r- cos 3 x in 5x-sin 3 2 sin r cos 4z )y=x-1,则回(1-)tmn2=1m-ymn(2(1+)=1-yt2y=m2 2 cos E+a -a sIn al (10) E cOSa (11) lim √1+x2+x)-(√1+x2-x) 2Cn(1+x2)2x+2C(1+x2)2x2+… 2(1+2)2+2C2a+2)9x+…] 2)由[=王 且0≤ <1, 则limx ={2=()} 5.若limf(x)=A,limg(x)=B,并且存在6>0,当0<|x-xol<6时有f(x)≥g(x),证明A≥B 若当0<|x-x<时f(x)>g(x),是否一定成立A>B 证明 (1)用反证法。假设A<B,则由lim∫(a)=A,img(x)=B及性质1,得30>0,使当0<|x-rol< b时,有g(x)>f(x)。这与已知:36>0,当0<|x-xo|<时,有∫(x)≥9(x)矛盾,故假设不成 立,即A≥B成立 ()成立。f(x)=2(x2+3x g(x)=x2+3x2x2,36>0,当0<团<6时,有f(x)>g(x) mf(x)=2,B=lm9(x)=1,故A>B成立 又A=x03+3x,g(x)=x2+x2x2,36>0,当0<1<6时,有∫(x)>g(x).又A= (i)不成立。∫(x)=-x imnf(x)=1.,B=img()=1,故有A=B 6.若在点xo的邻域内有g(x)≤f(x)≤h(x),并且g(x)和h(x)在xo的极限存在并且都等于A,证明limf(x)= A 证明:如果对任何xn,xn→x0,xn≠xo,并且可不妨假设rn∈O(x0,6)-{xo},有g(xn)≤f(xn)≤h(xn) 以及g(xn)→A,h(xn)→A(n ),由数列极限的性质得:f(xn)→A(n→∞),这就证明了f(x) A(
28 (1) limx→0 sin 2x − sin 3x x = limx→0 sin 2x x − limx→0 sin 3x x = 2 − 3 = −1 (2) limh→0 cos(x + h) − cos x h = limh→0 −2 sin 2x + h 2 sin h 2 h = limh→0 sin h 2 h sin 2x + h 2 = − sin x (3) lim x→+∞ ( √ x2 + 1 − x) = lim x→+∞ 1 √ x2 + 1 + x = 0 (4) lim x→−∞ ( √ x2 + 1 − x) = +∞ (5) limx→0 x 2 1 − cos x = limx→0 x 2 x 2 2 = 2 (6) lim x→+∞ q x + p x + √ x x + 1 = 0 (7) limx→0 cos x − cos 3x x2 = limx→0 2 sin x sin 2x x2 = 4 (8) limx→0 sin 5x − sin 3x sin 2x = limx→0 2 sin x cos 4x 2x = 1 (9) -y = x − 1ßKlimx→1 (1 − x) tan πx 2 = limy→0 −y tan π 2 (1 + y) = limy→0 −y cot π 2 y = limy→0 y cos π 2 y sin π 2 y = limy→0 y π 2 y = 2 π (10) limx→a sin x − sin a x − a = limx→a 2 cos x + a 2 sin x − a 2 x − a = cos a (11) limx→0 ( √ 1 + x2 + x) n − ( √ 1 + x2 − x) n x = limx→0 2C 1 n(1 + x 2 ) n−1 2 x + 2C 3 n(1 + x 2 ) n−3 2 x 2 + · · · x = limx→0 h 2n(1 + x 2 ) n−1 2 + 2C 3 n(1 + x 2 ) n−3 2 x + · · · i = 2n (12) du 1 x = 1 x − 1 x Ö0 6 1 x < 1ß Klimx→0 x 1 x = limx→0 1 − x 1 x = 1 − limx→0 x 1 x = 1 5. e limx→x0 f(x) = A, limx→x0 g(x) = BßøÖ3δ > 0ß0 < |x − x0| < δûkf(x) > g(x)ßy²A > B. qe0 < |x − x0| < δûf(x) > g(x)ߥƒò½§·A > B y²µ (1) ^áy{"bA < BßKd limx→x0 f(x) = A, limx→x0 g(x) = B95ü1ß∃δ0 > 0߶0 < |x − x0| < δ0ûßkg(x) > f(x)"˘ÜƵ∃δ > 0ß0 < |x − x0| < δûßkf(x) > g(x)gÒßbÿ§ ·ß=A > B§·" (2) ÿò½"~µ (i) §·"f(x) = 2(x 2 + 3x 4 ) x2 , g(x) = x 2 + 3x 4 x 2 , ∃δ > 0ß0 < |x| < δûßkf(x) > g(x)" qA = limx→x0 f(x) = 2, B = limx→x0 g(x) = 1ßA > B§·" (ii) ÿ§·"f(x) = x 2 + 3x 4 x2 , g(x) = x 2 + x 4 x 2 , ∃δ > 0ß0 < |x| < δûßkf(x) > g(x)"qA = limx→x0 f(x) = 1, B = limx→x0 g(x) = 1ßkA = B" 6. e3:x0çSkg(x) 6 f(x) 6 h(x)ßøÖg(x)⁄h(x)3x04Å3øÖ— uAßy² limx→x0 f(x) = A. y²µXJÈ?¤xn, xn → x0, xn 6= x0 ßøÖåÿîbxn ∈ O(x0, δ) − {x0} ßkg(xn) 6 f(xn) 6 h(xn) ±9g(xn) → A, h(xn) → A(n → ∞) ßdÍ4Å5üµf(xn) → A(n → ∞) ߢ“y² f(x) → A(x → x0)
若期=A=B0国明吗合 证明:考察(2 Bf()-Ag(=)Bf(r)-AB+ AB-Ag(r)I Bllf(ar)-Al+lAllg(r)-Bl 如a)-B= Bg(r) BG(e) Blog(a) 由于im∫(x)=A,limg(x)=B,故对ve>0,361>0,当0<x-xol<61时,有∫(x)-A<e:对上 有g(x)-Bl 又据乘法运算:mnB()=B2>2,则据性质3,得35>0,当0<|x-01<6时,有By()>2 取6=min{6,6,6},当0<|-ak<6时,有2)-4|41+1 2(4+|B) 于是,对v>035>0,当0<|-ak<时,有(口-a<24+12从而lm(=合 a> 8.(1)f(x)=1 求f(x)在x=1的左右极限 (2)f(a)=,xsin I x>0 x<0 求f(x)在x=0的左右极限。 (1)m0f(x)=,m (2)Himf(x)=im01+x2)=1 mo/(==im(asin 1)=0 9.说明下列函数在所示点的左右极限情形: 0<x≤1 1)y= 1<x<2(在x=1.5,2,1三点) 2<x<3 (2)y=x:sin-(在x=0点) =n(g) 2-1 (在x=0点) (4)y=-- ()D()={1 x为无理数(在任一点) y- x→1+0 (2)limo y=im y=0 (3)由于lim 则lim2=+∞,lim2h=0, 于是limy= 1. lim
29 7. e limx→x0 f(x) = A, limx→x0 g(x) = B 6= 0ßy² limx→x0 f(x) g(x) = A B . y²µ f(x) g(x) − A B = Bf(x) − Ag(x) Bg(x) = Bf(x) − AB + AB − Ag(x) BG(x) 6 |B||f(x) − A| + |A||g(x) − B| |B||g(x)| ß du limx→x0 f(x) = A, limx→x0 g(x) = BßÈ∀ε > 0, ∃δ1 > 0ß0 < |x − x0| < δ1ûßk|f(x) − A| < ε¶È˛ „ε > 0, ∃δ2 > 0ß0 < |x − x0| < δ2ûßk|g(x) − B| < ε q‚¶{$éµ limx→x0 Bg(x) = B 2 > B 2 2 ßK‚5ü3ß∃δ3 > 0ß0 < |x − x0| < δ3ûßkBg(x) > B 2 2 δ = min{δ1, δ2, δ3}ß0 < |x − x0| < δûßk f(x) g(x) − A B < (|A| + |B|)ε B2 2 = 2(|A| + |B|) B2 ε u¥ßÈ∀ε > 0, ∃δ > 0ß0 < |x − x0| < δûßk f(x) g(x) − A B < 2(|A| + |B|) B2 εßl limx→x0 f(x) g(x) = A B . 8. (1) f(x) = 0 x > 1 1 x = 1 x 2 + 2 x < 1 ¶f(x)3x = 1Üm4Å" (2) f(x) = x sin 1 x x > 0 1 + x 2 x < 0 ¶f(x)3x = 0Üm4Å" )µ (1) lim x→1−0 f(x) = lim x→1−0 (x 2 + 2) = 3, lim x→1+0 f(x) = 0 (2) lim x→−0 f(x) = lim x→−0 (1 + x 2 ) = 1, lim x→+0 f(x) = lim x→+0 (x sin 1 x ) = 0 9. `²eºÍ3§´:Üm4Åú/µ (1) y = 1 2x 0 < x 6 1 x 2 1 < x < 2 (3x = 1.5, 2, 1n:) 2x 2 < x < 3 (2) y = x · sin 1 x £3x = 0:§ (3) y = 2 1 x + 1 2 1 x − 1 £3x = 0:§ (4) y = 1 x − 1 x £3x = 1 n :§ (5) D(x) = 1 xèknÍ 0 xèÃnÍ (3?ò:) (6) y = (x − 1)(−1)[x] x2 − 1 (3x = −1§ )µ (1) lim x→1.5−0 y = lim x→1.5+0 y = 2.25, lim x→2−0 y = lim x→2−0 x 2 = 4, lim x→2+0 y = lim x→2+0 (2x) = 4 lim x→1−0 y = lim x→1−0 1 2x = 1 2 , lim x→1+0 y = lim x→1+0 x 2 = 1 (2) lim x→+0 y = lim x→+0 y = 0 (3) du lim x→+0 1 x = +∞, lim x→−0 1 x = −∞ß K lim x→+0 2 1 x = +∞, lim x→−0 2 1 x = 0ß u¥ lim x→+0 y = lim x→=0 2 1 x + 1 2 1 x − 1 = lim x→+0 1 + 2 2 1 x − 1 = 1, lim x→+−0 y = lim x→−0 2 1 x + 1 2 1 x − 1 = −1
(4) t2 (5)此函数在任一点的左右极限不存在 设x为R上任一点,由有理数和无理数在数轴上的稠密性,可知有理序列{x)}→x0+0,无理序 列{x2} 故limD(x(1)=1,imD(x(2)=0,从而此函数在任一点的右极限不存在 同理,此函数在任一点的左极限也不存在 从而此函数在任一点的左右极限不存在 (x-1)(-1)1_(-1) 且Iim{]=-1,lin x→-1-0 则1+0y=-∞,1+0y=-∞ 10.讨论下列极限: (1)lim sinz (2) lim e sin r (3 lim a arctan r (4) lim r tan an(x≠n丌+言) (1)由于1m=0且nx是有界量,故mx=0 (2)由于im2c2=+∞,若取xn=2nm-+∞(n→∞),则 een sinN=csmn2nx=0→0(n ∞):若取xn=+2m→+∞(n→∞),则 En In=e+2msin(+2nx)=e+ +∞(n→∞),故 lim e sin:不存在,从而 lim e sin r不存在 (3)由于 lin arctan=-7, lim rarctanzs 则 lim r arctan x=+∞, lim r arctan r=+∞,从而 lim r arctan=+∞ x→+ (4)取xn=m→x(n-∞),有 m In tan.n=1mntn=0:另取xn=了+mx→∞(n ),有 lim En tan En=lim(x+n丌)tan(x+n)=lim(x+nr)=+∞,故 lim r tanx(x≠ n丌+言)不存在 -ax-b)=0,求常数a和b :由于吗(2#-m-)+D=出++期+=b a+b=0 12.从条件im(vx2-x+1-a1x-b1)=0,lim(vx2-x+1-a2x-b2)=0,求常数a1,b1,a2,b2 由于1m。(-x+1-axb)=lm1-a6)2-(2+2b)x+1-6=0,则11+2b=0 0 x2-x+1+a1x+b1 又据条件可得:若a1=1,则lm(a-x+1-a1x-b1)=+x,从而=,1 13.若lim[f(x)-(kx+b=0,则称直线y=kx+b是曲线y=f(x)当x→+∞的渐近线利用这一方程推出渐 线存在的必要且充分的条件 证明:若曲线存在渐近线,则有
30 (4) lim x→ 1 n +0 y = lim x→ 1 n +0 1 x − 1 x = n − (n − 1) = 1 lim x→ 1 n −0 y = lim x→ 1 n −0 1 x − 1 x = n − n = 0 (5) dºÍ3?ò:Üm4Åÿ3" x0èR˛?ò:ßdknÍ⁄ÃnÍ3Ͷ˛»ó5ßåknS{x (1) n } → x0 + 0ßÃnS {x (2) n } → x0 + 0ß lim x (1) n →x0+0 D x (1) = 1, lim x (2) n →x0+0 D x (2) = 0ßl dºÍ3?ò:m4Åÿ3 ”nßdºÍ3?ò:Ü4Åèÿ3 l dºÍ3?ò:Üm4Åÿ3" (6) y = (x − 1)(−1)[x] x2 − 1 = (−1)[x] x + 1 Ö lim x→−1+0 [x] = −1, lim x→−1−0 [x] = −2 K lim x→−1+0 y = −∞, lim x→−1+0 y = −∞ 10. ?ÿe4ŵ (1) limx→∞ sin x x (2) limx→∞ e x sin x (3) limx→∞ x arctan x (4) limx→∞ x tan x(x 6= nπ + π 2 ) )µ (1) du limx→∞ 1 x = 0Ösin x¥k.˛ß limx→∞ sin x x = 0 (2) du lim x→+∞ e x = +∞ßexn = 2nπ → +∞(n → ∞)ßKe xn sin xn = e 2nπ sin 2nπ = 0 → 0(n → ∞)¶exn = π 2 + 2nπ → +∞(n → ∞)ßKe xn sin xn = e π 2 +2nπ sin π 2 + 2nπ = e π 2 +2nπ → +∞(n → ∞)ß lim x→+∞ e x sin xÿ3ßl limx→∞ e x sin xÿ3. (3) du lim x→−∞ arctan x = − π 2 , lim x→+∞ x arctan x = π 2 ß K lim x→−∞ x arctan x = +∞, lim x→+∞ x arctan x = +∞ßl limx→∞ x arctan x = +∞ (4) xn = nπ → ∞(n → ∞)ßk limn→∞ xn tan xn = limn→∞ nπ tan nπ = 0¶,xn = π 4 + nπ → ∞(n → ∞)ßk limn→∞ xn tan xn = limn→∞ π 4 + nπ tan π 4 + nπ = limn→∞ π 4 + nπ = +∞ß limx→∞ x tan x(x 6= nπ + π 2 )ÿ3. 11. l^á limx→∞ x 2 + 1 x + 1 − ax − b = 0߶~Ía⁄b. )µdu limx→∞ x 2 + 1 x + 1 − ax − b = limx→∞ (x 2 + 1) − ax(x + 1) − b(x + 1) x + 1 = limx→∞ (1 − a)x 2 − (a + b)x − b + 1 x + 1 = 0ßKk 1 − a = 0 a + b = 0 ßl a = 1 b = −1 12. l^á lim x→−∞ ( √ x2 − x + 1 − a1x − b1) = 0, lim x→−∞ ( √ x2 − x + 1 − a2x − b2) = 0߶~Ía1, b1, a2, b2. )µdu lim x→−∞ ( √ x2 − x + 1−a1x−b1) = lim x→−∞ (1 − a 2 1)x 2 − (1 + 2a1b1)x + 1 − b 2 1 √ x2 − x + 1 + a1x + b1 = 0ßK 1 − a 2 1 = 0 1 + 2a1b1 = 0 ß u¥ ( a1 = ±1 b1 = ∓ 1 2 . q‚^áåµea1 = 1ßK lim x→−∞ ( √ x2 − x + 1−a1x−b1) = +∞ßl ( a1 = −1 b1 = 1 2 ß”n ( a2 = 1 b2 = − 1 2 . 13. e lim x→+∞ [f(x) − (kx + b)] = 0ßK°ÜÇy = kx + b¥Çy = f(x)x → +∞ÏCÇ.|^˘òêßÌ—Ï CÇ37áÖø©^á. y²µeÇ3ÏCÇßKk lim x→+∞ [f(x) − (kx + b)] = 0. (1)