# Sequences of Functions
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/sequences-of-functions/
Fetched from algebrica.org post 17568; source modified 2026-04-12T20:24:14.
## Introduction
Imagine you have a list of different [functions](../functions.md), where each function in the list is linked to a number \\(n = 1, 2, 3… \\in \\mathbb{N} \\). So, for each \\(n\\), you get a different function, and that ordered list of functions is essentially what a [sequence](../sequences.md) of functions is. In formal terms, let \\( A \\subseteq \\mathbb{R} \\) be a non-empty subset and suppose that for each \\( n \\in \\mathbb{N} \\) we have a function \\( f\_n: A \\rightarrow \\mathbb{R} \\). We then say that \\( (f\_n) = (f\_1, f\_2, f\_3, \\dots) \\) is a **sequence of functions** on \\( A. \\)
* * *
Let’s consider a simple practical example. Let \\( (f\_n) \\) be a sequence of functions, with \\(n \\in \\mathbb{N}\_0\\) and \\( x \\in \\mathbb{R} \\), defined by:
\\[
f\_n(x) = \\frac{x}{n+1}
\\]
This is a family of functions where each function is linear, and the slope decreases as \\(n\\) increases. For \\(n = 0, 1, 2, 3, …\\), we have:
\\[
\\begin{aligned} f\_0(x) &= \\frac{x}{0+1} = x \\[0.5em] f\_1(x) &= \\frac{x}{1+1} = \\frac{x}{2} \\[0.5em] f\_2(x) &= \\frac{x}{2+1} = \\frac{x}{3} \\[0.5em] f\_3(x) &= \\frac{x}{3+1} = \\frac{x}{4} \\[0.5em] \\vdots \\end{aligned}
\\]
Thus, the sequence of functions is:
\\[
(f\_n) = \\left( x, \\frac{x}{2}, \\frac{x}{3}, \\frac{x}{4}, \\dots \\right)
\\]
Graphically, this situation can be observed as follows:
The graph shows how, as the index \\(n\\) increases, the slope of the line \\(f\_n(x)\\) progressively decreases. This reflects the fact that the function flattens toward the zero function \\(f(x) = 0\\) for every \\(x\\). In other words, the sequence of functions \\(f\_n(x)\\) converges pointwise to the zero function as \\(n\\) approaches infinity.
## Pointwise convergence
Let \\( \\lbrace f\_n(x) \\rbrace \\) be a sequence of functions defined on a common [domain](../determining-the-domain-of-a-function.md) \\( A \\subseteq \\mathbb{R} \\), with \\( n \\in \\mathbb{N} \\). We say that the sequence \\( \\lbrace f\_n(x) \\rbrace \\) converges pointwise on a set \\( C \\subseteq A \\) if, for every \\( x \\in C \\), the numerical sequence \\( \\lbrace f\_n(x) \\rbrace \\) converges. In this case, the [limit](../limits.md) function \\( f(x) \\) is defined as:
\\[
f(x) = \\lim\_{n \\to +\\infty} f\_n(x) \\quad \\forall \\, x \\in C
\\]
The set \\( C \\) is called the **pointwise convergence** set of the sequence \\( {f\_n(x)} \\).
* * *
Pointwise convergence can also be expressed as follows. Let \\((f\_n)\\) be a sequence of functions defined on a set \\(A\\). Then \\((f\_n)\\) converges pointwise to \\(f : A \\to \\mathbb{R}\\) if and only if \\(\\forall \\, x \\in A\\) and \\(\\forall \\varepsilon > 0\\) exists \\(K \\in \\mathbb{N} \\) such that:
\\[
|f\_n(x) - f(x)| < \\varepsilon \\quad \\forall \\; n \\geq K
\\]
##### In other words, for each fixed point \\( x \\), we can make \\( f\_n(x) \\) as close as we like to \\( f(x) \\) by choosing \\( n \\) large enough. The index \\( K \\) required to achieve the desired accuracy may vary depending on \\( x \\) and \\( \\varepsilon \\).
* * *
Let us consider the sequence of functions as previously discussed:
\\[
f\_n(x) = \\frac{x}{n+1}, \\quad x \\in \\mathbb{R}, \\quad n \\in \\mathbb{N}\_0
\\]
Let us examine what happens for each \\(x\\) as \\(n \\to \\infty\\). If we fix a generic \\(x\\), for example \\(x = 2\\), the associated sequence is:
\\[
\\begin{aligned} f\_0(2) &= 2 \\[0.5em] f\_1(2) &= 1 \\[0.5em] f\_2(2) &= \\frac{2}{3} \\[0.5em] f\_3(2) &= \\frac{2}{4} \\[0.5em] &\\vdots \\end{aligned}
\\]
This sequence of numbers tends to zero as \\(n \\to \\infty\\). In general, for every \\(x \\in \\mathbb{R}\\):
\\[
\\lim\_{n \\to \\infty} f\_n(x) = \\lim\_{n \\to \\infty} \\frac{x}{n+1} = 0
\\]
Therefore, the limit function is:
\\[
f(x) = 0 \\quad \\forall \\, x \\in \\mathbb{R}
\\]
By the uniqueness of limits of sequences of real numbers, the pointwise limit of a sequence of functions \\( (f\_n) \\) is unique.
## Example
Let’s study the behavior of the following sequence of functions in the interval \\(-1 < x < 1\\):
\\[
f\_n(x) = x^n
\\]
For a fixed value of \\(x\\) in this interval, we know that the [absolute value](../absolute-value.md) of \\(x\\) is less than \\(1\\), that is, \\(|x| < 1\\). This means we are considering [powers](../powers.md) of a number smaller than \\(1\\) in absolute value. By properties of [exponents](../exponential-function.md), when the base has an absolute value less than \\(1\\), the sequence \\(x^n\\) tends to zero as \\(n\\) tends to infinity:
\\[
\\lim\_{n \\to \\infty} x^n = 0
\\]
The sequence of powers of a real number \\(x\\) with \\(|x| < 1\\) converges to zero.
Therefore, for every \\(x\\) in the interval \\(-1 < x < 1\\), the sequence of functions \\(f\_n(x) = x^n\\) converges pointwise to the zero function.
\\[
f(x) = 0 \\quad \\forall \\, x \\in (-1,1)
\\]
## Consequences of pointwise convergence
Let \\( {f\_n} \\) be a sequence of functions \\( f\_n : A \\to \\mathbb{R} \\) that converges pointwise to a function \\( f : A \\to \\mathbb{R} \\). The following properties hold:
- If each \\( f\_n(x) \\geq 0 \\) for all \\( x \\in A \\), then \\( f(x) \\geq 0 \\) for all \\( x \\in A \\). In practice, if each function \\(f\_n(x)\\) is non-negative on \\(A\\), then the limit function \\(f(x)\\) will also be non-negative on \\(A\\). This reflects the fact that the limit of a sequence of non-negative real numbers cannot be negative.
- If each \\( f\_n \\) is non-decreasing on \\( A \\), then \\( f \\) is non-decreasing on \\( A \\). Consequently, if each function \\(f\_n\\) is non-decreasing on \\(A\\), then the limit function \\(f\\) will also be non-decreasing on \\(A\\). In other words, the property of monotonicity is preserved under pointwise convergence.
## Uniform convergence
Let \\( (f\_n) \\) be a sequence of functions defined on a set \\( A \\subseteq \\mathbb{R} \\). We say that \\( (f\_n) \\) **converges uniformly** on \\( A \\) to the function \\( f : A \\to \\mathbb{R} \\) if, for any \\( \\varepsilon > 0 \\), there exists a [natural number](../natural-numbers.md) \\( K \\) such that, for all \\( n \\geq K \\) and all \\( x \\in A \\), the following inequality holds:
\\[
|f\_n(x) - f(x)| < \\varepsilon
\\]
If \\( (f\_n) \\) converges uniformly to \\( f \\), then \\( (f\_n) \\) also converges pointwise to \\( f. \\)
* * *
Let us consider the sequence of functions:
\\[
f\_n(x) = \\frac{x}{n}, \\quad x \\in \[0,1\], \\quad n \\in \\mathbb{N}.
\\]
For each fixed \\(x\\) in the interval \\(\[0,1\]\\), we have:
\\[
\\lim\_{n \\to \\infty} f\_n(x) = 0
\\]
This means that the sequence \\(f\_n(x)\\) converges pointwise to the function (f(x) = 0). Let us now check whether the convergence is uniform on \\(\[0,1\]\\). We compute the difference between \\(f\_n(x)\\) and the limit function \\(f(x)\\):
\\[
|f\_n(x) - f(x)| = \\left| \\frac{x}{n} - 0 \\right| = \\frac{x}{n}
\\]
The maximum value of this difference on the interval \\(\[0,1\]\\) is:
\\[
\\sup\_{x \\in \[0,1\]} |f\_n(x) - f(x)| = \\frac{1}{n}
\\]
Given any \\(\\varepsilon > 0\\), we can choose \\(N\\) such that:
\\[
\\frac{1}{N} < \\varepsilon
\\]
Therefore, for all \\(n \\geq N\\) and for all \\(x \\in \[0,1\]\\), we have:
\\[
|f\_n(x) - f(x)| < \\varepsilon
\\]
This confirms that the sequence of functions \\(f\_n(x) = \\frac{x}{n}\\) converges uniformly to the limit function \\(f(x) = 0\\) on the interval \\(\[0,1\]\\).
