3 f(a)=1 令x→+∞两端取极限并注意到(1)式,得 既求出了k,再从(1)式求得 b= lim f(ar)-kr 反之,若(2)、(3)两式成立,立即可看出条件(1)成立 故曲线=)当一+时存在新近+4的分必要条件是极限==,= kr]=b均成立 14.若Imf(x)=A>0,证明存在x>0,使得当<-x成立:a<f(x)<3A A 证明:由于limf(x)=A>0,故对给定的=>0,丑X>0,当x<-X时,有|∫(x)-A|< 即<f(x)<A 15.若limf(x)=A,img(x)=B,证明limf(x)g(x)=AB 证明:由于im。f(x)=A,故对v>0.3X1>0,当x>x时,有()-4<e且BX2>0.M>0 x>X2时,有f(x)<A 又limg(x)=B,故对上述E>0,丑X3>0,当x>X3时,有(x)-1 取X {X1,X2,X3},对上述>0,当x>X时, 有∫f(x)9(x)-AB=|f(x)9(x)-f(x)B+f(x)B-AB≤|f(x)l(x)-B+|B川‖f(x)-4≤ME+|B|l (M+IBDe, ap lim f(r)g(er)=AB 16.证明im、f(x)=A的充要条件是:对任何数列xn→+∞,∫(xn)→A 证明 →由于limf(x)=A,故对ve>0,3X>0,当x>X时,有f(x)-4<e 又xn→+∞(n→∞),故对上述X>0,丑N∈2+,当n>N时,有xn>X,从而∫(xn)-A<ε,于 午用反证法。假设limf(x)≠A,则30>0,对vX>0,至少有一个x’,当x>X时,有f(x)-A|≥ Eo 特别地,取X为1,2,3,……,可得x1,x2,x3,…,使得 从左边可以看出xn→+∞(n→∞),而从右边看出limf(xn)≠A,与已知矛盾,则假设不成立, 故 ∫(x)=A 吻:“+0∫()=+∞的充要条件是:对任何数列n:zm>20m→x,有∫(an)→+∞ 17.证明 由于limf(x)=+∞,故对vG>0,35>0,当0<x-x<0时,有f(x)>G 又xn>xo,xn→x0(n→∞),故对上述6>0,丑N∈Z+,当n>N时,有0<xn-x0<b,从而f(xn)> G,于是limf(xn)=+∞ ←用反证法。假设limf(x)≠+∞,则3Go>0,对v6>0,至少有一个x,当0<x-r0<6时 特别地,取6为.,2,3,…,可得x1,x2,x3,…,使得 0<x1-x0<1时,有∫(x1)≤c0:0<x-x<时,有(x2)≤Go:0<x-x。<时, 有∫(x3)≤Go 从左边可以看出x>xo,xn→x0,而从右边看出lim∫(x)≠+∞,与已知矛盾,则假设不成立, 18.举出符合下列要求的f(x) (1)f(+0)=0,f(-0)=1
31 œ f(x) x = 1 x [f(x) − kx − b] + k + b x ß-x → +∞¸‡4Åø5ø(1)™ß lim x→+∞ f(x) x = k (2) . Q¶— kß2l(1)™¶ b = lim x→+∞ [f(x) − kx] (3) . áÉße(2)!(3)¸™§·ß·=åw—^á(1)§·. Çy = f(x)x → +∞û3ÏCÇy = kx + bø©7á^á¥4Å lim x→+∞ f(x) x = k! lim x→+∞ [f(x) − kx] = b˛§·. 14. e lim x→−∞ f(x) = A > 0ßy²3X > 0߶x < −X§·µ A 2 < f(x) < 3 2 A. y²µdu lim x→−∞ f(x) = A > 0ßÈâ½ε = A 2 > 0, ∃X > 0 ßx < −Xûßk|f(x) − A| < A 2 ß = A 2 < f(x) < 3 2 A. 15. e lim x→+∞ f(x) = A, lim x→+∞ g(x) = Bßy² lim x→+∞ f(x)g(x) = AB. y²µdu lim x→+∞ f(x) = AßÈ∀ε > 0, ∃X1 > 0ßx > X1ûßk|f(x) − A| < εÖ∃X2 > 0, M > 0ß x > X2ûßk|f(x)| < A. q lim x→+∞ g(x) = BßÈ˛„ε > 0, ∃X3 > 0ßx > X3ûßk|g(x) − B| < ε. X = max{X1, X2, X3}ßÈ˛„ε > 0ßx > Xûß k|f(x)g(x) − AB| = |f(x)g(x) − f(x)B + f(x)B − AB| 6 |f(x)||g(x) − B| + |B||f(x) − A| 6 Mε + |B|ε = (M + |B|)ε ß= lim x→+∞ f(x)g(x) = AB. 16. y² lim x→+∞ f(x) = Aøá^᥵È?¤Íxn → +∞, f(xn) → A. y²µ ⇒ du lim x→+∞ f(x) = AßÈ∀ε > 0, ∃X > 0ßx > Xûßk|f(x) − A| < ε. qxn → +∞(n → ∞)ßÈ˛„X > 0, ∃N ∈ Z +ßn > Nûßkxn > Xßl |f(xn) − A| < εßu ¥ limn→∞ f(xn) = A. ⇐ ^áy{"b lim x→+∞ f(x) 6= AßK∃ε0 > 0ßÈ∀X > 0ßñkòáx 0ßx 0 > Xûßk|f(x 0 ) − A| > ε0. AO/ßXè1, 2, 3, · · · ßåx 0 1, x0 2, x0 3, · · · ߶ x 0 1 > 1ûßk|f(x 0 1) − A| > ε0¶x 0 2 > 2ûßk|f(x 0 2) − A| > ε0¶x 0 3 > 3ûßk|f(x 0 3) − A| > ε0¶· · · lÜ>å±w—x 0 n → +∞(n → ∞)ß lm>w— limn→∞ f(x 0 n) 6= AßÜÆgÒßKbÿ§·ß lim x→+∞ f(x) = A 17. y² lim x→x0 +0 f(x) = +∞øá^᥵È?¤Íxn : xn > x0 , xn → x0ßkf(xn) → +∞. y²µ ⇒ du lim x→x0 +0 f(x) = +∞ßÈ∀G > 0, ∃δ > 0ß0 < x − x0 < δûßkf(x) > G. qxn > x0 , xn → x0 (n → ∞)ßÈ˛„δ > 0, ∃N ∈ Z +ßn > Nûßk0 < xn − x0 < δßl f(xn) > Gßu¥ limn→∞ f(xn) = +∞. ⇐ ^áy{"b lim x→x0 +0 f(x) 6= +∞ßK∃G0 > 0ßÈ∀δ > 0ßñkòáx 0ß0 < x0 − x0 < δûß kf(x 0 ) 6 G0. AO/ßδè1, 1 2 , 1 3 , · · · ßåx 0 1, x0 2, x0 3, · · · ߶ 0 < x0 1 − x0 < 1ûßkf(x 0 1) 6 G0¶0 < x0 2 − x0 < 1 2 ûßkf(x 0 2) 6 G0¶0 < x0 3 − x0 < 1 3 ûß kf(x 0 3) 6 G0¶· · · lÜ>å±w—x 0 n > x0 , x0 n → x0ß lm>w— lim x→x0 +0 f(x) 6= +∞ßÜÆgÒßKbÿ§·ß lim x→x0 +0 f(x) = +∞ 18. fi—Œ‹eá¶f(x) (1) f(+0) = 0, f(−0) = 1
(2)f(+0)不存在,也非∞,f(-0)=0 (3)f(+∞)=0,f(-∞)不存在 (4)f(+∞)=f(-∞)=A(常数) (5)f(x0+0)和f(x-0)都不存在 (6)f(xo+0)=+∞,f(xo-0)=-∞ (7)f(xa+0)=1,f(xo-0)=+∞ (8)f(+∞)不存在,也非∞,f(-∞)=-∞0 2)f(x)= (4)f(x)=4x+1 (5)f(x)=si (6)f(x)=-1 (7)f(x)=1+e-0 x≥0
32 (2) f(+0)ÿ3ßèö∞, f(−0) = 0 (3) f(+∞) = 0, f(−∞)ÿ3 (4) f(+∞) = f(−∞) = A£~ͧ (5) f(x0 + 0)⁄f(x0 − 0)—ÿ3 (6) f(x0 + 0) = +∞, f(x0 − 0) = −∞ (7) f(x0 + 0) = 1, f(x0 − 0) = +∞ (8) f(+∞)ÿ3ßèö∞, f(−∞) = −∞ )µ (1) f(x) = 0 x > 0 1 x 6 0 (2) f(x) = ( sin 1 x x > 0 0 x 6 0 (3) f(x) = e −x (4) f(x) = Ax + 1 x (5) f(x) = sin 1 x − x0 (6) f(x) = 1 x − x0 (7) f(x) = 1 + e − 1 x−x0 (8) f(x) = sin x x > 0 x x < 0
§3.连续函数 1.按定义证明下列函数在定义域内连续 (2 (3)y=|x (4)y=sin r 证明 (1)设x为(0,+∞)内任一点,a-a。< 故y=√a在ro点连续 又由x在(0,+∞)中的任意性, 全在(0,+∞)内连续. 当xo=0时,对上述ε>0,取6 0<x-x<时,有x-ya。<v<E,故f(+0)=0 从而y=√a在[0,+∞)内连续 (2)设,为0.+x)内任一点,不妨设一x1<,则>2,n>五,于是-1=-< 若x为(∞、0内任一点,不妨设p一动<-2则<2,xn>2,于题 设x0为(-∞,0)U(0,+∞)内任一点 对ve>0,取δ >0,当|-ro<b时 在x点连续 又由x在(-∞,0)U(0,+∞)内的任意性,得y=在(-∞,0)U(0,+∞)内连续 3)设x为(-∞,+∞)内任一点,||-|xl≤|r-xo 对ve>0,取6=>0,当x-xl<6时,有-{l‖l≤|x-x|<ε,故y=lx|在x点连续 又由x在(-∞,+)内的任意性,得y=|x|在(-∞,+∞)内连续 (4)设x为(0,+∞)内任一点,不妨设-x|<2,则x 于是 r+telcos arco 若为一x内任一点,不一<一,则<会>号,于一如< 设x0为(-∞,0)U(0,+∞)内任一点, 对>0,取6=mn{,}>0,当-xn<硎时,有m1-sm1 r-Iol < er 故y=sin-在x点连续 又由x0在(-∞,0)U(0.+∞)内的任意性,得y=snx在(-,0)U0,+∞)内连续 2.利用连续函数的运算,求下列函数的连续范围
33 §3. ÎYºÍ 1. U½¬y²eºÍ3½¬çSÎYµ (1) y = √ x (2) y = 1 x (3) y = | x| (4) y = sin 1 x y²µ (1) x0è(0, +∞)S?ò:ß| √ x − √x0 | < |x − x0 | √ x + √x0 6 |x − x0 | √x0 È∀ε > 0ßδ = √x0 εß|x − x0 | < δûßk| √ x − √x0 | < |x − x0 | √x0 < εßy = √ x3x0:ÎY. qdx03(0, +∞)•?ø5ßKy = √ x3(0, +∞)SÎY. x0 = 0ûßÈ˛„ε > 0ßδ = ε 2ß0 < x − x0 < δûßk| √ x − √x0 | < √ x < εßf(+0) = 0 = f(0)ß l y = √ x3[0, +∞)SÎY. (2) x0è(0, +∞)S?ò:ßÿî|x − x0 | < x0 2 ßKx > x0 2 , xx0 > x 2 0 2 ßu¥| 1 x − 1 x0 | = |x − x0 | xx0 < |x − x0 | x 2 0 2 ex0è(−∞, 0)S?ò:ßÿî|x − x0 | < − x0 2 ßKx < x0 2 , xx0 > x 2 0 2 ßu¥| 1 x − 1 x0 | = |x − x0 | xx0 < |x − x0 | x 2 0 2 x0è(−∞, 0) S (0, +∞)S?ò:ß È∀ε > 0ßδ = min |x0 | 2 , x 2 0 2 ε > 0ß|x − x0 | < δûßk 1 x − 1 x0 = |x − x0 | xx0 > εßy = 1 x 3x0:ÎY qdx03(−∞, 0) S (0, +∞)S?ø5ßy = 1 x 3(−∞, 0) S (0, +∞)SÎY. (3) x0è(−∞, +∞)S?ò:ß||x| − |x0 || 6 |x − x0 |. È∀ε > 0ßδ = ε > 0ß|x − x0 | < δûßk||x| − |x0 || 6 |x − x0 | < εßy = |x|3x0:ÎY qdx03(−∞, +∞)S?ø5ßy = |x|3(−∞, +∞)SÎY. (4) x0è(0, +∞)S?ò:ßÿî|x − x0 | < x0 2 ßKx > x0 2 , xx0 > x 2 0 2 ßu¥ sin 1 x − sin 1 x0 = 2 sin x + x0 2xx0 cos x − x0 2xx0 6 |x − x0 | xx0 < |x − x0 | x 2 0 2 ex0è(−∞, 0)S?ò:ßÿî|x − x0 | < − x0 2 ßKx < x0 2 , xx0 > x 2 0 2 ßu¥ sin 1 x − sin 1 x0 6 |x − x0 | xx0 < |x − x0 | x 2 0 2 x0è(−∞, 0) S (0, +∞)S?ò:ß È∀ε > 0ßδ = min |x0 | 2 , x 2 0 2 ε > 0ß|x − x0 | < δûßk sin 1 x − sin 1 x0 6 |x − x0 | xx0 < εß y = sin 1 x 3x0:ÎY qdx03(−∞, 0) S (0, +∞)S?ø5ßy = sin 1 x 3(−∞, 0) S (0, +∞)SÎY. 2. |^ÎYºÍ$é߶eºÍÎYâåµ (1) y = tan x
(3)y=sec r+csca (4)y= (1+x) (1)因y=tanx= Cos T 则当sr≠0时,y=tmz连续,故y=tam的连续范围为(-2+kx2+kx)k∈ 连续范围为(-∞,0)∪(0,+∞);若n≤0 -连续,即它的连续范围 为 ∞,+∞0 (3因wx的连续范围为(-)<x<(+)=0士士2…,cmx的连续范国为<x< (k+1)r(k=0.,±1,±2,…), 故y=8+c的连续范围为(kx-2)U(k元A+2)(=0士1士2…) (4)当osx>0时,y=√cs 连续,故y 的连续范围为(-2+2kx,2+2x) (5)因mn(1+x)当x>-1时连续, 当x≠0.,x≠2时连续,故 的连续范围为(-1,0)U(0,2)U(2,+∞) ()因=回mx=m同m,则当mx≠1,cx≠0,xgZ/0)时,y=回tm正连续 故=H如的连续范围为∈(x=2+2)且要2/(∈ 3.研究下列函数的连续性,并画出其图形 若x≠2 1)y= (2)y= sinT 4 (1)因limy=lim limn(x+2)=4,且当x=2时,y=4,故函数在x=2连续 2-4-2=x+2显然连续, 当x≠2时,y= 若x≠2 在(-∞,+∞)内连续 (2)当x≠0时,y= 显然连续。又 =1=f(0),故函数在x=0连 x≠0 在(-∞,+∞)内连续
34 (2) y = 1 xn (3) y = sec x + csc x (4) y = 1 √ cos x (5) y = ln(1 + x) x2 − 2x (6) y = [x] tan x 1 + sin x )µ (1) œy = tan x = sin x cos x ßKcos x 6= 0ûßy = tan xÎYßy = tan xÎYâåè − π 2 + kπ, π 2 + kπ (k ∈ Z). (2) en > 0ßKy = 1 xn ÎYâåè(−∞, 0) S (0, +∞)¶en 6 0ßKy = 1 xn ÎYß=ßÎYâå è(−∞, +∞). (3) œsec xÎYâåè k − 1 2 π < x < k + 1 2 π(k = 0, ±1, ±2, · · ·)ßcsc xÎYâåèkπ < x < (k + 1)π(k = 0, ±1, ±2, · · ·)ß y = sec x + csc xÎYâåè kπ − π 2 S kπ, kπ + π 2 ((k = 0, ±1, ±2, · · ·). (4) cos x > 0ûßy = 1 √ cos x ÎYßy = 1 √ cos x ÎYâåè − π 2 + 2kπ, π 2 + 2kπ . (5) œln(1+x)x > −1ûÎYß 1 x2 − 2x x 6= 0, x 6= 2ûÎYßy = ln(1 + x) x2 − 2x ÎYâåè(−1, 0) S (0, 2) S (2, +∞). (6) œy = [x] tan x 1 + sin x = [x] sin x (1 + sin x) cos x ßKsin x 6= 1, cos x 6= 0, x /∈ Z/{0}ûßy = [x] tan x 1 + sin x ÎYß y = [x] tan x 1 + sin x ÎYâåèx ∈ kπ − π 2 , kπ + π 2 Öx /∈ Z/{0}(k ∈ Z). 3. ÔƒeºÍÎY5ßøx—Ÿ„/. (1) y = x 2 − 4 x − 2 , ex 6= 2 4, x = 2 (2) y = sin x x , x 6= 0 1, x = 0 (3) y == sin x |x| , x 6= 0 1, x = 0 (4) y=[x] )µ (1) œlimx→2 y = limx→2 x 2 − 4 x − 2 = limx→2 (x + 2) = 4ßÖx = 2ûßy = 4ߺÍ3x = 2ÎY x 6= 2ûßy = x 2 − 4 x − 2 = x + 2w,ÎYß y = x 2 − 4 x − 2 , ex 6= 2 4, x = 2 3(−∞, +∞)SÎY. (2) x 6= 0ûßy = sin x x = sin x x ½y = − sin x x w,ÎY"qlimx→0 sin x x = 1 = f(0)ߺÍ3x = 0Î Yß u¥y = sin x x , x 6= 0 1, x = 0 3(−∞, +∞)SÎY
1=m:y=-m需=1,,y===1,故不存在,又当x>0时,y= ix,当x<时,y=x_sinx,显然连续,故此函数在除0外连续,即在(-∞,O)U(,+∞)内 连续 (4)因职0=即=k,职。=,职回=k-1∈2),则Hmy不存在,故x=kk∈ Z)为y=[]的间断点,但在间断点处右连续 当k<<k+1(k∈Z)时,y=回显然连续,故此函数在除k(k∈Z)外连续 4.若∫(x)连续,‖f(x)和f2(x)是否也连续?又若∫(x)或尸2(x)连续,f(x)是否连续? (1)设∫(x)在其定义域Ⅰ上连续,x为Ⅰ上任一点 因f(x)在r点连续,故对ve>0,36>0,当-x<时,有f(x)-f(xo)<E 而f(x)-|f(x0川≤|∫f(x)-f(x0)<E,即对ve>0,306>0,当x-x。|<6时,有∫f(x)-f(x0)<e 故∫f(x)在x点连续 又由x在I上的任意性,知f(x)在Ⅰ上也连续 同样2(x)-f2(xo)=f(x)-f(xo川∫(x)+f(xo)=|f(x)-f(xo川f(x)-f(xa)+2f(xo)≤ f(x)-∫(xr川|(f(x)-∫(x)+2f(xa)<ε(e+2f(x0),故f2(x)在x点连续 又由x在上的任意性,知f2(x)在Ⅰ上也连续 (2)反过来,若f(x)或f2(x)连续,f(x)不一定连续 (i)不连续。例:f(x) x≥0 1,z<0|f(x)=1和f2(x)=1均在(-∞,+∞)内连续,但f(x)在x 0点不连续 (i)连续。例:∫(x)=x,则∫(x)、|∫(x)、f2(x)在(-∞,+∞)内均连续 5.(1)函数f(x)当x=xo时连续,而函数g(x)当x=x0时不连续,问此二函数的和在x0点是否连续? (2)当x=xo时函数∫(x)和g(x)二者都不连续,问此二函数的和f(x)+g(x)在已知点x0是否必为不连续? (1)用反证法。假设f(x)+g(x)在x点连 因f(x)当x=xr时连续,则由连续函数性质,得g(x)=[f(x)+g(x)-f(x)当xo时连续与已知矛盾 故假设不成立,即f(x)+g(x)在x点连续 (2)不一定。 (i)连续:例:f(x)= x≥0 x<0,9(x) x<0 在x=0都不连续,但f(x)+ (i)不连续:例:f(x)=g(x)=-在x=0都不连续,f(x)+g(x)=二在x=0不连续 6.(1)函数f(x)在x连续,而函数g(x)在x不连续; (2)当x=x0时函数f(x)和g(x)二者都不连续,问此二函数的乘积f(x)9(x)在已知点x0是否必不连续? )连续:例:(x)=0在x=0连续,g(2)={0在x=0不连续,但(l()=0在x= (i)不连续:例:∫(x)=x在x=0连续,g(x)=立在x=0不连续,f(x)g(x)=在x=0不连续 2)不一定。 ()连续:例:f(x)={-1,x<0,s(x)={;.x≥0 x<0在x=0都不连续,但f(x)(x) 1在x=0连续 (1)不连续:例:f(x)=)=在x=0都不连续,f(a)()=在x=0不连续 7.若∫(x)在{a,∞)连续,并且lim∫(x)存在,证明f(x)在{a,∞)有界 证明:由于limf(x)存在,不妨设limf(x)=A 则对e=1,丑X>0,当x>X时,有∫(x)-A|<E=1成立,从而得∫(x)=|f(x)-A+A≤|f(a)-4|+
35 (3) œ lim x→+0 y = lim x→+0 sin x |x| = 1, lim x→−0 y = lim x→−0 sin x |x| = −1 ßlimx→0 yÿ3"qx > 0ûßy = sin x |x| = sin x x ßx < 0ûßy = sin x |x| = − sin x x ßw,ÎYßdºÍ3ÿ0 ÎYß=3(−∞, 0) S (0, +∞)S ÎY. (4) œ lim x→k+0 y = lim x→k+0 [x] = k, lim x→k−0 y = lim x→k−0 [x] = k − 1(k ∈ Z)ßKlimx→k yÿ3ßx = k(k ∈ Z)èy = [x]m‰:ß3m‰:?mÎY k < x < k + 1(k ∈ Z)ûßy = [x]w,ÎYßdºÍ3ÿk(k ∈ Z) ÎY. 4. ef(x)ÎYß|f(x)|⁄f 2 (x)¥ƒèÎYºqe|f(x)|½f 2 (x)ÎYßf(x)¥ƒÎYº )µ (1) f(x)3Ÿ½¬çI˛ÎYßx0èI˛?ò: œf(x)3x0:ÎYßÈ∀ε > 0, ∃δ > 0ß|x − x0 | < δûßk|f(x) − f(x0 )| < ε ||f(x)|−|f(x0 )|| 6 |f(x)−f(x0 )| < εß=È∀ε > 0, ∃δ > 0ß|x−x0 | < δûßk|f(x)−f(x0 )| < εß |f(x)|3x0:ÎY qdx03I˛?ø5ß|f(x)|3I˛èÎY ”|f 2 (x) − f 2 (x0 )| = |f(x) − f(x0 )||f(x) + f(x0 )| = |f(x) − f(x0 )||f(x) − f(x0 ) + 2f(x0 )| 6 |f(x) − f(x0 )|(|f(x) − f(x0 )| + 2f(x0 )) < ε(ε + 2f(x0 ))ßf 2 (x)3x0:ÎY qdx03I˛?ø5ßf 2 (x)3I˛èÎY (2) áL5ße|f(x)|½f 2 (x)ÎYßf(x)ÿò½ÎY. (i) ÿÎY"~µf(x) = 1, x > 0 −1, x < 0 ß|f(x)| = 1⁄f 2 (x) = 1˛3(−∞, +∞)SÎYßf(x)3x = 0:ÿÎY¶ (ii) ÎY"~µf(x) = x,Kf(x)!|f(x)|!f 2 (x)3(−∞, +∞)S˛ÎY" 5. (1) ºÍf(x)x = x0ûÎYß ºÍg(x)x = x0ûÿÎYßØdºÍ⁄3x0:¥ƒÎYº (2) x = x0ûºÍf(x)⁄g(x)ˆ—ÿÎYßØdºÍ⁄f(x) + g(x)3Æ:x0¥ƒ7èÿÎYº ): (1) ^áy{"bf(x) + g(x)3x0:ÎY" œf(x)x = x0ûÎYßKdÎYºÍ5üßg(x) = [f(x) + g(x)] − f(x)x0ûÎYÜÆgÒ" bÿ§·ß=f(x) + g(x)3x0:ÎY. (2) ÿò½" (i) ÎYµ~µf(x) = 1, x > 0 −1, x < 0 , g(x) = −1, x > 0 1, x < 0 3x = 0—ÿÎYßf(x) + g(x) = 03x = 0ÎY. (ii) ÿÎYµ~µf(x) = g(x) = 1 x 3x = 0—ÿÎYßf(x) + g(x) = 2 x 3x = 0ÿÎY. 6. (1) ºÍf(x)3x0ÎYß ºÍg(x)3x0ÿÎY¶ (2) x = x0ûºÍf(x)⁄g(x)ˆ—ÿÎYßØdºÍ¶»f(x)g(x)3Æ:x0¥ƒ7ÿÎYº ): (1) ÿò½" (i) ÎYµ~µf(x) = 03x = 0ÎYßg(x) = 1, x > 0 0, x < 0 3x = 0ÿÎYßf(x)g(x) = 03x = 0ÎY. (ii) ÿÎYµ~µf(x) = x3x = 0ÎYßg(x) = 1 x2 3x = 0ÿÎYßf(x)g(x) = 1 x 3x = 0ÿÎY. (2) ÿò½" (i) ÎYµ~µf(x) = 1, x > 0 −1, x < 0 , g(x) = −1, x > 0 1, x < 0 3x = 0—ÿÎYßf(x)g(x) = −13x = 0ÎY. (ii) ÿÎYµ~µf(x) = g(x) = 1 x 3x = 0—ÿÎYßf(x)g(x) = 1 x2 3x = 0ÿÎY. 7. ef(x)3[a,∞)ÎYßøÖ limx→∞ f(x)3ßy²f(x)3[a,∞)k.. y²µ du limx→∞ f(x)3ßÿî limx→∞ f(x) = A KÈε = 1, ∃X > 0ßx > Xûßk|f(x) − A| < ε = 1§·ßl |f(x)| = |f(x) − A + A| 6 |f(x) − A| +