1 dt f x dx 1 1 dt 2 1 d 4 2 t 2 tg Do đó lim S n 2 n n 5 n 5 5 5 n 1 sin 1 sin

1 dt f x dx 1 1 dt 2 1 d 4 2 t 2 tg do đó lim s n 2

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1 dt f x dx 1 1 dt 2 1 d 4 2 t 2 tg . Do đó: lim S n = 2 n n 5 n 5 5 5 n 1 + sin 1 + sin 2 1 + 2 2 n sin n n 1 ( ) 1 = = = = . 0 0 1 + 2 0 cos 2 cos 2 4 t 2 . 0 = t 4 t 4
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n 2.5.5. Tính gi i h n d a vào vi c gi i ph ng trình sai ươ phân Bài toán 1 Cho dãy s u n xác đ nh b i:
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 2 ( n + 2 ) 2 u n + 2 ( n + 1 )( n + 3 ) u n + 1 + ( n + 1 )( n + 2 ) u n = n ( n + 3 ) 2 ( n + 2 ) n + 3 , ∀ n > 2 u 1 = 0 u 2 = − 3 Ch ng minh r ng: lim u n = 1 n 2 L i gi i Tr c h t ta tìm s h ng t ng quát ướ ế c a dãy ( u n ) b ng cách gi i ph ng sai phân: ươ 2 ( n + 2 ) 2 u n + 2 ( n + 1 ) ( n + 3 ) u n + 1 + ( n + 1 )( n + 2 ) u n = n ( n + 3 ) 2 ( n + 2 ) n + 3 , ∀ n > 2 (∗) v i đ i u k i n b a n đ u : u 1 = 0 60
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v à u 2 = 3 . N h â n c h a i v ế c a ( * ) v i n + 3 t a đ ư c : n ( 2 ( n + 2 ) u n + n + 1 n + 1 u n = 2 , n > 2 (∗ ∗) Đ t v n = n + 1 u n . Khi đó ta đ c ượ ph n ươ g trình: v n 2 2 v n 1 + n = 2 , n > 2 (∗ ∗ ∗) v 1 = 0; v 2 = 9 / 2. Ta có ph n ươ g trình đ c tr ng: ư t 2 2 t + 1 = 0 nghi m kép t=1. ta gi i đ c: ượ v n = 13 15 n + n 2 . Suy ra: u n v 12 n 1 3 1 2 1 5 n n 2 Σ 1 3 n 1 5 n 2 n 3 Do đó: l i m u n = 1 n 2 2.5.6. S d ng dãy ph đ tính gi i h n * S d ng dãy con đ xét s h i t c a dãy C s c a ơ ở ph ng pháp ươ là đ nh lý: M i dãy con c a dãy h i t đ u h i t ng c l i : ượ N u m i dãy con ế c a dãy ( a n ) đ u h i t thì chúng ph i h i t đ n ế cùng m t gi i h n a và s a đó cũng chính là gi i h n c a dãy ( a n ) . Nh v y: ư N u ế limx 2 n = limx 2 n + 1 = a thì limx n = a . T ng quát: Cho s nguyên m 2, n u: ế limx mn + i = a , m i i=0,1,2,...,m-1 61 + n n + + . n = n + n = 12 n + + = 2 ( n + 1 ) 2 ( n + 1 ) + n
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thì limx n = a . Bài toán 1. Cho dãy s ( x n ) xác đ nh b i công th c: x 0 = x 1 = 1 3 x n + 2 = x n + x n + 1 Ch ng minh r ng dãy ( x n ) h i t . L i gi i Xét dãy s ( a n ) đ c ượ xác đ nh b i: a 0 = 1 , a n + 2 a n 1 = 3 , d th y ( a n ) là dãy gi m d n v 0. Ta ch ng t m ax { x 2 n , x 2 n + 1 } ≤ a n , v i m i n (1). Th t v y, (1) đúng v i n=0 và n=1. Gi s (1) đúng ả ử v i n và chú ý r ng ( a n ) dãy s gi m nên ta có: 3 x 2 n + 2 = x 2 n + x 2 n + 1 2 a n suy ra x 2 n + 2 a n + 1 3 x 2 n + 3 = x 2 n + 1 + x 2 n + 2 a n + a n + 1 2 a n suy ra x 2 n + 3 a n + 1 . Nh v y (1) đúng v i n+1, theo nguyên lí quy n p thì (1) đ c ch ng minh. D ư ượ th y x n > 0 v i m i n , và t (1) theo nguyên lý k p có limx 2 n = lim 2 n + 1 = 0 suy ra limx n = 0.
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