Bài tập toán cao cấp part 4 pot

7.3. H`am liˆen tu . c 47 nguyˆen) sao cho ta . id´o h `a m b ˘a ` ng h˘a ` ng sˆo ´ . Do vˆa . y n´o liˆen tu . cta . i x 0 . Nˆe ´ u x 0 = n l`a sˆo ´ nguyˆen th`ı [n − 0] = n −1, [n +0]=n.T`u . d ´o suy r˘a ` ng x 0 = n l`a diˆe ’ m gi´an doa . nkiˆe ’ uI. V´ı du . 8. Kha ’ o s´at su . . liˆen tu . c v`a phˆan loa . id iˆe ’ m gi´an doa . ncu ’ a c´ac h`am 1) f(x)= x 2 x , 2) f(x)=e − 1 x , 3) f(x)=    x nˆe ´ u x  1 lnx nˆe ´ u x>1. Gia ’ i 1) H`am f(x)=x nˆe ´ u x = 0 v`a khˆong x´ac d i . nh khi x =0. V`ı ∀a ta c´o lim x→a x = a nˆen khi a =0: lim x→a f(x)=a = f(a) v`a do vˆa . y h`am f(x)liˆen tu . c ∀x = 0. Ta . id iˆe ’ m x = 0 ta c´o gi´an doa . n khu . ’ du . o . . cv`ıtˆo ` nta . i lim x→0 f(x) = lim x→0 x =0. 2) H`am f(x)=e − 1 x l`a h`am so . cˆa ´ pv`ı n´o l`a ho . . pcu ’ a c´ac h`am y = −x −1 v`a f = e y .Hiˆe ’ n nhiˆen l`a h`am f(x) x´ac di . nh ∀x =0v`a do d´o n´o liˆen tu . c ∀x =0. V`ı h`am f(x) x´ac di . nh trong lˆan cˆa . ndiˆe ’ m x = 0 v`a khˆong x´ac di . nh ta . ich´ınh diˆe ’ m x =0nˆen diˆe ’ m x =0l`adiˆe ’ m gi´an d oa . n. Ta t´ınh f(0 + 0) v`a f(0 − 0). Ta x´et d˜ay vˆo c`ung b´et`uy ´y (x n ) sao cho x n > 0 ∀n.V`ı lim x→∞  − 1 x n  = −∞ nˆen lim x→∞ e − 1 x n =0. T`u . d´o suy r˘a ` ng lim x→0+0 e − 1 x =0. Bˆay gi`o . ta x´et d˜ay vˆo c`ung b´e bˆa ´ tk`y(x  n ) sao cho x  0 < 0 ∀n.V`ı lim n→∞  − 1 x  n  =+∞ nˆen lim x→0 e − 1 x  n =+∞.Dod´o lim x→0−0 e − 1 x =+∞ t´u . cl`af(0 −0)=+∞. Nhu . vˆa . y gi´o . iha . n bˆen tr´ai cu ’ a h`am f(x)ta . id iˆe ’ m x = 0 khˆong tˆo ` n ta . idod´odiˆe ’ m x =0l`adiˆe ’ m gi´an doa . nkiˆe ’ uII. 48 Chu . o . ng 7. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am sˆo ´ 3) Ta ch´u . ng minh r˘a ` ng f(x)liˆen tu . cta . id iˆe ’ m x = a = 1. Ta lˆa ´ y ε<|a −1|, ε>0. Khi d´o ε-lˆan cˆa . ncu ’ adiˆe ’ m x = a khˆong ch´u . ad iˆe ’ m x =1nˆe ´ u ε<|a − 1|. Trong ε-lˆan cˆa . n n`ay h`am f(x) ho˘a . ctr`ung v´o . i h`am ϕ(x)=x nˆe ´ u a<1 ho˘a . ctr`ung v´o . i h`am ϕ(x)=lnx nˆe ´ u a>1. V`ı c´ac h`am so . cˆa ´ pco . ba ’ n n`ay liˆen tu . cta . idiˆe ’ m x = a nˆen h`am f(x) liˆen tu . cta . id iˆe ’ m x = a =1. Ta kha ’ o s´at t´ınh liˆen tu . ccu ’ a h`am f(x)ta . idiˆe ’ m x = a =1. Dˆe ’ l`am viˆe . cd ´o ta cˆa ` n t´ınh c´ac gi´o . iha . nmˆo . t ph´ıa cu ’ a f(x)ta . id iˆe ’ m x = a =1. Ta c´o f(1 + 0) = lim x→1+0 f(x) = lim x→1+0 lnx =0, f(1 − 0) = lim x→1−0 f(x) = lim x→1−0 x = lim x→1 x =1. Nhu . vˆa . y f(1 + 0) = f(1 −0) v`a do d´o h`am f(x) c´o gi´an doa . nkiˆe ’ u Ita . i x = a =1. B ` AI T ˆ A . P Kha ’ o s´at t´ınh liˆen tu . c v`a phˆan loa . id iˆe ’ m gi´an doa . ncu ’ a h`am 1. f(x)= |2x − 3| 2x − 3 (DS. H`am x´ac di . nh v`a liˆen tu . c ∀x = 3 2 ;ta . i x 0 = 3 2 h`am c´o gi´an doa . nkiˆe ’ uI) 2. f(x)=    1 x nˆe ´ u x =0 1nˆe ´ u x =0. (D S. H`am liˆen tu . c ∀x ∈ R) 3. C´o tˆo ` nta . i hay khˆong gi´a tri . a d ˆe ’ h`am f(x)liˆen tu . cta . i x 0 nˆe ´ u: 1) f(x)=    4 · 3 x nˆe ´ u x<0 2a + x khi x  0. (DS. H`am f liˆen tu . c ∀x ∈ R nˆe ´ u a =2) 7.3. H`am liˆen tu . c 49 2) f(x)=    x sin 1 x ,x=0; a, x =0,x 0 =0. . (DS. a =0) 3) f(x)=    1+x 1+x 3 ,x= −1 a, x = −1,x 0 = −1. (DS. a = 1 3 ) 4) f(x)=    cos x, x  0; a(x −1),x>0; x 0 =0. (DS. a = −1) 4. f(x)= |sin x| sin x (DS. H`am c´o gi´an doa . nta . i x = kπ, k ∈ Z v`ı: f(x)=    1nˆe ´ u sin x>0 −1nˆe ´ u sin x<0) 5. f(x)=E(x) − E(−x) (D S. H`am c´o gi´an doa . nkhu . ’ du . o . . cta . i x = n, x ∈ Z v`ı: f(x)=    −1nˆe ´ u x = n 0nˆe ´ u x = n.) 6. f(x)=    e 1/x khi x =0 0 khi x =0. (D S. Ta . idiˆe ’ m x = 0 h`am c´o gi´an doa . nkiˆe ’ uII;f(−0) = 0, f(+0) = ∞) T`ım diˆe ’ m gi´an doa . n v`a t´ınh bu . ´o . c nha ’ ycu ’ a c´ac h`am: 7. f(x)=x + x +2 |x +2| (D S. x = −2l`adiˆe ’ m gi´an doa . nkiˆe ’ uI,δ(−2) = 2) 50 Chu . o . ng 7. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am sˆo ´ 8. f(x)= 2|x − 1| x 2 −x 3 (DS. x =0l`adiˆe ’ m gi´an doa . nkiˆe ’ uII,x =1l`adiˆe ’ m gi´an doa . nkiˆe ’ u I, δ(1) = −4) H˜ay bˆo ’ sung c´ac h`am sau dˆay ta . idiˆe ’ m x =0dˆe ’ ch´ung tro . ’ th`anh liˆen tu . c 9. f(x)= tgx x (DS. f(0) = 1) 10. f(x)= √ 1+x − 1 x (D S. f(0) = 1 2 ) 11. f(x)= sin 2 x 1 − cos x (DS. f(0) = 2) 12. Hiˆe . ucu ’ a c´ac gi´o . iha . nmˆo . tph´ıa cu ’ a h`am f(x): d = lim x→x 0 +0 f(x) − lim x→x 0 −0 f(x) d u . o . . cgo . il`abu . ´o . c nha ’ y cu ’ a h`am f(x)ta . idiˆe ’ m x 0 .T`ım diˆe ’ m gi´an doa . n v`a bu . ´o . c nha ’ y cu ’ a h`am f(x)nˆe ´ u: 1) f(x)=    − 1 2 x 2 nˆe ´ u x  2, x nˆe ´ u x>2. (D S. x 0 =2l`adiˆe ’ m gi´an doa . nkiˆe ’ uI;d =4) 2) f(x)=          2 √ x nˆe ´ u0 x  1; 4 − 2x nˆe ´ u1<x 2, 5; 2x − 7nˆe ´ u2, 5  x<+∞. (D S. x 0 =2, 5l`adiˆe ’ m gi´an doa . nkiˆe ’ uI;d = −1) 3) f(x)=    2x +5 nˆe ´ u −∞<x<−1, 1 x nˆe ´ u − 1  x<+∞. (DS. x 0 =0l`adiˆe ’ m gi´an doa . nkiˆe ’ uII;diˆe ’ m x 0 = −1l`adiˆe ’ m gi´an doa . nkiˆe ’ uI,d = −4) 7.4. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am nhiˆe ` ubiˆe ´ n 51 7.4 Gi´o . iha . nv`aliˆen tu . ccu ’ ah`am nhiˆe ` u biˆe ´ n 1. Gia ’ su . ’ u = f(M)=f(x, y) x´ac d i . nh trˆen tˆa . pho . . p D. Gia ’ su . ’ M 0 (x 0 ,y 0 )l`adiˆe ’ mcˆo ´ di . nh n`ao d´ocu ’ am˘a . t ph˘a ’ ng v`a x → x 0 , y → y 0 , khi d´o d iˆe ’ m M(x, y) → M 0 (x 0 ,y 0 ). Diˆe ` u n`ay tu . o . ng du . o . ng v´o . i khoa ’ ng c´ach ρ(M,M 0 )gi˜u . ahaidiˆe ’ m M v`a M 0 dˆa ` ndˆe ´ n0.Talu . u´yr˘a ` ng ρ(M,M 0 )=[(x − x 0 ) 2 +(y − y 0 ) 2 ] 1/2 . Ta c´o c´ac d i . nh ngh˜ıa sau dˆay: i) Di . nh ngh˜ıa gi´o . iha . n (theo Cauchy) Sˆo ´ b d u . o . . cgo . i l`a gi´o . iha . ncu ’ a h`am f(M) khi M → M 0 (hay ta . i diˆe ’ m M 0 )nˆe ´ u ∀ε>0, ∃δ = δ(ε) > 0:∀M ∈{D :0<ρ(M, M 0 ) <δ(ε)} ⇒|f(M) −b| <ε. ii) D i . nh ngh˜ıa gi´o . iha . n (theo Heine) Sˆo ´ b du . o . . cgo . i l`a gi´o . iha . ncu ’ a h`am f(M)ta . idiˆe ’ m M 0 nˆe ´ udˆo ´ iv´o . i d˜ay diˆe ’ m {M n } bˆa ´ tk`yhˆo . itu . dˆe ´ n M 0 sao cho M n ∈ D, M n = M 0 ∀n ∈ N th`ı d˜ay c´ac gi´a tri . tu . o . ng ´u . ng cu ’ a h`am {f(M n )} hˆo . itu . dˆe ´ n b. K´yhiˆe . u: i) lim M→M 0 f(M)=b, ho˘a . c ii) lim x → x 0 y → y 0 f(x, y)=b Hai di . nh ngh˜ıa gi´o . iha . ntrˆendˆay tu . o . ng du . o . ng v´o . i nhau. Ch´u´y.Ta nhˆa ´ nma . nh r˘a ` ng theo di . nh ngh˜ıa, gi´o . iha . ncu ’ a h`am khˆong phu . thuˆo . c v`ao phu . o . ng M dˆa ` nt´o . i M 0 .Dod´onˆe ´ u M → M 0 theo c´ac hu . ´o . ng kh´ac nhau m`a f(M)dˆa ` nd ˆe ´ n c´ac gi´a tri . kh´ac nhau th`ı khi M → M 0 h`am f(M) khˆong c´o gi´o . iha . n. 52 Chu . o . ng 7. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am sˆo ´ iii) Sˆo ´ b du . o . . cgo . i l`a gi´o . iha . ncu ’ a h`am f(M) khi M →∞nˆe ´ u ∀ε>0, ∃R>0:∀M ∈{D : ρ(M, 0) >R}⇒|f(M) − b| <ε. D ˆo ´ iv´o . i h`am nhiˆe ` ubiˆe ´ n, c`ung v´o . i gi´o . iha . n thˆong thu . `o . ng d ˜anˆeuo . ’ trˆen (gi´o . iha . nk´ep !), ngu . `o . i ta c`on x´et gi´o . iha . nl˘a . p. Ta s˜e x´et kh´ai niˆe . m n`ay cho h`am hai biˆe ´ n u = f(M)=f(x, y). Gia ’ su . ’ u = f(x, y) x´ac di . nh trong h`ınh ch˜u . nhˆa . t Q = {(x, y):|x −x 0 | <d 1 , |y − y 0 | <d 2 } c´o thˆe ’ tr `u . ra ch´ınh c´ac diˆe ’ m x = x 0 , y = y 0 . Khi cˆo ´ di . nh mˆo . t gi´a tri . y th`ı h`am f(x, y) tro . ’ th`anh h`am mˆo . tbiˆe ´ n. Gia ’ su . ’ d ˆo ´ iv´o . i gi´a tri . cˆo ´ d i . nh y bˆa ´ tk`y tho ’ a m˜an diˆe ` ukiˆe . n0< |y − y 0 | <d 2 tˆo ` nta . i gi´o . iha . n lim x→x 0 y cˆo ´ di . nh f(x, y)=ϕ(y). Tiˆe ´ p theo, gia ’ su . ’ lim y→y 0 ϕ(y)=b tˆo ` nta . i. Khi d´o ngu . `o . i ta n´oi r˘a ` ng tˆo ` nta . i gi´o . iha . nl˘a . p cu ’ a h`am f(x, y)ta . idiˆe ’ m M 0 (x 0 ,y 0 ) v`a viˆe ´ t lim y→y 0 lim x→x 0 f(x, y)=b, trong d´o gi´o . iha . n lim x→x 0 y cˆo ´ di . nh 0<|y−y 0 |<d 2 f(x, y)go . i l`a gi´o . iha . n trong. Tu . o . ng tu . . ,ta c´o thˆe ’ ph´at biˆe ’ ud i . nh ngh˜ıa gi´o . iha . nl˘a . p kh´ac lim x→x 0 lim y→y 0 f(x, y) trong d´o g i ´o . iha . n lim y→y 0 x cˆo ´ di . nh 0<|x−x 0 |<d 1 f(x, y) l`a gi´o . iha . n trong. Mˆo ´ i quan hˆe . gi˜u . a gi´o . iha . n k´ep v`a c´ac gi´o . iha . nl˘a . pd u . o . . cthˆe ’ hiˆe . n trong d i . nh l´y sau dˆay: 7.4. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am nhiˆe ` ubiˆe ´ n 53 Gia ’ su . ’ ta . id iˆe ’ m M 0 (x 0 ,y 0 ) gi´o . iha . nk´ep v`a c´ac gi´o . iha . n trong cu ’ a c´ac gi´o . iha . nl˘a . pcu ’ a h`am tˆo ` nta . i. Khi d ´o c´ac gi´o . iha . nl˘a . ptˆo ` nta . iv`a lim x→x 0 lim y→y 0 f(x, y) = lim y→y 0 lim x→x 0 = lim x→x 0 y→y 0 f(x, y). T`u . d i . nh l´y n`ay ta thˆa ´ yr˘a ` ng viˆe . c thay dˆo ’ ith´u . tu . . trong c´ac gi´o . i ha . n khˆong pha ’ i bao gi`o . c˜ung d u . o . . cph´ep. D ˆo ´ iv´o . i h`am nhiˆe ` ubiˆe ´ ntac˜ung c´o nh˜u . ng d i . nh l´y vˆe ` c´ac t´ınh chˆa ´ t sˆo ´ ho . ccu ’ a gi´o . iha . ntu . o . ng tu . . c´ac di . nh l´yvˆe ` gi´o . iha . ncu ’ a h`am mˆo . t biˆe ´ n. 2. T`u . kh´ai niˆe . m gi´o . iha . n ta s˜e tr`ınh b`ay kh´ai niˆe . mvˆe ` t´ınh liˆen tu . c cu ’ a h`am nhiˆe ` ubiˆe ´ n. H`am u = f(M)d u . o . . cgo . il`aliˆen tu . c ta . id iˆe ’ m M 0 nˆe ´ u: i) f(M) x´ac di . nh ta . ich´ınh diˆe ’ m M 0 c˜ung nhu . trong mˆo . t lˆan cˆa . n n`ao d´ocu ’ adiˆe ’ m M 0 . ii) Gi´o . iha . n lim M→M 0 f(M)tˆo ` nta . i. iii) lim M→M 0 f(M)=f(M 0 ). Su . . liˆen tu . cv`u . adu . o . . cdi . nh ngh˜ıa go . i l`a su . . liˆen tu . c theo tˆa . pho . . p biˆe ´ nsˆo ´ . H`am f(M)liˆen tu . c trong miˆe ` n D nˆe ´ u n´o liˆen tu . cta . imo . idiˆe ’ mcu ’ a miˆe ` nd ´o. Diˆe ’ m M 0 du . o . . cgo . il`ad iˆe ’ m gi´an doa . n cu ’ a h`am f(M)nˆe ´ udˆo ´ iv´o . i d iˆe ’ m M 0 c´o ´ıt nhˆa ´ tmˆo . t trong ba diˆe ` ukiˆe . n trong di . nh ngh˜ıa liˆen tu . c khˆong tho ’ a m˜an. Diˆe ’ m gi´an doa . ncu ’ a h`am nhiˆe ` ubiˆe ´ nc´othˆe ’ l`a nh˜u . ng diˆe ’ m cˆo lˆa . p, v`a c˜ung c´o thˆe ’ l`a ca ’ mˆo . tdu . `o . ng (du . `o . ng gi´an doa . n). Nˆe ´ u h`am f(x, y)liˆen tu . cta . id iˆe ’ m M 0 (x 0 ,y 0 ) theo tˆa . pho . . pbiˆe ´ nsˆo ´ th`ı n´o liˆen tu . c theo t`u . ng biˆe ´ nsˆo ´ .Diˆe ` u kh˘a ’ ng di . nh ngu . o . . cla . i l`a khˆong d´ung. C˜ung nhu . dˆo ´ iv´o . i h`am mˆo . tbiˆe ´ n, tˆo ’ ng, hiˆe . u v`a t´ıch c´ac h`am liˆen tu . c hai biˆe ´ nta . id iˆe ’ m M 0 l`a h`am liˆen tu . cta . idiˆe ’ md´o; thu . o . ng cu ’ a hai h`am liˆen tu . cta . i M 0 c˜ung l`a h`am liˆen tu . cta . i M 0 nˆe ´ uta . idiˆe ’ m M 0 h`am 54 Chu . o . ng 7. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am sˆo ´ mˆa ˜ usˆo ´ kh´ac 0. Ngo`ai ra, di . nh l´y vˆe ` t´ınh liˆen tu . ccu ’ a h`am ho . . pvˆa ˜ n d ´ung trong tru . `o . ng ho . . p n`ay. Nhˆa . nx´et. Tu . o . ng tu . . nhu . trˆen ta c´o thˆe ’ tr`ınh b`ay c´ac kh´ai niˆe . mco . ba ’ n liˆen quan d ˆe ´ n gi´o . iha . n v`a liˆen tu . ccu ’ a h`am ba biˆe ´ n, C ´ AC V ´ IDU . V´ı du . 1. Ch´u . ng minh r˘a ` ng h`am f(x, y)=(x + y) sin 1 x sin 1 y l`a vˆo c`ung b´e ta . id iˆe ’ m O(0, 0). Gia ’ i. Theo d i . nh ngh˜ıa vˆo c`ung b´e (tu . o . ng tu . . nhu . d ˆo ´ iv´o . i h`am mˆo . t biˆe ´ n) ta cˆa ` nch´u . ng minh r˘a ` ng lim x→0 y→0 f(x, y)=0. Ta´apdu . ng d i . nh ngh˜ıa gi´o . iha . n theo Cauchy. Ta cho sˆo ´ ε>0t`uy ´yv`ad ˘a . t δ = ε 2 . Khi d´o n ˆe ´ u ρ  M(x, y),O(0, 0)  =  x 2 + y 2 <δth`ı |x| <δ,|y| <δ. Do d ´o |f(x, y) − 0| =    (x + y) sin 1 x sin 1 y     |x| + |y| < 2δ = ε. Diˆe ` ud´och´u . ng to ’ r˘a ` ng lim x→0 y→0 f(x, y)=0. V´ı du . 2. T´ınh c´ac gi´o . iha . n sau d ˆay: 1) lim x→0 y→2  1+xy  2 x 2 + xy , 2) lim x→0 y→2  x 2 +(y − x) 2 +1− 1 x 2 +(y − 2) 2 , 3) lim x→0 y→0 x 4 + y 4 x 2 + y 2 . 7.4. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am nhiˆe ` ubiˆe ´ n 55 Gia ’ i. 1) Ta biˆe ’ udiˆe ˜ n h`am du . ´o . idˆa ´ u gi´o . iha . ndu . ´o . ida . ng   1+xy  1 xy  2y x + y . V`ı t = xy → 0 khi  x → 0 y → 0  nˆen lim x→0 y→2  1+xy  1 xy = lim t→0  1+t  1 t = e. Tiˆe ´ p theo v`ı lim x→0 y→2 2 x + y = 2 (theo di . nh l´y thˆong thu . `o . ng vˆe ` gi´o . iha . n cu ’ athu . o . ng), do d´o gi´o . iha . ncˆa ` n t`ım b˘a ` ng e 2 . 2) Ta t`ım gi´o . iha . nv´o . id iˆe ` ukiˆe . n M(x, y) → M 0 (0, 2). Khoa ’ ng c´ach gi˜u . a hai diˆe ’ m M v`a M 0 b˘a ` ng ρ =  x 2 +(y − 2) 2 . Do d´o lim x→0 y→2 f(x, y) = lim ρ→0  ρ 2 +1− 1 ρ 2 = lim ρ→0 (ρ 2 +1)−1 ρ 2 (  ρ 2 +1+1) = lim ρ→0 1  ρ 2 +1+1 = 1 2 · 3) Chuyˆe ’ n sang to . adˆo . cu . . ctac´ox = ρ cos ϕ, y = ρ sin ϕ.Tac´o x 4 + y 4 x 2 + y 2 = ρ 4 (cos 4 ϕ + sin 4 ϕ) ρ 2 (cos 2 ϕ + sin 2 ϕ) = ρ 2 (cos 4 ϕ + sin 4 ϕ). V`ı cos 4 ϕ + sin 4 ϕ  2nˆen lim x→0 y→0 x 4 + y 4 x 2 + y 2 = lim ρ→0 ρ 2 (cos 4 ϕ + sin 4 ϕ)=0. 56 Chu . o . ng 7. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am sˆo ´ V´ı du . 3. 1) Ch´u . ng minh r˘a ` ng h`am f 1 (x, y)= x −y x + y khˆong c´o gi´o . iha . nta . idiˆe ’ m(0, 0). 2) H`am f 2 (x, y)= xy x 2 + y 2 c´o gi´o . iha . nta . idiˆe ’ m(0, 0) hay khˆong ? Gia ’ i. 1) H`am f 1 (x, y) x´ac di . nh kh˘a ´ pno . i ngoa . itr`u . du . `o . ng th˘a ’ ng x + y =0. Tach´u . ng minh r˘a ` ng h`am khˆong c´o gi´o . iha . nta . i(0, 0). Ta lˆa ´ y hai d˜ay d iˆe ’ mhˆo . itu . dˆe ´ ndiˆe ’ m(0, 0): M n =  1 n , 0  → (0, 0),n→∞, M  n =  0, 1 n  → (0, 0),n→∞. Khi d ´othudu . o . . c lim n→∞ f 1 (M n ) = lim n→∞ 1 n −0 1 n +0 =1; lim n→∞ f 1 (M  n ) = lim n→∞ 0 − 1 n 0+ 1 n = −1. Nhu . vˆa . y hai d˜ay diˆe ’ m kh´ac nhau c`ung hˆo . itu . dˆe ´ ndiˆe ’ m(0, 0) nhu . ng hai d˜ay gi´a tri . tu . o . ng ´u . ng cu ’ a h`am khˆong c´o c`ung gi´o . iha . n. Do d ´o theo di . nh ngh˜ıa h`am khˆong c´o gi´o . iha . nta . i(0, 0). 2) Gia ’ su . ’ diˆe ’ m M(x, y)dˆa ` ndˆe ´ ndiˆe ’ m(0, 0) theo du . `o . ng th˘a ’ ng y = kx qua gˆo ´ cto . adˆo . . Khi d´o ta c´o lim x→0 y→0 (y=kx) xy x 2 + y 2 = lim x→0 kx 2 x 2 + k 2 x 2 = k 1+k 2 · [...]... cˆp 1 61 a - a ´ Dao h`m cˆp cao 62 a Vi phˆn 75 a 8.2.1 8.2.2 8.3 ´ Vi phˆn cˆp 1 75 a a ´ Vi phˆn cˆp cao 77 a a ` a ’ ’ C´c dinh l´ co ban vˆ h`m kha vi Quy a e y ´ t˘c l’Hospital Cˆng th´.c Taylor 84 a o u 8.3.1 ’ ` a ’ C´c d inh l´ co ban vˆ h`m kha vi 84 a e y 8.3.2 ´ o ’ a Khu c´c dang vˆ dinh Quy... w = 2 w = 3 w = 4 w = √ xy (DS |y| 0, y (DS x a2 − x2 − y 2 1 x2 + y 2 − a2 |x|) 0 ho˘c x a (DS x2 + y 2 0, y 0) a2 ) (DS x2 + y 2 > a2) x2 y 2 x2 y 2 − 2 (DS 2 + 2 1) a2 b a b 6 w = ln(z 2 − x2 − y 2 − 1) (DS x2 + y 2 − z 2 < −1) x √ ’ ’ u a o a 7 w = arcsin + xy (DS Hai nu.a b˘ng vˆ han th˘ng d´.ng 2 {0 x 2, 0 y < +∞} v` {−2 x 0, −∞ < y 0}) a 5 w = 1− 8 w = x2 + y 2 − 1 + ln (4 − x2 − y 2) a o... − 1 + ln (4 − x2 − y 2) a o (DS V`nh tr`n 1 x2 + y 2 < 4) ` a o a 9 w = sin π(x2 + y 2 ) (DS Tˆp ho.p c´c v`nh dˆng tˆm a a 2 2 2 2 1; 2 x + y 3; ) 0 x +y 10 w = ln(1 + z − x2 − y 2 ) ` ’ a a (DS Phˆn trong cua mˆt paraboloid z = x2 + y 2 − 1) ’ a a a ınh a o Trong c´c b`i to´n sau dˆy (11-18) h˜y t´ c´c gi´.i han cua h`m a a a ´ ` ’ a 7 .4 Gi´.i han v` liˆn tuc cua h`m nhiˆu biˆn o a e e e...´ ` ’ a 7 .4 Gi´.i han v` liˆn tuc cua h`m nhiˆu biˆn o a e e e ’ ’ ´ ` a a a e o a Nhu vˆy khi dˆn dˆn diˆm (0, 0) theo c´c du.`.ng th˘ng kh´c nhau e a o.ng u.ng v´.i c´c gi´ tri k kh´c nhau) ta thu du.o.c c´c gi´ tri gi´.i (tu ´ o a a a a o a han kh´c nhau, t´.c l` h`m d˜ cho khˆng c´ gi´.i han tai (0, 0) a u a a a o o o ’ a ınh e ’ a a V´ du 4 Khao s´t t´ liˆn tuc cua c´c... t´ c´c gi´.i han cua h`m a a a ´ ` ’ a 7 .4 Gi´.i han v` liˆn tuc cua h`m nhiˆu biˆn o a e e e sin xy x→0 xy (DS 1) sin xy x→0 x (DS 0) 11 lim y→0 12 lim y→0 xy 13 lim √ x→0 xy + 1 − 1 (DS 2) y→0 14 lim x→0 y→0 x2 + y 2 x2 + y 2 + 1 − 1 (DS 2) ’ ˜ ’ ’ a a Chı dˆ n Su dung khoang c´ch ρ = x2 + y 2 ho˘c nhˆn - chia a a i dai lu.o.ng liˆn ho.p v´.i mˆ u sˆ ˜ o e o a ´ v´ o y 2 2 15 lim 1 + xy... liˆn tuc Nˆu h`m f (x) kha e e o e a ’ ` vi th` n´ liˆn tuc Diˆu kh˘ng dinh ngu.o.c lai l` khˆng d´ng ı o e e a u a o ´ Chu.o.ng 8 Ph´p t´ vi phˆn h`m mˆt biˆn e ınh a a o e 62 - ´ Dao h`m cˆp cao a a 8.1.2 ´ ´ Dao h`m f (x) du.o.c goi l` dao h`m cˆp 1 (hay dao h`m bˆc nhˆt) a a a a a a a ´ ’ a a a a u Dao h`m cua f (x) du.o.c goi l` dao h`m cˆp hai (hay dao h`m th´ a o.c k´ hiˆu . ϕ.Tac´o x 4 + y 4 x 2 + y 2 = ρ 4 (cos 4 ϕ + sin 4 ϕ) ρ 2 (cos 2 ϕ + sin 2 ϕ) = ρ 2 (cos 4 ϕ + sin 4 ϕ). V`ı cos 4 ϕ + sin 4 ϕ  2nˆen lim x→0 y→0 x 4 + y 4 x 2 + y 2 = lim ρ→0 ρ 2 (cos 4 ϕ + sin 4 ϕ)=0. 56. lim x→0 y→2  1+xy  2 x 2 + xy , 2) lim x→0 y→2  x 2 +(y − x) 2 +1− 1 x 2 +(y − 2) 2 , 3) lim x→0 y→0 x 4 + y 4 x 2 + y 2 . 7 .4. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am nhiˆe ` ubiˆe ´ n 55 Gia ’ i. 1) Ta biˆe ’ udiˆe ˜ n. doa . nkiˆe ’ uII;diˆe ’ m x 0 = −1l`adiˆe ’ m gi´an doa . nkiˆe ’ uI,d = 4) 7 .4. Gi´o . iha . n v`a liˆen tu . ccu ’ a h`am nhiˆe ` ubiˆe ´ n 51 7 .4 Gi´o . iha . nv`aliˆen tu . ccu ’ ah`am nhiˆe ` u biˆe ´ n 1.