This is a beta course, so its structure, chapters, and examples may continue to change.
Operation Rules for Sequence Limits
The operation rules for sequence limits are similar to those for function limits, but since sequences are defined only at discrete points, they have their own peculiarities.
Basic Operation Rules Relation to Function Limits A sequence can be viewed as a function defined on the set of positive integers, so a sequence limit is a special case of a function limit.
If lim x → ∞ f ( x ) = A \lim_{x \to \infty} f(x) = A lim x → ∞ f ( x ) = A lim n → ∞ f ( n ) = A \lim_{n \to \infty} f(n) = A lim n → ∞ f ( n ) = A
But the converse does not always hold: lim n → ∞ f ( n ) = A \lim_{n \to \infty} f(n) = A lim n → ∞ f ( n ) = A lim x → ∞ f ( x ) = A \lim_{x \to \infty} f(x) = A lim x → ∞ f ( x ) = A
Example
The function f ( x ) = sin ( π x ) f(x) = \sin(\pi x) f ( x ) = sin ( π x )
Sequence limit: lim n → ∞ sin ( n π ) = 0 \lim_{n \to \infty} \sin(n\pi) = 0 lim n → ∞ sin ( nπ ) = 0 sin ( n π ) = 0 \sin(n\pi) = 0 sin ( nπ ) = 0 n n n
Function limit: lim x → ∞ sin ( π x ) \lim_{x \to \infty} \sin(\pi x) lim x → ∞ sin ( π x ) [ − 1 , 1 ] [-1, 1] [ − 1 , 1 ]
Worked Examples Example 1
Compute lim n → ∞ 2 n + 1 n \lim_{n \to \infty} \frac{2n + 1}{n} lim n → ∞ n 2 n + 1
Solution : lim n → ∞ 2 n + 1 n = lim n → ∞ ( 2 + 1 n ) = 2 + 0 = 2 \lim_{n \to \infty} \frac{2n + 1}{n} = \lim_{n \to \infty} \left(2 + \frac{1}{n}\right) = 2 + 0 = 2 lim n → ∞ n 2 n + 1 = lim n → ∞ ( 2 + n 1 ) = 2 + 0 = 2
Example 2
Compute lim n → ∞ n 2 + 3 n n 2 + 1 \lim_{n \to \infty} \frac{n^2 + 3n}{n^2 + 1} lim n → ∞ n 2 + 1 n 2 + 3 n
Solution : lim n → ∞ n 2 + 3 n n 2 + 1 = lim n → ∞ 1 + 3 n 1 + 1 n 2 = 1 + 0 1 + 0 = 1 \lim_{n \to \infty} \frac{n^2 + 3n}{n^2 + 1} = \lim_{n \to \infty} \frac{1 + \frac{3}{n}}{1 + \frac{1}{n^2}} = \frac{1 + 0}{1 + 0} = 1 lim n → ∞ n 2 + 1 n 2 + 3 n = lim n → ∞ 1 + n 2 1 1 + n 3 = 1 + 0 1 + 0 = 1
Example 3
Compute lim n → ∞ 1 n ⋅ 2 n ⋅ 3 n \lim_{n \to \infty} \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} lim n → ∞ n 1 ⋅ n 2 ⋅ n 3
Solution : lim n → ∞ 1 n ⋅ 2 n ⋅ 3 n = lim n → ∞ 6 n 3 = 0 \lim_{n \to \infty} \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} = \lim_{n \to \infty} \frac{6}{n^3} = 0 lim n → ∞ n 1 ⋅ n 2 ⋅ n 3 = lim n → ∞ n 3 6 = 0
Limits of Special Sequences Common Sequence Limits
lim n → ∞ 1 n = 0 \lim_{n \to \infty} \frac{1}{n} = 0 lim n → ∞ n 1 = 0
lim n → ∞ 1 n k = 0 \lim_{n \to \infty} \frac{1}{n^k} = 0 lim n → ∞ n k 1 = 0 k > 0 k > 0 k > 0
lim n → ∞ a n = 0 \lim_{n \to \infty} a^n = 0 lim n → ∞ a n = 0 ∣ a ∣ < 1 |a| < 1 ∣ a ∣ < 1
lim n → ∞ a n = ∞ \lim_{n \to \infty} a^n = \infty lim n → ∞ a n = ∞ a > 1 a > 1 a > 1
lim n → ∞ n n = 1 \lim_{n \to \infty} \sqrt[n]{n} = 1 lim n → ∞ n n = 1
lim n → ∞ a n = 1 \lim_{n \to \infty} \sqrt[n]{a} = 1 lim n → ∞ n a = 1 a > 0 a > 0 a > 0
Practice Problems Exercise 1
Compute the limit lim n → ∞ 3 n + 2 2 n + 1 \lim_{n \to \infty} \frac{3n + 2}{2n + 1} lim n → ∞ 2 n + 1 3 n + 2
Reference Answer
sequence limit
Idea : Divide both the numerator and the denominator by n n n
Detailed steps :
3 n + 2 2 n + 1 = 3 + 2 n 2 + 1 n \frac{3n + 2}{2n + 1} = \frac{3 + \frac{2}{n}}{2 + \frac{1}{n}} 2 n + 1 3 n + 2 = 2 + n 1 3 + n 2
lim n → ∞ 3 n + 2 2 n + 1 = lim n → ∞ 3 + 2 n 2 + 1 n = 3 + 0 2 + 0 = 3 2 \lim_{n \to \infty} \frac{3n + 2}{2n + 1} = \lim_{n \to \infty} \frac{3 + \frac{2}{n}}{2 + \frac{1}{n}} = \frac{3 + 0}{2 + 0} = \frac{3}{2} lim n → ∞ 2 n + 1 3 n + 2 = lim n → ∞ 2 + n 1 3 + n 2 = 2 + 0 3 + 0 = 2 3
Answer : The limit is 3 2 \frac{3}{2} 2 3
Exercise 2
Compute the limit lim n → ∞ n 2 + 2 n + 1 n 2 − n \lim_{n \to \infty} \frac{n^2 + 2n + 1}{n^2 - n} lim n → ∞ n 2 − n n 2 + 2 n + 1
Reference Answer
sequence limit
Idea : Divide both the numerator and the denominator by n 2 n^2 n 2
Detailed steps :
n 2 + 2 n + 1 n 2 − n = 1 + 2 n + 1 n 2 1 − 1 n \frac{n^2 + 2n + 1}{n^2 - n} = \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{1 - \frac{1}{n}} n 2 − n n 2 + 2 n + 1 = 1 − n 1 1 + n 2 + n 2 1
lim n → ∞ n 2 + 2 n + 1 n 2 − n = lim n → ∞ 1 + 2 n + 1 n 2 1 − 1 n = 1 + 0 + 0 1 − 0 = 1 \lim_{n \to \infty} \frac{n^2 + 2n + 1}{n^2 - n} = \lim_{n \to \infty} \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{1 - \frac{1}{n}} = \frac{1 + 0 + 0}{1 - 0} = 1 lim n → ∞ n 2 − n n 2 + 2 n + 1 = lim n → ∞ 1 − n 1 1 + n 2 + n 2 1 = 1 − 0 1 + 0 + 0 = 1
Answer : The limit is 1.
Exercise 3
Compute the limit lim n → ∞ ( 1 n 2 + 2 n 2 + ⋯ + n n 2 ) \lim_{n \to \infty} \left(\frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{n}{n^2}\right) lim n → ∞ ( n 2 1 + n 2 2 + ⋯ + n 2 n )
Reference Answer
sequence limit
Idea : First sum the terms, then take the limit.
Detailed steps :
1 n 2 + 2 n 2 + ⋯ + n n 2 = 1 + 2 + ⋯ + n n 2 = n ( n + 1 ) 2 n 2 = n + 1 2 n \frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{n}{n^2} = \frac{1 + 2 + \cdots + n}{n^2} = \frac{\frac{n(n+1)}{2}}{n^2} = \frac{n+1}{2n} n 2 1 + n 2 2 + ⋯ + n 2 n = n 2 1 + 2 + ⋯ + n = n 2 2 n ( n + 1 ) = 2 n n + 1
lim n → ∞ n + 1 2 n = lim n → ∞ 1 + 1 n 2 = 1 2 \lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1}{2} lim n → ∞ 2 n n + 1 = lim n → ∞ 2 1 + n 1 = 2 1
Answer : The limit is 1 2 \frac{1}{2} 2 1
Summary Symbols Used in This Article
符号 类型 读音/说明 在本文中的含义 n n n 数学符号 positive integer The index of a term in a sequence x n x_n x n 数学符号 sequence The n n n lim \lim lim 数学符号 limit The limit of a sequence → \to → 数学符号 tends to A variable tending to some value ∞ \infty ∞ 数学符号 infinity Infinity
中英对照
中文术语 英文术语 音标 说明 数列 sequence /ˈsiːkwəns/ A collection of numbers arranged in order 数列极限 limit of sequence /ˈlɪmɪt əv ˈsiːkwəns/ The value a sequence tends to 正整数 positive integer /ˈpɒzətɪv ˈɪntɪdʒə/ An integer greater than zero 离散 discrete /dɪˈskriːt/ Discontinuous, separated
