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The World of Limits in Advanced Mathematics

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

Operation rules for sequence limits

Let lim⁡n→∞xn=A\lim_{n \to \infty} x_n = A and lim⁡n→∞yn=B\lim_{n \to \infty} y_n = B, where AA and BB are both finite numbers, then:

  1. Addition rule: lim⁡n→∞(xn±yn)=A±B\lim_{n \to \infty} (x_n \pm y_n) = A \pm B

  2. Multiplication rule: lim⁡n→∞(xn⋅yn)=A⋅B\lim_{n \to \infty} (x_n \cdot y_n) = A \cdot B

  3. Division rule: lim⁡n→∞xnyn=AB\lim_{n \to \infty} \frac{x_n}{y_n} = \frac{A}{B} (where B≠0B \neq 0)

  4. Power rule: lim⁡n→∞xnk=Ak\lim_{n \to \infty} x_n^k = A^k (where kk is a positive integer)

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, then lim⁡n→∞f(n)=A\lim_{n \to \infty} f(n) = A.

But the converse does not always hold: lim⁡n→∞f(n)=A\lim_{n \to \infty} f(n) = A does not imply lim⁡x→∞f(x)=A\lim_{x \to \infty} f(x) = A.

Example

The function f(x)=sin⁡(πx)f(x) = \sin(\pi x):

  • Sequence limit: lim⁡n→∞sin⁡(nπ)=0\lim_{n \to \infty} \sin(n\pi) = 0 (since sin⁡(nπ)=0\sin(n\pi) = 0 for every integer nn)
  • Function limit: lim⁡x→∞sin⁡(πx)\lim_{x \to \infty} \sin(\pi x) does not exist (since the function oscillates between [−1,1][-1, 1])

Worked Examples

Example 1

Compute lim⁡n→∞2n+1n\lim_{n \to \infty} \frac{2n + 1}{n}

Solution: lim⁡n→∞2n+1n=lim⁡n→∞(2+1n)=2+0=2\lim_{n \to \infty} \frac{2n + 1}{n} = \lim_{n \to \infty} \left(2 + \frac{1}{n}\right) = 2 + 0 = 2

Example 2

Compute lim⁡n→∞n2+3nn2+1\lim_{n \to \infty} \frac{n^2 + 3n}{n^2 + 1}

Solution: lim⁡n→∞n2+3nn2+1=lim⁡n→∞1+3n1+1n2=1+01+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

Example 3

Compute lim⁡n→∞1n⋅2n⋅3n\lim_{n \to \infty} \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n}

Solution: lim⁡n→∞1n⋅2n⋅3n=lim⁡n→∞6n3=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

Limits of Special Sequences

Common Sequence Limits

  1. lim⁡n→∞1n=0\lim_{n \to \infty} \frac{1}{n} = 0

  2. lim⁡n→∞1nk=0\lim_{n \to \infty} \frac{1}{n^k} = 0 (k>0k > 0)

  3. lim⁡n→∞an=0\lim_{n \to \infty} a^n = 0 (∣a∣<1|a| < 1)

  4. lim⁡n→∞an=∞\lim_{n \to \infty} a^n = \infty (a>1a > 1)

  5. lim⁡n→∞nn=1\lim_{n \to \infty} \sqrt[n]{n} = 1

  6. lim⁡n→∞an=1\lim_{n \to \infty} \sqrt[n]{a} = 1 (a>0a > 0)

Practice Problems

Exercise 1

Compute the limit lim⁡n→∞3n+22n+1\lim_{n \to \infty} \frac{3n + 2}{2n + 1}.

sequence limit

Approach: Divide both the numerator and the denominator by nn.

Steps:

  1. 3n+22n+1=3+2n2+1n\frac{3n + 2}{2n + 1} = \frac{3 + \frac{2}{n}}{2 + \frac{1}{n}}

  2. lim⁡n→∞3n+22n+1=lim⁡n→∞3+2n2+1n=3+02+0=32\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}

Answer: The limit is 32\frac{3}{2}.

Exercise 2

Compute the limit lim⁡n→∞n2+2n+1n2−n\lim_{n \to \infty} \frac{n^2 + 2n + 1}{n^2 - n}.

sequence limit

Approach: Divide both the numerator and the denominator by n2n^2.

Steps:

  1. n2+2n+1n2−n=1+2n+1n21−1n\frac{n^2 + 2n + 1}{n^2 - n} = \frac{1 + \frac{2}{n} + \frac{1}{n^2}}{1 - \frac{1}{n}}

  2. lim⁡n→∞n2+2n+1n2−n=lim⁡n→∞1+2n+1n21−1n=1+0+01−0=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

Answer: The limit is 1.

Exercise 3

Compute the limit lim⁡n→∞(1n2+2n2+⋯+nn2)\lim_{n \to \infty} \left(\frac{1}{n^2} + \frac{2}{n^2} + \cdots + \frac{n}{n^2}\right).

sequence limit

Approach: First sum the terms, then take the limit.

Steps:

  1. 1n2+2n2+⋯+nn2=1+2+⋯+nn2=n(n+1)2n2=n+12n\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}

  2. lim⁡n→∞n+12n=lim⁡n→∞1+1n2=12\lim_{n \to \infty} \frac{n+1}{2n} = \lim_{n \to \infty} \frac{1 + \frac{1}{n}}{2} = \frac{1}{2}

Answer: The limit is 12\frac{1}{2}.

Summary

Symbols Used in This Article

SymbolTypePronunciation/ExplanationMeaning in This Article
nnMathematical symbolpositive integerThe index of a term in a sequence
xnx_nMathematical symbolsequenceThe nn-th term of a sequence
lim⁡\limMathematical symbollimitThe limit of a sequence
→\toMathematical symboltends toA variable tending to some value
∞\inftyMathematical symbolinfinityInfinity
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