极限入门题分析

Exercises and Solutions

• Evaluate the limit:
  lim⁡n→∞(1+12+13+⋯+1n)1n\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+12+13+⋯+1n)1n≤lim⁡n→∞n1n=11 \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}} = 11≤n→∞lim​(1+21​+31​+⋯+n1​)n1​≤n→∞lim​nn1​=1
  By the Squeeze Theorem:
  lim⁡n→∞(1+12+13+⋯+1n)1n=1\lim_{n \rightarrow \infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right)^{\frac{1}{n}} = 1n→∞lim​(1+21​+31​+⋯+n1​)n1​=1
  

---

• Evaluate the limit:
  lim⁡n→∞(1n+1+1n+2+⋯+1n+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=1n1n+n≤lim⁡n→∞∑i=1n1n+i≤lim⁡n→∞∑i=1n1n=11 = \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} = 11=n→∞lim​i=1∑n​n+n​1​≤n→∞lim​i=1∑n​n+i​1​≤n→∞lim​i=1∑n​n1​=1
  By the Squeeze Theorem:
  lim⁡n→∞∑i=1n1n+i=1\lim_{n \rightarrow \infty} \sum_{i=1}^n \frac{1}{n + \sqrt{i}} = 1n→∞lim​i=1∑n​n+i​1​=1
  

---

• Evaluate the limit:
  lim⁡n→∞∑k=n2(n+1)21k\lim_{n \rightarrow \infty} \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{k}}n→∞lim​k=n2∑(n+1)2​k​1​
  Solution:
  ∑k=n2(n+1)21(n+1)2≤∑k=n2(n+1)21k≤∑k=n2(n+1)21n2\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)2​k​1​≤k=n2∑(n+1)2​n2​1​
  Taking the limit:
  lim⁡n→∞2≤lim⁡n→∞∑k=n2(n+1)21k≤lim⁡n→∞(2+2n)\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→∞lim​2≤n→∞lim​k=n2∑(n+1)2​k​1​≤n→∞lim​(2+n2​)
  By the Squeeze Theorem:
  lim⁡n→∞∑k=n2(n+1)21k=2\lim_{n \rightarrow \infty} \sum_{k = n^2}^{(n + 1)^2} \frac{1}{\sqrt{k}} = 2n→∞lim​k=n2∑(n+1)2​k​1​=2
  

---

• Evaluate the limit:
  lim⁡n→∞1⋅3⋅5⋅⋯⋅(2n−1)2⋅4⋅6⋅⋯⋅(2n)\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→∞lim​2⋅4⋅6⋅⋯⋅(2n)1⋅3⋅5⋅⋯⋅(2n−1)​
  Solution:
  Since:
  12(2n−2)!!(2n−1)!!≤(2n−1)!!(2n)!!≤(2n−2)!!(2n−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:
  14n≤((2n−1)!!(2n)!!)2≤12n\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⋅⋯⋅(2n−1)2⋅4⋅6⋅⋯⋅(2n)=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)} = 0n→∞lim​2⋅4⋅6⋅⋯⋅(2n)1⋅3⋅5⋅⋯⋅(2n−1)​=0
  

---

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

---

• Evaluate the limit:
  lim⁡n→∞n3+2n2−3n+12n3−n+3\lim_{n \rightarrow \infty} \frac{n^3 + 2n^2 - 3n + 1}{2n^3 - n + 3}n→∞lim​2n3−n+3n3+2n2−3n+1​
  Solution:
  lim⁡n→∞n3+2n2−3n+12n3−n+3=12\lim_{n \rightarrow \infty} \frac{n^3 + 2n^2 - 3n + 1}{2n^3 - n + 3} = \frac{1}{2}n→∞lim​2n3−n+3n3+2n2−3n+1​=21​
  

---

• Evaluate the limit:
  lim⁡n→∞3n+n33n+1+(n+1)3\lim_{n \rightarrow \infty} \frac{3^n + n^3}{3^{n + 1} + (n + 1)^3}n→∞lim​3n+1+(n+1)33n+n3​
  Solution:
  lim⁡n→∞3n+n33n+1+(n+1)3=13\lim_{n \rightarrow \infty} \frac{3^n + n^3}{3^{n + 1} + (n + 1)^3} = \frac{1}{3}n→∞lim​3n+1+(n+1)33n+n3​=31​
  

---

• Evaluate the limit:
  lim⁡n→∞(n2+1n−1)sin⁡nπ2\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→∞(n2+1n−1)sin⁡nπ2=0\lim_{n \rightarrow \infty} \left(\sqrt[n]{n^2 + 1} - 1\right) \sin \frac{n\pi}{2} = 0n→∞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→∞lim​n​(n+1​−n​)
  Solution:
  lim⁡n→∞n(n+1−n)=12\lim_{n \rightarrow \infty} \sqrt{n} \left(\sqrt{n + 1} - \sqrt{n}\right) = \frac{1}{2}n→∞lim​n​(n+1​−n​)=21​
  

---

• Evaluate the limit:
  lim⁡n→∞n(n2+14−n+1)\lim_{n \rightarrow \infty} \sqrt{n} \left(\sqrt[4]{n^2 + 1} - \sqrt{n + 1}\right)n→∞lim​n​(4n2+1​−n+1​)
  Solution:
  lim⁡n→∞n(n2+14−n+1)=−12\lim_{n \rightarrow \infty} \sqrt{n} \left(\sqrt[4]{n^2 + 1} - \sqrt{n + 1}\right) = -\frac{1}{2}n→∞lim​n​(4n2+1​−n+1​)=−21​
  

---

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

---

• Evaluate the limit:
  lim⁡n→∞(1−122)(1−132)⋯(1−1n2)\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→∞(1−122)(1−132)⋯(1−1n2)=12\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→∞nln⁡nn\lim_{n \rightarrow \infty} \sqrt[n]{n \ln n}n→∞lim​nnlnn​
  Solution:
  lim⁡n→∞nln⁡nn=0\lim_{n \rightarrow \infty} \sqrt[n]{n \ln n} = 0n→∞lim​nnlnn​=0
  

---

• Evaluate the limit:
  lim⁡n→∞(12+322+⋯+2n−12n)\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→∞(12+322+⋯+2n−12n)=3\lim_{n \rightarrow \infty} \left(\frac{1}{2} + \frac{3}{2^2} + \cdots + \frac{2n - 1}{2^n}\right) = 3n→∞lim​(21​+223​+⋯+2n2n−1​)=3
  

---

• Given lim⁡n→∞an=a\lim_{n \rightarrow \infty} a_n = alimn→∞​an​=a and lim⁡n→∞bn=b\lim_{n \rightarrow \infty} b_n = blimn→∞​bn​=b, prove:
  lim⁡n→∞a1bn+a2bn−1+⋯+anb1n=ab\lim_{n \rightarrow \infty} \frac{a_1 b_n + a_2 b_{n - 1} + \cdots + a_n b_1}{n} = abn→∞lim​na1​bn​+a2​bn−1​+⋯+an​b1​​=ab
  Proof:
  Let an=a+αna_n = a + \alpha_nan​=a+αn​ and bn=b+βnb_n = b + \beta_nbn​=b+βn​. Then:
  ∑i=1naibn+1−in=ab+bn∑i=1nαi+an∑i=1nβi+1n∑i=1nαiβn−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=1n​ai​bn+1−i​​=ab+nb​i=1∑n​αi​+na​i=1∑n​βi​+n1​i=1∑n​αi​βn−i+1​
  Let ∣αn∣,∣βn∣≤M|\alpha_n|, |\beta_n| \leq M∣αn​∣,∣βn​∣≤M. For n>N1n > N_1n>N1​ and n>N2n > N_2n>N2​:
  ∣1n∑i=1nαiβn−i+1∣<∣1n∑i=1Nαiβn−i+1∣+∣1n∑i=N+1nβn−i+1∣ε<NM2n+Mε\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∣∣∣∣∣​n1​i=1∑n​αi​βn−i+1​∣∣∣∣∣​<∣∣∣∣∣​n1​i=1∑N​αi​βn−i+1​∣∣∣∣∣​+∣∣∣∣∣​n1​i=N+1∑n​βn−i+1​∣∣∣∣∣​ε<nNM2​+Mε
  Therefore:
  lim⁡n→∞∑i=1naibn+1−in=ab\lim_{n \rightarrow \infty} \frac{\sum_{i=1}^n a_i b_{n + 1 - i}}{n} = abn→∞lim​n∑i=1n​ai​bn+1−i​​=ab