Limits of sequences: finite limits, infinite limits, and subsequences
Distinguish finite and infinite limits from nonexistence, and apply limit laws with the required hypotheses.
Content updated: October 22, 2026
A limit describes the behavior of an entire tail of a sequence. A sequence may approach a number, tend to positive infinity, tend to negative infinity, or have no limit; increasing magnitude alone does not imply a limit of positive infinity.
Objectives: distinguish finite limits, infinite limits, and nonexistence; apply limit laws under their proper conditions.
Prerequisites: sequences and boundedness (English lesson); geometric sequences (English lesson).
On this page: Idea · Method · Examples · Practice · Further learning
1. The tail of a sequence and the definition
We write $u_n\to L$ if, for every $\varepsilon>0$, there is an integer $N$ such that $|u_n-L|<\varepsilon$ for every $n\ge N$. The crucial point is that all terms from some index onward are close to $L$, rather than just one term. The full notation is $\lim_{n\to\infty}u_n=L$; this lesson sometimes suppresses the index tending to infinity when it is clear.
We write $u_n\to+\infty$ if, for every $A>0$, we have $u_n>A$ from some index onward. We write $u_n\to-\infty$ if, for every $A>0$, we have $u_n<-A$ from some index onward. The infinity symbols are not real numbers. A sequence may have neither a finite limit nor either of these infinite limits.
2. Basic limits and limit laws
For a positive integer $k$, $1/n^k\to0$. If $|q|<1$, then $q^n\to0$. To explain the second statement, if $0<r=|q|<1$, write $1/r=1+c$ with $c>0$. Bernoulli's inequality $(1+c)^n\ge1+nc$, proved by induction, implies $|q^n|\le1/(1+nc)\to0$. For $q=0$, the conclusion is immediate from the terms themselves. If $q>1$, Bernoulli's inequality also implies $q^n\ge1+n(q-1)\to+\infty$.
We use the following standard limit laws: if $u_n\to a$ and $v_n\to b$ are finite limits and $C$ is fixed, then the sum, difference, and product tend to $a+b,a-b,ab$, respectively, and $Cu_n\to Ca$. If also $b\ne0$, the quotient tends to $a/b$; in this case, its denominator differs from $0$ for all sufficiently large indices. Do not apply these finite-limit operations to forms such as $\infty-\infty$ or $0\cdot\infty$.
3. Problems and worked examples
Problem 1. The sequence 1/n
At $n=1,2,\ldots,30$, the values are $1,1/2,1/3,\ldots,1/30$. Plotting $(n;1/n)$ produces points above the horizontal axis, with heights decreasing toward $0$. For a proof, given $\varepsilon>0$, choose $N>1/\varepsilon$; then $n\ge N$ implies $0<1/n<\varepsilon$. Hence $1/n\to0$; a finite table only illustrates the conclusion.
Problem 2. The sequence 2 to the power n
At $n=1,2,\ldots,10$, the terms are $2,4,8,\ldots,1024$. Since $2^n\ge n+1$, for every threshold $A$ we can choose $N>A$ so that every $n\ge N$ satisfies $2^n>A$. Thus $2^n\to+\infty$.
Problem 3. Alternating signs with limit 0
The sequence $u_n=(-1/2)^n$ begins $-1/2,1/4,-1/8,1/16,\ldots$. Its signs alternate, but $|u_n|=2^{-n}\to0$, so $u_n\to0$. A sequence need not be increasing or decreasing to have a finite limit.
Problem 4. Alternating signs and increasing magnitude
For $u_n=(-2)^n$, the even-indexed terms satisfy $u_{2k}=4^k\to+\infty$, whereas the odd-indexed terms satisfy $u_{2l+1}=-2\cdot4^l\to-\infty$, with $l\ge0$. If the sequence had a finite or infinite limit, every subsequence would have the same limit, since its terms still lie in the tail of the original sequence. These two subsequences have opposite behavior, so $u_n$ has no limit. Do not write its limit as an infinity symbol without a sign.
Example 1. Basic limits
By the results in Section 2, $1/n^2\to0$, $1/n^4\to0$, $(2/3)^n\to0$, and $3^n\to+\infty$. The conditions $|2/3|<1$ and $3>1$ determine the two different conclusions.
Example 2. Four finite-limit calculations
- $\lim(1/n^2+1/n^4)=0+0=0$.
- $\lim(1/n^2+1)=0+1=1$.
- $\lim 1/(3n^2)=(1/3)\cdot0=0$.
-
$$\lim\frac{3n^2-1}{n^2}=\lim\left(3-\frac1{n^2}\right)=3.$$
4. Practice exercises
- Evaluate $\lim(2+3/n)$.
- Does $(-1)^n$ have a limit? Examine even and odd indices.
- Evaluate $\lim(2n^2+n)/(n^2+1)$ and check the denominator condition.
Show answers and hints
- Since $3/n\to0$, the limit is $2$.
- No. The even subsequence is constantly $1$, and the odd subsequence is constantly $-1$, so their limits differ.
- Divide numerator and denominator by $n^2$ to obtain $(2+1/n)/(1+1/n^2)\to2$. The original denominator is positive, and the normalized denominator tends to $1\ne0$.
5. Common mistakes
- Treating a finite table of values as a proof for every index.
- Confusing $|u_n|\to\infty$ with $u_n\to+\infty$.
- Adding or subtracting infinity symbols as though they were real numbers.
- Applying the quotient law without checking that the denominator's limit differs from $0$.
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