极限入门题解析

Exercises and Solutions

Evaluate the limit :
lim ⁡ n → ∞ ( 1 + 1 2 + 1 3 + ⋯ + 1 n ) 1 n \lim_{n \rightarrow \infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)^{\frac{1}{n}} n→∞lim(1+21+31+⋯+n1)n1
Solution :
1 ≤ lim ⁡ n → ∞ ( 1 + 1 2 + 1 3 + ⋯ + 1 n ) 1 n ≤ lim ⁡ n → ∞ n 1 n = 1 1 \leq \lim_{n \rightarrow \infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)^{\frac{1}{n}} \leq \lim_{n \rightarrow \infty} n^{\frac{1}{n}} = 1 1≤n→∞lim(1+21+31+⋯+n1)n1≤n→∞limnn1=1
By the Squeeze Theorem:
lim ⁡ n → ∞ ( 1 + 1 2 + 1 3 + ⋯ + 1 n ) 1 n = 1 \lim_{n \rightarrow \infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)^{\frac{1}{n}} = 1 n→∞lim(1+21+31+⋯+n1)n1=1

Evaluate the limit :
lim ⁡ n → ∞ ( 1 n + 1 + 1 n + 2 + ⋯ + 1 n + n ) \lim_{n \rightarrow \infty} \left(\frac{1}{n + \sqrt{1}} + \frac{1}{n + \sqrt{2}} + \cdots + \frac{1}{n + \sqrt{n}}\right) n→∞lim(n+1 1+n+2 1+⋯+n+n 1)
Solution :
1 = lim ⁡ n → ∞ ∑ i = 1 n 1 n + n ≤ lim ⁡ n → ∞ ∑ i = 1 n 1 n + i ≤ lim ⁡ n → ∞ ∑ i = 1 n 1 n = 1 1 = \lim_{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n + \sqrt{n}} \leq \lim_{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n + \sqrt{i}} \leq \lim_{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n} = 1 1=n→∞limi=1∑nn+n 1≤n→∞limi=1∑nn+i 1≤n→∞limi=1∑nn1=1
By the Squeeze Theorem:
lim ⁡ n → ∞ ∑ i = 1 n 1 n + i = 1 \lim_{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n + \sqrt{i}} = 1 n→∞limi=1∑nn+i 1=1

Evaluate the limit :
lim ⁡ n → ∞ ∑ k = n 2 ( n + 1 ) 2 1 k \lim_{n \rightarrow \infty} \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{k}} n→∞limk=n2∑(n+1)2k 1
Solution :
∑ k = n 2 ( n + 1 ) 2 1 ( n + 1 ) 2 ≤ ∑ k = n 2 ( n + 1 ) 2 1 k ≤ ∑ k = n 2 ( n + 1 ) 2 1 n 2 \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{(n + 1)^2}} \leq \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{k}} \leq \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{n^2}} k=n2∑(n+1)2(n+1)2 1≤k=n2∑(n+1)2k 1≤k=n2∑(n+1)2n2 1
Taking the limit:
lim ⁡ n → ∞ 2 ≤ lim ⁡ n → ∞ ∑ k = n 2 ( n + 1 ) 2 1 k ≤ lim ⁡ n → ∞ ( 2 + 2 n ) \lim_{n \rightarrow \infty} 2 \leq \lim_{n \rightarrow \infty} \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{k}} \leq \lim_{n \rightarrow \infty} \left(2 + \frac{2}{n}\right) n→∞lim2≤n→∞limk=n2∑(n+1)2k 1≤n→∞lim(2+n2)
By the Squeeze Theorem:
lim ⁡ n → ∞ ∑ k = n 2 ( n + 1 ) 2 1 k = 2 \lim_{n \rightarrow \infty} \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{k}} = 2 n→∞limk=n2∑(n+1)2k 1=2

Evaluate the limit :
lim ⁡ n → ∞ 1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ ( 2 n ) \lim_{n \rightarrow \infty} \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} n→∞lim2⋅4⋅6⋅⋯⋅(2n)1⋅3⋅5⋅⋯⋅(2n−1)
Solution :
Since:
1 2 ( 2 n − 2 ) ! ! ( 2 n − 1 ) ! ! ≤ ( 2 n − 1 ) ! ! ( 2 n ) ! ! ≤ ( 2 n − 2 ) ! ! ( 2 n − 1 ) ! ! \frac{1}{2} \frac{(2n - 2)!!}{(2n - 1)!!} \leq \frac{(2n - 1)!!}{(2n)!!} \leq \frac{(2n - 2)!!}{(2n - 1)!!} 21(2n−1)!!(2n−2)!!≤(2n)!!(2n−1)!!≤(2n−1)!!(2n−2)!!
Then:
1 4 n ≤ ( ( 2 n − 1 ) ! ! ( 2 n ) ! ! ) 2 ≤ 1 2 n \frac{1}{4n} \leq \left(\frac{(2n - 1)!!}{(2n)!!}\right)^2 \leq \frac{1}{2n} 4n1≤((2n)!!(2n−1)!!)2≤2n1
By the Squeeze Theorem, the limit exists:
lim ⁡ n → ∞ 1 ⋅ 3 ⋅ 5 ⋅ ⋯ ⋅ ( 2 n − 1 ) 2 ⋅ 4 ⋅ 6 ⋅ ⋯ ⋅ ( 2 n ) = 0 \lim_{n \rightarrow \infty} \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n - 1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot (2n)} = 0 n→∞lim2⋅4⋅6⋅⋯⋅(2n)1⋅3⋅5⋅⋯⋅(2n−1)=0

Evaluate the limit :
lim ⁡ n → ∞ 3 n 2 + 4 n − 1 n 2 + 1 \lim_{n \rightarrow \infty} \frac{3n^2 + 4n - 1}{n^2 + 1} n→∞limn2+13n2+4n−1
Solution :
lim ⁡ n → ∞ 3 n 2 + 4 n − 1 n 2 + 1 = 3 \lim_{n \rightarrow \infty} \frac{3n^2 + 4n - 1}{n^2 + 1} = 3 n→∞limn2+13n2+4n−1=3

Evaluate the limit :
lim ⁡ n → ∞ n 3 + 2 n 2 − 3 n + 1 2 n 3 − n + 3 \lim_{n \rightarrow \infty} \frac{n^3 + 2n^2 - 3n + 1}{2n^3 - n + 3} n→∞lim2n3−n+3n3+2n2−3n+1
Solution :
lim ⁡ n → ∞ n 3 + 2 n 2 − 3 n + 1 2 n 3 − n + 3 = 1 2 \lim_{n \rightarrow \infty} \frac{n^3 + 2n^2 - 3n + 1}{2n^3 - n + 3} = \frac{1}{2} n→∞lim2n3−n+3n3+2n2−3n+1=21

Evaluate the limit :
lim ⁡ n → ∞ 3 n + n 3 3 n + 1 + ( n + 1 ) 3 \lim_{n \rightarrow \infty} \frac{3^n + n^3}{3^{n + 1} + (n + 1)^3} n→∞lim3n+1+(n+1)33n+n3
Solution :
lim ⁡ n → ∞ 3 n + n 3 3 n + 1 + ( n + 1 ) 3 = 1 3 \lim_{n \rightarrow \infty} \frac{3^n + n^3}{3^{n + 1} + (n + 1)^3} = \frac{1}{3} n→∞lim3n+1+(n+1)33n+n3=31

Evaluate the limit :
lim ⁡ n → ∞ ( n 2 + 1 n − 1 ) sin ⁡ n π 2 \lim_{n \rightarrow \infty} \left(\sqrt[n]{n^2 + 1} - 1\right) \sin \frac{n\pi}{2} n→∞lim(nn2+1 −1)sin2nπ
Solution :
lim ⁡ n → ∞ ( n 2 + 1 n − 1 ) sin ⁡ n π 2 = 0 \lim_{n \rightarrow \infty} \left(\sqrt[n]{n^2 + 1} - 1\right) \sin \frac{n\pi}{2} = 0 n→∞lim(nn2+1 −1)sin2nπ=0

Evaluate the limit :
lim ⁡ n → ∞ n ( n + 1 − n ) \lim_{n \rightarrow \infty} \sqrt{n} \left(\sqrt{n + 1} - \sqrt{n}\right) n→∞limn (n+1 −n )
Solution :
lim ⁡ n → ∞ n ( n + 1 − n ) = 1 2 \lim_{n \rightarrow \infty} \sqrt{n} \left(\sqrt{n + 1} - \sqrt{n}\right) = \frac{1}{2} n→∞limn (n+1 −n )=21

Evaluate the limit :
lim ⁡ n → ∞ n ( n 2 + 1 4 − n + 1 ) \lim_{n \rightarrow \infty} \sqrt{n} \left(\sqrt[4]{n^2 + 1} - \sqrt{n + 1}\right) n→∞limn (4n2+1 −n+1 )
Solution :
lim ⁡ n → ∞ n ( n 2 + 1 4 − n + 1 ) = − 1 2 \lim_{n \rightarrow \infty} \sqrt{n} \left(\sqrt[4]{n^2 + 1} - \sqrt{n + 1}\right) = -\frac{1}{2} n→∞limn (4n2+1 −n+1 )=−21

Evaluate the limit :
lim ⁡ n → ∞ 1 n ! n \lim_{n \rightarrow \infty} \sqrt[n]{\frac{1}{n!}} n→∞limnn!1
Solution :
lim ⁡ n → ∞ 1 n ! n = 0 \lim_{n \rightarrow \infty} \sqrt[n]{\frac{1}{n!}} = 0 n→∞limnn!1 =0

Evaluate the limit :
lim ⁡ n → ∞ ( 1 − 1 2 2 ) ( 1 − 1 3 2 ) ⋯ ( 1 − 1 n 2 ) \lim_{n \rightarrow \infty} \left(1 - \frac{1}{2^2}\right) \left(1 - \frac{1}{3^2}\right) \cdots \left(1 - \frac{1}{n^2}\right) n→∞lim(1−221)(1−321)⋯(1−n21)
Solution :
lim ⁡ n → ∞ ( 1 − 1 2 2 ) ( 1 − 1 3 2 ) ⋯ ( 1 − 1 n 2 ) = 1 2 \lim_{n \rightarrow \infty} \left(1 - \frac{1}{2^2}\right) \left(1 - \frac{1}{3^2}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{1}{2} n→∞lim(1−221)(1−321)⋯(1−n21)=21

Evaluate the limit :
lim ⁡ n → ∞ n ln ⁡ n n \lim_{n \rightarrow \infty} \sqrt[n]{n \ln n} n→∞limnnlnn
Solution :
lim ⁡ n → ∞ n ln ⁡ n n = 0 \lim_{n \rightarrow \infty} \sqrt[n]{n \ln n} = 0 n→∞limnnlnn =0

Evaluate the limit :
lim ⁡ n → ∞ ( 1 2 + 3 2 2 + ⋯ + 2 n − 1 2 n ) \lim_{n \rightarrow \infty} \left(\frac{1}{2} + \frac{3}{2^2} + \cdots + \frac{2n - 1}{2^n}\right) n→∞lim(21+223+⋯+2n2n−1)
Solution :
lim ⁡ n → ∞ ( 1 2 + 3 2 2 + ⋯ + 2 n − 1 2 n ) = 3 \lim_{n \rightarrow \infty} \left(\frac{1}{2} + \frac{3}{2^2} + \cdots + \frac{2n - 1}{2^n}\right) = 3 n→∞lim(21+223+⋯+2n2n−1)=3

Given lim ⁡ n → ∞ a n = a \lim_{n \rightarrow \infty} a_n = a limn→∞an=a and lim ⁡ n → ∞ b n = b \lim_{n \rightarrow \infty} b_n = b limn→∞bn=b, prove :
lim ⁡ n → ∞ a 1 b n + a 2 b n − 1 + ⋯ + a n b 1 n = a b \lim_{n \rightarrow \infty} \frac{a_1 b_n + a_2 b_{n - 1} + \cdots + a_n b_1}{n} = ab n→∞limna1bn+a2bn−1+⋯+anb1=ab
Proof :
Let a n = a + α n a_n = a + \alpha_n an=a+αn and b n = b + β n b_n = b + \beta_n bn=b+βn. Then:
∑ i = 1 n a i b n + 1 − i n = a b + b n ∑ i = 1 n α i + a n ∑ i = 1 n β i + 1 n ∑ i = 1 n α i β n − i + 1 \frac{\sum_{i=1}^n a_i b_{n + 1 - i}}{n} = ab + \frac{b}{n} \sum_{i=1}^n \alpha_i + \frac{a}{n} \sum_{i=1}^n \beta_i + \frac{1}{n} \sum_{i=1}^n \alpha_i \beta_{n - i + 1} n∑i=1naibn+1−i=ab+nbi=1∑nαi+nai=1∑nβi+n1i=1∑nαiβn−i+1
Let ∣ α n ∣ , ∣ β n ∣ ≤ M |\alpha_n|, |\beta_n| \leq M ∣αn∣,∣βn∣≤M. For n > N 1 n > N_1 n>N1 and n > N 2 n > N_2 n>N2:
∣ 1 n ∑ i = 1 n α i β n − i + 1 ∣ < ∣ 1 n ∑ i = 1 N α i β n − i + 1 ∣ + ∣ 1 n ∑ i = N + 1 n β n − i + 1 ∣ ε < N M 2 n + M ε \left| \frac{1}{n} \sum_{i=1}^n \alpha_i \beta_{n - i + 1} \right| < \left| \frac{1}{n} \sum_{i=1}^N \alpha_i \beta_{n - i + 1} \right| + \left| \frac{1}{n} \sum_{i=N+1}^n \beta_{n - i + 1} \right| \varepsilon < \frac{NM^2}{n} + M\varepsilon ∣∣∣∣∣n1i=1∑nαiβn−i+1∣∣∣∣∣<∣∣∣∣∣n1i=1∑Nαiβn−i+1∣∣∣∣∣+∣∣∣∣∣n1i=N+1∑nβn−i+1∣∣∣∣∣ε<nNM2+Mε
Therefore:
lim ⁡ n → ∞ ∑ i = 1 n a i b n + 1 − i n = a b \lim_{n \rightarrow \infty} \frac{\sum_{i=1}^n a_i b_{n + 1 - i}}{n} = ab n→∞limn∑i=1naibn+1−i=ab