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1. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 10 tháng 06 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 BTVN NGÀY 10-06 Tính các giới hạn sau đây. 4 − x2 1− Bµi i 1:lm x→ 2 πx cos 4 s ns ns nx i i i 2 − Bµi2:lm i x→0 x 1− cosx cos2x 3− Bµi3:lm i x→0 x2 1− cosxcos2x.. .cos2010x 4 − Bµi4:lmi x→0 x2 l ( s nx + cosx) n i 5− Bµi5:lmi x→∞ x 3 esinx−sin x − cos2x 6 − Bµi6:lm i x→0 x2 x  x + 3 7 − Bµi7:lm  i  x→+∞ x + 1   8 − Bµi :lm 8 i x→+∞ ( 3 x3 + 3x2 − x2 − x + 1 ) t − s nx anx i 9 − Bµi9:lm i x→∞ x3 1+ x2 − cosx 10 − Bµi10:lm i x→0 x2 1+ tanx − 1+ s nx i 11− Bµi11:lm i x→0 x3 x3 + x2 − 2 12 − Bµi12:lm i x→∞ s n( − 1) i x ………………….Hết………………… BT Viên môn Toán hocmai.vn Trịnh Hào Quang Hocmai.vn – Ngôi trường chung của học trò Việt 1
2. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 28 tháng 02 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 HDG CÁC BTVN • BTVN NGÀY 09-06: *Bµi i 1:lm ( 1+ x) ( 1+ 2x) ( 1+ 3x) − 1 x→0 x = lm i ( 1+ x) ( 1+ 2x) ( 1+ 3x) − ( 1+ x) ( 1+ 2x) + ( 1+ x) ( 1+ 2x) − ( 1+ x) + ( 1+ x) − 1 x→0 x ( 1+ x) ( 1+ 2x) ( ( 1+ 3x) − 1) + ( 1+ x) ( ( 1+ 2x) − 1) + x = lm i x→0 x 3x( 1+ x) ( 1+ 2x) + 2x( 1+ x) + x 3( 1+ x) ( 1+ 2x) + 2( 1+ x) + 1 = lm i = lm i = 1+ 2 + 3 = 6 x→0 x x→0 1 xm − 1 *Bµi2:lm i x→1 xn − 1 = lm i ( ( x − 1) xm −1 + xm −2 + ..+ x + 1 . ) = lm ( x i m −1 + xm −2 + .. x + 1 .+ ) =m x→1 ( x − 1) ( x n−1 + xn−2 + .. x + 1) .+ (x x→1 n−1 + xn−2 + .. x + 1) .+ n x100 − 2x + 1 *Bµi3:lm i x→1 x50 − 2x + 1 = lm i ( ) x100 − 1 − 2( − 1) x = lm i ( x − 1) ( x 99 + x98 + .. x + 1− 2 .+ ) = 98 = 49 x→1 (x − 1) − 2( − 1) 50 x ( x − 1) ( x x→1 49 + x48 + .. x + 1− 2) 48 24 .+ ( x − x − 2) 20 2 *Bµi4:lm i ( x − 12x + 16) x→2 10 3 ( x − 2) ( x − 2) ( x − 2) 20 20 20 ( + 1) ( + 1) ( + 1) 20 20 20 10 x x x  3 = lm i = lm i = lm i =  ( (x − 2) (x + 4)) ( (x − 2) (x + 4)) x→ 2 ( − 2) ( + 4) 10 10 20 10 x→2 2 x→2 2 x x  2 x + 9 + x + 16 − 7 *Bµi5:lm i x→0 x x + 9 − 3+ x + 16 − 4 x x = lm i = lmi + lm i x→0 x x→0 x ( x+ 9 + 3 ) x→0 x ( x + 16 + 4 ) 1 1 1 1 7 = lm i + lm i = + = x→0 ( x+ 9 + 3 ) x→0 ( x + 16 + 4 ) 6 8 24 Page 2 of 8
3. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 28 tháng 02 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 2 1+ x − 3 8 − x *Bµi6:lm i x→0 x = lm i 2 ( 1+ x − 1 − ) ( 3 8− x − 2 ) = lim 2( 1+ x − 1 ) − lim ( 3 8− x − 2 ) x→0 x x→0 x x x→0 2x −x 1 13 = lm i − lm i = 1+ = x→0 x (   ) 1+ x + 1 x→0 x 3 ( 8 − x) 2 + 23 8 − x + 4   12 12 2x + 1 − 3 1+ 3x *Bµi7:lm i x→0 x2 = lm i ( 2x + 1 − ( + 1) − x ) ( 3 1+ 3x − ( + 1) x ) = lim ( 2x + 1− (x + 1) ) 2 x→0 x2 x ( 2x + 1 + ( + 1) x→0 2 x ) − lm i ( 1+ 3x − (x + 1) ) 3 = lm i −x2 x→0 x2  3 ( 1+ 3x) + ( + 1)3 1+ 3x + ( + 1)    2 x x 2   x→0 2 x 2x + 1 + ( + 1) x ( ) −x2 ( + 3) x 1 3 − lm i = − −1= − x→0 2  x  3 ( 1+ 3x) + ( + 1)3 1+ 3x + ( + 1)  2 2 2 2 x x    = 1+ 4x. 1+ 6x. 1+ 8x. 1+ 10x − 1 3 4 5 *Bµi i 8:lm x→0 x 1+ 4x. 1+ 6x. 1+ 8x. 1+ 10x − 1+ 4x. 1+ 6x. 1+ 8x 3 4 5 3 4 lm i x→0 x 1+ 4x. 1+ 6x. 1+ 8x − 1+ 4x. 1+ 6x. 1+ 4x. 1+ 6x − 1+ 4x + 1+ 4x − 1 3 4 3 + 3 + lm i x→0 x = lm i 1+ 4x. 1+ 6x. 1+ 8x. 5 1+ 10x − 1 3 4 ( ) + lim ( 1+ 4x. 1+ 6x. 4 1+ 8x − 1 3 ) x→0 x x→0 x + lm i 1+ 4x ( 3 1+ 6x − 1 ) + lim 1+ 4x − 1 x→0 x x→0 x 1+ 4x − 1 4x 4 X Ðt:I = lm i = lm i = =2 2 x→0 x x→0 x ( 1+ 4x + 1 ) 2 n 1+ 2nx − 1 Còng nh­ vËy t cã:I = lm a n i = 2 ⇒ I= I + I + I + I = 8 5 4 3 2 x→0 x Page 3 of 8
4. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 28 tháng 02 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 2x + 1 − 3 x2 + 1 *Bµi9:lm i x→0 s nx i = lm i ( 2x + 1 − 1 − ) ( 3 x2 + 1 − 1 ) = lim ( 2x + 1 − 1 ) − lim ( 3 ) x2 + 1 − 1 x→0 s nx i x→0 s nx i x→0 s nx i 2 x = lm i − lm i = 2− 0 = 2 x→0 ( 2x + 1 + 1 s nx i x ) x→0 3 2   ( 2 ) 2 (  s nx x + 1 + 3 x2 + 1 + 1  x i ) x + 2 − 3 x + 20 *Bµi10:lm i x→7 4 x+ 9 − 2 t − 7 − 3 t + 11 4 4 § Ætt= x + 9 ⇒ x = t − 9 ⇒ I= lm 4 i 4 → t 2 t− 2 t −7−3 4 t + 11 − 3 3 4 t − 16 4 = lm i − lm i = lm i → t 2 t− 2 → t 2 t− 2 → t 2 ( t− 2) t4 − 7 + 3 ( ) − lm i t − 16 4 = lm i ( t + 4) ( t+ 2) 2 → t 2 ( t− 2)  3 ( t4 + 11)   2  + 33 t + 11 + 9 4  → t 2 ( t −7+3 4 ) − lm i ( t + 4) ( t+ 2) 2 16 32 176 = + = 3  3 27 27 ( t + 11) + 3 t + 11 + 9 → t 2 2 4 3 4    1+ 4x − 3 1+ 6x *Bµi11:lm i x→0 x2 1+ 4x − ( + 2x) 1 3 1+ 6x − ( + 2x) 1 1+ 4x − ( + 2x) 1 2 = lm i − lm i = lm i x→0 x2 x→0 x2 x 1+ 4x + ( + 2x) x→0 2 1 ( ) 1+ 6x − ( + 2x) 1 3 −x2 − lm i = lm i x→0  3 1+ 6x 2 + ( + 2x)3 1+ 6x + ( + 2x)  x→0 x2 1+ 4x + ( + 2x) x  (  2 ) 1 1 2   1 ( ) −4x ( + 2x) 2 3 1 12 7 − lm i =− + = x→0 2  2 2 3 2 x  3 ( 1+ 6x) + ( + 2x)3 1+ 6x + ( + 2x)  2 1 1   • BTVN NGÀY 10-06: Page 4 of 8
5. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 28 tháng 02 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 4 − x2 1− Bµi i 1:lm x→2 πx cos 4 t t+ 4) ( t t+ 4) ( § Æt:t= x − 2 ⇒ x = t+ 2 ⇒ I= lm i = − lm i → t 0  πt 1  → t 0 πt cos +  sn i  4 2 4 t t+ 4) ( t t+ 4) ( ( + 4) 16 t = − lm i = − lm i = − lm i =− → t 0 πt → t 0 πt → t 0 π π sn i . . πt . 4 4 4 4 πt 4 s ns ns nx i i i 2 − Bµi2:lm i x→0 x  s n( s ns nx) s ns nx s nx  i i i i i i = lm  i . .  =1  s ns nx i i s nx i x  x→0 1− cosx cos2x 3− Bµi3:lm i x→0 x2 = lm i 1− cosx + cosx − cosx cos2x = lm i 1− cosx + lm i ( cosx 1− cos2x ) x→0 x2 x→0 x2 x→0 x2 x 2s n2 2 + lm cosx( 1− cos2x) = 1 + lm 2cosx. i x = 1 + 1 = 3 i s n2 = lm i i i x→0  x 4.   2 x→0 ( x2 1+ cos2x ) ( 2 x→0 1+ cos2x x2 2 ) 2  2 1− cosxcos2x.. .cos2010x 4 − Bµi4:lm i x→0 x2 1− cosx + cosx − cosxcos2x + .. cosxcos2x.. .+ .cos2010x = lm i 2 x→0 x 1− cosx cosx.1− cos2x) ( = lm i 2 + lm i + .. I . + 2010 x→0 x x→0 x2 nx 2s n2 i 1− cosnx 2 2 = n ⇒ I= I + I + .. I = 1 1+ 22 + 32 + .. 20102 X ÐtI = n x2 = 2 2 1 2 . + 2010 2 (.+ ) 4  nx  . n2  2    Page 5 of 8
6. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 28 tháng 02 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 2010( 2010 + 1) 2. ( 2010 + 1) = 12 l ( s nx + cosx) n i 5− Bµi5:lm i x→∞ x l ( s nx + cosx) l ( s nx + cosx) s n2x 2 n i n i i = lm i = lm i . x→0 2x x→0 s n2x i 2x l ( s nx + cosx) n i l ( 1+ t n ) = 1 V íit= si vµ lm si = 1 n2x M µ:lmi = lm i n2x i x→0 s n2x i → t 0 t x→0 2x ⇒ I= 1. = 1 1 ecosx−cos3x − cos2x 6 − Bµi6:lm i x→0 x2  ecosx−cos3x − 1 1− cos  2x = lm  i 2 + 2  x→0  x x  ecosx−cos3x − 1  ecosx−cos3x − 1 cosx − cos3x  *) cã:lm Ta i = lm  i .  x→0 x2 x→0 cosx − cos3x x2   ecosx−cos3x − 1  1− cos3x 1− cosx  ecosx−cos3x − 1 et − 1 = lm i  − .D o lm i = lm i =1 x→0 cosx − cos3x  x2 x2   x→0 cosx − cos3x → t 0 t  1− cos3x 1− cosx  3 1 2 2 lm  i − = − =4 x→0  x2 x2  2 2  1− cos2x *) Ætkh¸c:lm M i 2 = 2 ⇒ I= 4 + 2 = 6 x→0 x x  x + 3 7 − Bµi7:lm  i  x→+∞ x + 1   x  2  2 1 = lm  1+ i  .§ Æt: = ⇒ x = 2t− 1; → +∞ ⇒ t→ +∞ x x→+∞  x + 1 x+1 t − 2t 1 2t −1  1  1  1 ⇒ I= lm  1+  i = lm  1+  .lm  1+  = e2 i i t→+∞  t t→+∞  t t→+∞  t Page 6 of 8
7. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 28 tháng 02 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 8 − Bµi i 8:lm x→+∞ ( 3 x3 + 3x2 − x2 − x + 1 ) = lm i x→+∞ (( 3 ) ( x − x + 1 − x) ) = A − B x3 + 3x2 − x − 2 *) = lm ( x + 3x − x) = lm 2 3 3 2 3x A i i ( ) x→+∞ x→+∞ 2 x + 3x + x x + 3x + x 3 3 2 3 3 2 2 3 = lm i =1 x→+∞ 2  3 3 3 3  1+  + 1+ + 1  x x 1 −1+ B = lm i ( x2 − x + 1 − x = lm i ) −x + 1 = lm i x =− 1 x→+∞ x→+∞ ( x − x +1+ x 2 ) x→+∞    1 1  1− + 2 + 1 x x  2   t − s nx anx i 9 − Bµi9:lm i x→0 x3  1  si nx x si  nx − 1 ( − cosx) 1 2s n2 i  cosx  = lm x 2 1 = lm i i = lm i 2 = x→0 x3 x→0 x2.cosx x→0  x 2 4.  .  2 cosx   1+ x2 − cosx 10 − Bµi10:lm i x→0 x2     −2s n2 x  i 1+ x2 − 1 cosx − 1  1   2 = 1+ 1 =1 = lm  i −  = lm  i  − l→0  i m x→0   x→0  1+ x2 + 1 x  x  2 2 2 2 2  x x   4.      2  Page 7 of 8
8. TRUNG TÂM HOCMAI.ONLINE Hà Nội, ngày 28 tháng 02 năm 2010 P.2512 – 34T – Hoàng Đạo Thúy Tel: (094)-2222-408 1+ t anx − 1+ s nx i 11− Bµi11:lm i 3 x→0 x t − s nx anx i si 1− cosx) nx( = lm i = lmi x→0 x3 ( 1+ t anx + 1+ s nx i ) x→0 3 x ( 1+ t anx + 1+ s nx cosx i ) x s n2 i si nx 2 . 2 2 x  x 4.  2   1 1 1 = lm i = . = x→0 ( 1+ t i ) anx + 1+ s nx cosx 2 2 4 x3 + x2 − 2 12 − Bµi12:lm i x→1 s n( − 1) i x x3 − 1+ x2 − 1 ( − 1) x2 + x + 1)+ ( x − 1) ( + 1) x ( x ( 2 + x + 1)+ ( + 1) x x = lm i = lm i = lm i x→1 s n( − 1) i x x→1 s n( − 1) i x x→1 s n( − 1) i x x−1 s n( − 1) i x D olm i = 1⇒ I= 5 x→1 x−1 ………………….Hết………………… BT Viên môn Toán hocmai.vn Trịnh Hào Quang Page 8 of 8
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