Bài tập toán cao cấp part 6 doc

8.2. Vi phˆan 79 C ´ AC V ´ IDU . V´ı du . 1. T´ınh vi phˆan df nˆe ´ u 1) f(x) = ln(arctg(sin x)); 2) f(x)=x √ 64 −x 2 +64arcsin x 8 . Gia ’ i. 1) ´ Ap du . ng c´ac t´ınh chˆa ´ tcu ’ a vi phˆan ta c´o df = d[arctg(sin x)] arctg(sin x) = d(sin x) (1 + sin 2 x)arctg(sin x) = cos xdx (1 + sin 2 x)arctg(sin x) · 2) df = d[x √ 64 − x 2 ]+d  64arcsin x 8  = xd √ 64 − x 2 + √ 64 −x 2 dx +64d  arcsin x 8  = x d(64 − x 2 ) 2 √ 64 − x 2 + √ 64 −x 2 dx +64· d  x 8   1 − x 2 64 = −x 2 dx √ 64 −x 2 + √ 64 − x 2 dx +64 dx √ 64 − x 2 =2 √ 64 − x 2 dx, |x| < 8.  V´ı du . 2. T´ınh vi phˆan cˆa ´ p2cu ’ a c´ac h`am 1) f(x)=xe −x ,nˆe ´ u x l`a biˆe ´ ndˆo . clˆa . p; 2) f(x)=sinx 2 nˆe ´ u a) x l`a biˆe ´ ndˆo . clˆa . p, b) x l`a h`am cu ’ amˆo . tbiˆe ´ ndˆo . clˆa . pn`aod´o . Gia ’ i. 1) Phu . o . ng ph´ap I. Theo d i . nh ngh˜ıa vi phˆan cˆa ´ p 2 ta c´o d 2 f = d[df ]=d[xde −x + e −x dx] = d(−xe −x dx + e −x dx)=−d(xe −x )dx + d(e −x )dx = −(xde −x + e −x dx)dx −e −x dx 2 = xe −x dx 2 − e −x dx 2 −e −x dx 2 =(x − 2)e −x dx 2 . 80 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n Phu . o . ng ph´ap II.T´ınh d a . o h`am cˆa ´ p hai f  (x) ta c´o f  (x)=(xe −x )  =(e −x − xe −x )  = −e −x − e −x + xe −x =(x − 2)e −x v`a theo cˆong th´u . c (8.6) ta c´o d 2 f =(x − 2)e −x dx 2 . 2) a) Phu . o . ng ph´ap I. Theo di . nh ngh˜ıa vi phˆan cˆa ´ p hai ta c´o d 2 f = d[d sin x 2 ]=d[2x cos x 2 dx]=d[2x cos x 2 ]dx =  2 cos x 2 dx +2x(−sin x 2 )2xdx  dx = (2 cos x 2 − 4x 2 sin x 2 )dx 2 . Phu . o . ng ph´ap II.T´ınh da . o h`am cˆa ´ p hai f  xx ta c´o f  x =2x cos x 2 ,f  xx = 2 cos x 2 − 4x 2 sin x 2 v`a theo (8.6) ta thu du . o . . c d 2 f = (2 cos x 2 − 4x 2 sin x 2 )dx 2 . b) Nˆe ´ u x l`a biˆe ´ n trung gian th`ı n´oi chung d 2 x =0v`adod´o ta c´o d 2 f = d(2x cos x 2 dx)=(2x cos x 2 )d 2 x +[d(2x cos x 2 )]dx =2x cos x 2 d 2 x + (2 cos x 2 − 4x 2 sin x 2 )dx 2 .  V´ı d u . 3. ´ Ap du . ng vi phˆan d ˆe ’ t´ınh gˆa ` nd´ung c´ac gi´a tri . : 1) 5  2 −0, 15 2+0, 15 ; 2) arcsin 0, 51; 3) sin 29 ◦ . Gia ’ i. Cˆong th´u . cco . ba ’ nd ˆe ’ ´u . ng du . ng vi phˆan d ˆe ’ t´ınh gˆa ` nd´ung l`a ∆f(x 0 ) ≈ df (x 0 ) ⇒ f(x 0 +∆x) − f(x 0 ) ≈ f  (x 0 )∆x ⇒ f(x 0 +∆x) ≈ f(x 0 )+f  (x 0 )∆x 8.2. Vi phˆan 81 T`u . d ´o , d ˆe ’ t´ınh gˆa ` nd´ung c´ac gi´a tri . ta cˆa ` n thu . . chiˆe . nnhu . sau: 1 + Chı ’ ra biˆe ’ uth´u . c gia ’ it´ıchd ˆo ´ iv´o . i h`am m`a gi´a tri . gˆa ` nd ´ung cu ’ a n´o cˆa ` n pha ’ i t´ınh. 2 + Cho . ndiˆe ’ m M 0 (x 0 ) sao cho gi´a tri . cu ’ a h`am v`a cu ’ ada . o h`am cˆa ´ p 1cu ’ a n´o ta . idiˆe ’ mˆa ´ y c´o thˆe ’ t´ınh m`a khˆong d `ung ba ’ ng. 3 + Tiˆe ´ pdˆe ´ n l`a ´ap du . ng cˆong th´u . cv`u . a nˆeu. 1) T´ınh gˆa ` nd ´ung 5  2 −0, 15 2+0, 15 Sˆo ´ d ˜a cho l`a gi´a tri . cu ’ a h`am y = 5  2 − x 2+x ta . idiˆe ’ m x =0, 15. Ta d˘a . t x 0 =0;∆x =0, 15. Ta c´o y  = −4 5  2 −x 2+x 5(4 − x 2 ) = − 4y 5(4 −x 2 ) ⇒ y  (x 0 )=y  (0) = − 1 5 · Do d ´ov`ıy(0) = 1 nˆen y(0, 15) ≈ y(0) + y  (0)∆x =1− 1 5 · (0, 15) = 1 −0, 03 = 0, 97. 2) T´ınh gˆa ` nd ´ung arcsin 0, 51. X´et h`am y = arcsin x.Sˆo ´ cˆa ` nt´ınh l`a gi´a tri . cu ’ a h`am ta . idiˆe ’ m 0, 51; t´u . cl`ay(0, 51). D˘a . t x 0 =0, 5; ∆x =0, 01. Khi d´o ta c´o arcsin(x 0 +∆x ≈ arcsinx 0 + (arcsinx)  x=x 0 ∆x ⇒ arcsin(0, 5+0, 01) ≈ arcsin0, 5 + (arcsinx)    x=0,5 · 0, 01 = π 6 + 1  1 −(0, 5) 2 × (0, 01). 82 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n C´o thˆe ’ t´ınh gˆa ` nd´ung  1 −(0, 5) 2 = √ 0, 75 ≈ 0, 88 v`a do d´o arcsin0, 51 ≈ π 6 +0, 011 ≈ 0, 513. 3) Sˆo ´ sin 29 ◦ l`a gi´a tri . cu ’ a h`am y = sin x khi x = π 180 ×29. Ta d˘a . t x 0 = π 180 ×30 = π 6 ; y  π 6  = 1 2 ,y= cos x ⇒ y   π 6  = cos π 6 = √ 3 2 · D ˘a . t∆x = x − x 0 = 29π 180 − π 6 = − π 180 .Dod´o sin 29 ◦ ≈ y  π 6  + y   π 6  · ∆x = 1 2 + √ 3 2  − π 180  ≈ 0, 48.  B ` AI T ˆ A . P T´ınh vi phˆan df nˆe ´ u: 1. f(x) = arctg 1 x .(DS. df = −dx 1+x 2 ) 2. f(x)=2 tg 2 x .(DS. 2 tg 2 x ln2 ·2tgx · dx cos 2 x ) 3. f(x) = arccos(2 x ). (DS. − 2 x ln2dx √ 1 −e 2x ) 4. f(x)=x 3 lnx.(DS. x 2 (1 + 3lnx) dx) 5. f(x) = cos 2 ( √ x). (DS. −2 cos √ x ·sin √ x · dx 2 √ x ) 6. f(x)=(1+x 2 )arcotgx.(DS. (2xarccotgx − 1)dx) 7. f(x)= arctgx √ 1+x 2 .(DS. 1 − xarctgx (1 + x 2 ) 3/2 dx) 8. f(x) = sin 3 2x.(DS. 3 sin 2x sin 4xdx) 9. f(x) = ln(sin √ x). (DS. cotg √ x 2 √ x dx) 8.2. Vi phˆan 83 10. f(x)=e − 1 cos x .(DS. −tgx ·e − 1 cos x cos x dx) 11. f(x)=2 −x 2 .(DS. −2xe −x 2 ln2dx) 12. f(x) = arctg √ x 2 + 1. (DS. 2xdx 2+x 2 ) 13. f(x)= √ xarctg √ x.(DS. 1 2 √ x  arctg √ x + √ x 1+x  dx) 14. f(x)= x 2 arcsinx .(DS. x  2arcsinx − x √ 1 −x 2  (arcsinx) 2 dx). T´ınh vi phˆan cˆa ´ ptu . o . ng ´u . ng cu ’ a c´ac h`am sau 15. f(x)=4 −x 2 ; d 2 f ?(DS. 4 −x 2 2ln4(2x 2 ln4 −1)(dx) 2 ) 16. f(x)=  ln 2 x − 4. d 2 f ?(DS. 4lnx −4 −ln 3 x x 2  (lnx − 4) 3 (dx) 2 ) 17. f(x) = sin 2 x. d 3 f ?(DS. −4 sin 2x(dx) 3 ) 18. f(x)= √ x −1, d 4 f ?(DS. −15 16(x −1) 7/2 (dx) 4 ) 19. f(x)=xlnx, d 5 f ?(DS. − 6 x 4 (dx) 5 , x>0) 20. f(x)=x sin x; d 10 f ?(DS. (10 cos x − x sin x)(dx) 10 ) Su . ’ du . ng cˆong th´u . cgˆa ` nd ´ung ∆f ≈ df (khi f  (x) =0)dˆe ’ t´ınh gˆa ` nd´ung c´ac gi´a tri . sau 21. y = √ 3, 98. (DS. 1,955) 22. y = 3 √ 26, 19. (DS. 2,97) 23. y =  (2, 037) 2 − 3 (2, 037) 2 +5 .(DS. 0,35) 24. y = cos 31 ◦ .(DS. 0,85) 84 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n 25. y = tg45 ◦ 10  .(DS. 0,99) 26. y = ln(10, 21). (DS. 1,009) 27. y = sin 31 ◦ .(DS. 0,51) 28. y = arcsin0, 54. (D S. 0,57) 29. y = arctg(1, 05). (DS. 0,81) 30. y =(1, 03) 5 .(DS. 1,15) 8.3 C´ac di . nh l´yco . ba ’ nvˆe ` h`am kha ’ vi. Quy t˘a ´ c l’Hospital. Cˆong th´u . cTay- lor 8.3.1 C´ac di . nh l´yco . ba ’ nvˆe ` h`am kha ’ vi D - i . nh l´y Rˆon (Rolle). Gia ’ su . ’ : i) f(x) liˆen tu . ctrˆen doa . n [a, b]. ii) f(x) c´o da . o h`am h˜u . uha . n trong (a, b). iii) f(a)=f(b). Khi d ´otˆo ` nta . idiˆe ’ m ξ : a<ξ<bsao cho f(ξ)=0. D - i . nh l´y Lagr˘ang (Lagrange). Gia ’ su . ’ : i) f(x) liˆen tu . ctrˆen d oa . n [a, b]. ii) f(x) c´o da . o h`am h˜u . uha . n trong (a, b). Khi d´ot`ımdu . o . . c ´ıt nhˆa ´ tmˆo . tdiˆe ’ m ξ ∈ (a, b) sao cho f(b) −f(a) b −a = f  (ξ) (8.12) hay l`a f(b)=f(a)+f  (ξ)(b −a). (8.13) Cˆong th´u . c (8.12) go . i l`a cˆong th´u . csˆo ´ gia h˜u . uha . n. 8.3. C´ac di . nh l´y co . ba ’ nvˆe ` h`am kha ’ vi 85 D - i . nh l´y Cˆosi (Cauchy). Gia ’ su . ’ : i) f(x) v`a ϕ(x) liˆen tu . ctrˆen d oa . n [a, b]. ii) f(x) v`a ϕ(x) c´o d a . o h`am h˜u . uha . n trong (a, b). iii) [f  (x)] 2 +[ϕ  (x)] 2 =0, ngh˜ıa l`a c´ac da . o h`am khˆong dˆo ` ng th`o . i b˘a ` ng 0. iv) ϕ(a) = ϕ(b). Khi d ´ot`ımdu . o . . cdiˆe ’ m ξ ∈ (a, b) sao cho: f(b) − f(a) ϕ(b) − ϕ(a) = f  (ξ) ϕ  (ξ) · (8.14) D i . nh l´y Lagrange l`a tru . `o . ng ho . . p riˆeng cu ’ adi . nh l´y Cauchy v`ı khi ϕ(x)=x th`ı t`u . (8.14) thu du . o . . c (8.13). Di . nh l´y Rˆon c˜ung l`a tru . `o . ng ho . . p riˆeng cu ’ adi . nh l´y Lagrange v´o . idiˆe ` ukiˆe . n f(a)=f(b). C ´ AC V ´ IDU . V´ı du . 1. Gia ’ su . ’ P ( x)=(x + 3)(x + 2)(x − 1). Ch´u . ng minh r˘a ` ng trong khoa ’ ng (−3, 1) tˆo ` nta . i nghiˆe . mcu ’ aphu . o . ng tr`ınh P  (ξ)=0. Gia ’ i. D ath´u . c P(x) c´o nghiˆe . mta . ic´acdiˆe ’ m x 1 = −3, x 2 = −2, x 3 = 1. Trong c´ac khoa ’ ng (−3, −2) v`a (−2, 1) h`am P (x) kha ’ vi v`a tho ’ a m˜an c´ac diˆe ` ukiˆe . ncu ’ adi . nh l´y Rˆon v`a: P ( −3) = P(−2)=0, P ( −2) = P(1) = 0. Do d ´o theo di . nh l´y Rˆon, t`ım du . o . . cd iˆe ’ m ξ 1 ∈ (−3, −2); ξ 2 ∈ (−2, 1) sao cho: P  (ξ 1 )=P  (ξ 2 )=0. Bˆay gi`o . la . i ´ap du . ng d i . nh l´y Rˆon cho doa . n[ξ 1 ,ξ 2 ] v`a h`am P  (x), ta la . it`ımdu . o . . cdiˆe ’ m ξ ∈ (ξ 1 ,ξ 2 ) ⊂ (−3, 1) sao cho P  (ξ)=0. 86 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n V´ı du . 2. H˜ay x´et xem h`am f(x) = arcsinx trˆen doa . n[−1, +1] c´o tho ’ a m˜an di . nh l´y Lagrange khˆong ? Nˆe ´ u tho ’ a m˜an th`ı h˜ay t`ım diˆe ’ m ξ (xem (8.12)). Gia ’ i. H`am f(x) x´ac d i . nh v`a liˆen tu . ctrˆen[−1, +1]. Ta t`ım f  (x). f  (x)= 1 √ 1 −x 2 → f  (x) < ∞,x∈ (−1, 1) (Lu . u´yr˘a ` ng khi x = ±1da . o h`am khˆong tˆo ` nta . inhu . ng diˆe ’ ud´o khˆong a ’ nh hu . o . ’ ng dˆe ´ nsu . . tho ’ am˜andiˆe ` ukiˆe . ncu ’ adi . nh l´y Lagrange !). Nhu . vˆa . y h`am f tho ’ am˜andi . nh l´y Lagrange. Tat`ımd iˆe ’ m ξ. Ta c´o: arcsin1 − arcsin( −1) 1 − (−1) = 1  1 −ξ 2 ⇒ π 2 −  − π 2  2 = 1  1 −ξ 2 ⇒  1 −ξ 2 = 2 π ⇒ ξ 1,2 = ±  1 − 4 π 2 Nhu . vˆa . y trong tru . `o . ng ho . . p n`ay cˆong th´u . c (8.12) tho ’ a m˜an dˆo ´ iv´o . i hai d iˆe ’ m. V´ı du . 3. H˜ay kha ’ o s´at xem c´ac h`am f(x)=x 2 − 2x +3v`a ϕ(x)= x 3 − 7x 2 +20x − 5 c´o tho ’ a m˜an diˆe ` ukiˆe . ndi . nh l´y Cauchy trˆen doa . n [1, 4] khˆong ? Nˆe ´ uch´ung tho ’ a m˜an d i . nh l´y Cauchy th`ı h˜ay t`ım diˆe ’ m ξ. Gia ’ i. i) Hiˆe ’ n nhiˆen ca ’ f(x)v`aϕ(x)liˆen tu . c khi x ∈ [1, 4]. ii) f(x)v`aϕ(x)c´od a . o h`am h˜u . uha . n trong (1, 4). iii) D iˆe ` ukiˆe . nth´u . iii) c˜ung tho ’ a m˜an v`ı: g  (x)=3x 2 − 14x +20> 0,x∈ R. iv) Hiˆe ’ n nhiˆen ϕ(1) = ϕ(4). 8.3. C´ac di . nh l´y co . ba ’ nvˆe ` h`am kha ’ vi 87 Do d´o f(x)v`aϕ(x) tho ’ a m˜an di . nh l´y Cauchy v`a ta c´o f(4) −f(1) ϕ(4) −ϕ(1) = f  (ξ) ϕ  (ξ) hay 11 − 2 27 − 9 = 2ξ −2 3ξ 2 − 14ξ +20 ,ξ∈ (1, 4). T`u . d ´othudu . o . . c ξ 1 =2,ξ 2 =4v`ao . ’ d ˆay chı ’ c´o ξ 1 =2l`adiˆe ’ m trong cu ’ a(1, 4). Do d´o: ξ =2. V´ı d u . 4. Di . nh l´y Cauchy c´o ´ap du . ng du . o . . c cho c´ac h`am f(x)=cosx, ϕ(x)=x 3 trˆen doa . n[−π/2,π/2] hay khˆong ? Gia ’ i. Hiˆe ’ n nhiˆen f(x)v`aϕ(x) tho ’ a m˜an c´ac diˆe ` ukiˆe . n i), ii) v`a iv) cu ’ adi . nh l´y Cauchy. Tiˆe ´ p theo ta c´o: f  (x)=−sin x; ϕ  (x)=3x 2 v`a ta . i x = 0 ta c´o: f  (0) = −sin 0 = 0; ϕ  (0) = 0 v`a nhu . vˆa . y [ϕ  (0)] 2 +[f  (0)] 2 = 0. Do d´odiˆe ` ukiˆe . n iii) khˆong du . o . . c tho ’ a m˜an. Ta x´et vˆe ´ tr´ai cu ’ a (8.14): f(b) − f(a) ϕ(b) −ϕ(a) = cos(π/2) −cos(−π/2) (π/2) 3 −(−π/2) 3 =0. Bˆay gi`o . ta x´et vˆe ´ pha ’ icu ’ a (8.14). Ta c´o: f  (ξ) ϕ  (ξ) = − sin ξ 3ξ 2 · Nhu . ng d ˆo ´ iv´o . ivˆe ´ pha ’ i n`ay ta c´o: lim ξ→0  − sin ξ 3ξ 2  = lim ξ→0 sin ξ ξ · lim ξ→0  − 1 3ξ  = ∞. Diˆe ` ud´och´u . ng to ’ r˘a ` ng c´ac h`am d˜a cho khˆong tho ’ a m˜an di . nh l´y Cauchy. B ` AI T ˆ A . P 1. H`am y =1− 3 √ x 2 trˆen doa . n[−1, 1] c´o tho ’ am˜andiˆe ` ukiˆe . ncu ’ adi . nh l´y Rˆon khˆong ? Ta . i sao ? (Tra ’ l`o . i: Khˆong) 88 Chu . o . ng 8. Ph´ep t´ınh vi phˆan h`am mˆo . tbiˆe ´ n 2. H`am y =3x 2 − 5 c´o tho ’ am˜andi . nh l´y Lagrange trˆen doa . n[−2, 0] khˆong ? Nˆe ´ u n´o tho ’ a m˜an, h˜ay t`ım gi´a tri . trung gian ξ. (Tra ’ l`o . i: C´o) 3. Ch´u . ng minh r˘a ` ng h`am f(x)=x +1/x tho ’ a m˜an d i . nh l´y Lagrange trˆen d oa . n[1/2, 2]. T`ım ξ.(DS. ξ =1) 4. Ch´u . ng minh r˘a ` ng c´ac h`am f(x) = cos x, ϕ(x) = sin x tho ’ a m˜an d i . nh l´y Cauchy trˆen doa . n[0,π/2]. T`ım ξ ?(DS. ξ = π/4) 5. Ch´u . ng minh r˘a ` ng h`am f(x)=e x v`a ϕ(x)=x 2 /(1 + x 2 ) khˆong tho ’ a m˜an d i . nh l´y Cauchy trˆen doa . n[−3, 3]. 6. Trˆen du . `o . ng cong y = x 3 h˜ay t`ım diˆe ’ m m`a ta . id´otiˆe ´ p tuyˆe ´ nv´o . i du . `o . ng cong song song v´o . i dˆay cung nˆo ´ idiˆe ’ m A(−1, −1) v´o . i B(2, 8). (D S. M(1, 1)) Chı ’ dˆa ˜ n. Du . . a v`ao ´y ngh˜ıa h`ınh ho . ccu ’ a cˆong th´u . csˆo ´ gia h˜u . uha . n. 8.3.2 Khu . ’ c´ac da . ng vˆo d i . nh. Quy t˘a ´ c Lˆopitan (L’Hospitale) Trong chu . o . ng II ta d˜a d ˆe ` cˆa . pdˆe ´ nviˆe . ckhu . ’ c´ac da . ng vˆo di . nh. Bˆay gi`o . ta tr`ınh b`ay quy t˘a ´ c Lˆopitan - cˆong cu . co . ba ’ ndˆe ’ khu . ’ c´ac da . ng vˆo di . nh Da . ng vˆo di . nh 0/0 Gia ’ su . ’ hai h`am f(x)v`aϕ(x) tho ’ a m˜an c´ac diˆe ` ukiˆe . n i) lim x→a f(x) = 0; lim x→a ϕ(x)=0. ii) f(x)v`aϕ(x) kha ’ vi trong lˆan cˆa . n n`ao d´o c u ’ adiˆe ’ m x = a v`a ϕ  (x) = 0 trong lˆan cˆa . nd´o, c´o thˆe ’ tr `u . ra ch´ınh d iˆe ’ m x = a. iii) Tˆo ` nta . i gi´o . iha . n(h˜u . uha . n ho˘a . cvˆoc`ung) lim x→a f  (x) ϕ  (x) = k. [...]... phai l` diˆu kiˆn cˆn ` ` ’ a e e a ch´ o u ` ˆ BAI TAP ´ ’ ´ Ap dung quy t˘c L’Hospital dˆ t´nh gi´.i han: a e ı o 16 x4 − 16 (DS ) 1 lim 3 2 − 6x − 16 x→2 x + 5x 13 m xm − am (DS am−n ) 2 lim n n x→a x − a n ´ Chu.o.ng 8 Ph´p t´nh vi phˆn h`m mˆt biˆn e ı a a o e 94 3 4 5 6 e2x − 1 (DS 2) lim x→0 sin x a2 1 − cos ax (DS 2 ) lim x→0 1 + cos bx b x −x e − e − 2x (DS 2) lim x→0 x − sin x ln(1... o u ’ u u o ´ quy t˘c L’Hospital ta thu du.o.c a 2x 2 ln(1 + x2 ) 2x = lim 1x+ x = lim x lim x x→0 e − 1 − x x→0 e − 1 x→0 (e − 1)(1 + x2 ) 2 2 = lim x = = 2 2 ) + (ex − 1)2x x→0 e (1 + x 1 V´ du 6 T´ ı ınh lim tgx π 2 cos x x→ 2 ’ Giai Ta c´ vˆ dinh dang “∞0 ” Nhu.ng o o tgx 2 cos x 2ln tgx = e2 cos xln tgx = e 1/ cos x ´ ´ v` o sˆ m˜ cua l˜y th`.a ta thu du.o.c vˆ dinh dang “∞/∞” Ap dung... (DS −1) x→∞ x x (DS +∞) 11 lim x→∞ ln(1 + x) ln sin x (DS 1) ln sin 5x x−a cotg(x − a) (DS 1/a) 13 lim arcsin x→a a (DS 0) 14 lim (π − 2arctgx)lnx 12 lim x→+0 x→∞ 15 lim (a1/x − 1)x, a > 0 x→∞ πx 16 lim (2 − x)tg 2 x→1 (DS lna) (DS e2/π ) x 1 − (DS −1) x→1 lnx lnx 18 lim (x − x2ln(1 + 1/x)) (Ds 1/2) 17 lim x→∞ 1 − cotg2 x (DS 2/3) x→0 x2 x (DS e) 20 lim x1/ln(e −1) 19 lim x→0 . d[x √ 64 − x 2 ]+d  64 arcsin x 8  = xd √ 64 − x 2 + √ 64 −x 2 dx +64 d  arcsin x 8  = x d (64 − x 2 ) 2 √ 64 − x 2 + √ 64 −x 2 dx +64 · d  x 8   1 − x 2 64 = −x 2 dx √ 64 −x 2 + √ 64 − x 2 dx. d˘a . t x 0 = π 180 ×30 = π 6 ; y  π 6  = 1 2 ,y= cos x ⇒ y   π 6  = cos π 6 = √ 3 2 · D ˘a . t∆x = x − x 0 = 29π 180 − π 6 = − π 180 .Dod´o sin 29 ◦ ≈ y  π 6  + y   π 6  · ∆x = 1 2 + √ 3 2  − π 180  ≈. T ˆ A . P ´ Ap du . ng quy t˘a ´ c L’Hospital d ˆe ’ t´ınh gi´o . iha . n: 1. lim x→2 x 4 − 16 x 3 +5x 2 −6x − 16 .(DS. 16 13 ) 2. lim x→a x m − a m x n − a n .(DS. m n a m−n ) 94 Chu . o . ng 8. 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