Why a Sequence Limit Must Agree with Its Recurrence
A recursive definition produces each term from the one before it. If the terms settle toward a limit, the recurrence offers a way to determine what that limit must be. The key idea is simple: far enough along a convergent sequence, consecutive terms are both close to the same limit. If the recurrence map is continuous, applying it to terms close to the limit produces values close to the map's value at the limit.
This gives a necessary equation for the limit. It does not, by itself, prove that the sequence converges. Nor does it necessarily determine which solution of the equation is the limit. Keeping those distinctions clear is essential when using recursive definitions.
A number \(L\in D\) is a fixed point of \(f\) if \(f(L)=L\). The word “fixed” describes what happens when the map is applied: the input \(L\) is returned unchanged.
The Limit Condition
To evaluate \(f(L)\), the limit \(L\) must belong to the domain of \(f\). For the result below, we also require that \(f\) be continuous at \(L\). Continuity at \(L\) means that for every \(\varepsilon>0\), there is a \(\delta>0\) such that, for every \(x\in D\), $$ |x-L|<\delta \quad\Longrightarrow\quad |f(x)-f(L)|<\varepsilon. $$ This condition says that inputs in the domain sufficiently close to \(L\) have outputs close to \(f(L)\).
The conditions are local: continuity is needed at the limit \(L\), not necessarily at every point in \(D\). In applications, it is common to know that \(f\) is continuous on an interval containing the sequence and its limit, which is more than enough.
Why the Theorem Holds
The recurrence relates \(x_{n+1}\) to \(f(x_n)\). We can take limits on both sides only after identifying the limit of each side. A shifted sequence \(x_{n+1}\) has the same limit as \(x_n\), while continuity ensures that \(f(x_n)\) tends to \(f(L)\).
Given \(\varepsilon>0\), convergence gives an \(N\) such that \(n\geq N\) implies \(|x_n-L|<\varepsilon\). Then \(n\geq N\) also implies \(n+1\geq N\), so \(|x_{n+1}-L|<\varepsilon\). Thus \(x_{n+1}\to L\).
Given \(\varepsilon>0\), continuity at \(L\) supplies a \(\delta>0\) such that \(x\in D\) and \(|x-L|<\delta\) imply \(|f(x)-f(L)|<\varepsilon\). Since \(x_n\to L\), eventually \(|x_n-L|<\delta\). Hence \(f(x_n)\to f(L)\).
For every \(n\), \(x_{n+1}=f(x_n)\). The left side tends to \(L\), and the right side tends to \(f(L)\). A convergent real sequence has only one limit, so \(L=f(L)\).
This is the complete reason the fixed-point equation follows. It is not a guess based on a few terms: it depends on convergence, the recurrence holding at every index, continuity at the limit, and the uniqueness of limits.
Worked Examples
Worked Example: A Rational Recurrence
Consider the sequence defined by \(x_1=2\) and $$ x_{n+1}=2-\frac{1}{x_n}. $$ We can identify its limit and verify the fixed-point condition.
The first terms are $$ x_1=2,\qquad x_2=2-\frac12=\frac32,\qquad x_3=2-\frac{2}{3}=\frac43,\qquad x_4=2-\frac{3}{4}=\frac54. $$ These values suggest the formula \(x_n=1+\frac1n\). Check it by induction. For \(n=1\), the formula gives \(1+1=2=x_1\). If \(x_n=1+\frac1n=\frac{n+1}{n}\), then $$ x_{n+1} =2-\frac{1}{(n+1)/n} =2-\frac{n}{n+1} =\frac{2n+2-n}{n+1} =\frac{n+2}{n+1} =1+\frac{1}{n+1}. $$ So the formula holds for every \(n\in\mathbb{N}\), and therefore \(x_n\to1\).
Here \(f(x)=2-\frac1x\), which is continuous at \(1\). The theorem requires the limit to satisfy $$ L=f(L)=2-\frac1L. $$ Using the limit \(L=1\), both sides equal \(1\). Alternatively, solving the equation gives \(L=2-\frac1L\), so \(L^2=2L-1\), hence \((L-1)^2=0\) and \(L=1\). Finally, direct substitution checks the fixed point: \(f(1)=2-1=1\).
Worked Example: A Squaring Recurrence
Let \(x_1=\frac12\) and \(x_{n+1}=x_n^2\). The recurrence map is \(f(x)=x^2\), a continuous function. The terms begin $$ x_1=\frac12,\qquad x_2=\frac14,\qquad x_3=\frac1{16},\qquad x_4=\frac1{256}. $$ In fact, \(x_n=(\frac12)^{2^{n-1}}\). The formula holds at \(n=1\). If it holds at \(n\), then $$ x_{n+1}=x_n^2 =\left(\left(\frac12\right)^{2^{n-1}}\right)^2 =\left(\frac12\right)^{2^n}, $$ which is the formula with \(n+1\) in place of \(n\). Since the positive powers \(2^{n-1}\) grow without bound and \(0<\frac12<1\), the terms tend to \(0\). Thus \(L=0\).
The fixed-point equation is $$ L=f(L)=L^2. $$ At the sequence's limit, \(0=0^2\), and direct substitution gives \(f(0)=0\). The equation also has the solution \(L=1\), but the explicit sequence here tends to \(0\). The theorem says that the limit must be a fixed point; the equation alone does not say that every fixed point is the limit.
Worked Example: Why Continuity Matters
Define \(f:[0,\infty)\to[0,\infty)\) by $$ f(x)=\frac{x}{1+x}\ \text{if }x>0, \qquad f(0)=1. $$ Starting from \(x_1=1\), use \(x_{n+1}=f(x_n)\). We claim that \(x_n=\frac1n\). This holds for \(n=1\). If \(x_n=\frac1n>0\), then $$ x_{n+1}=f\left(\frac1n\right) =\frac{1/n}{1+1/n} =\frac{1/n}{(n+1)/n} =\frac1{n+1}. $$ Thus \(x_n=\frac1n\) for every \(n\), and \(x_n\to0\).
But \(f(0)=1\), so the limit does not satisfy \(L=f(L)\): it would require \(0=1\). There is no contradiction with the theorem because \(f\) is not continuous at \(0\). For positive \(x\), \(f(x)=x/(1+x)\) tends to \(0\) as \(x\) approaches \(0\), whereas the defined value \(f(0)\) is \(1\). This example shows that continuity at the limit cannot simply be omitted.
A Common Mistake: Taking Limits Without Checking Continuity
A tempting calculation is to start from \(x_{n+1}=f(x_n)\), write “take limits,” and conclude \(L=f(L)\). That step is valid when the sequence converges to \(L\), \(L\) lies in the domain, and \(f\) is continuous at \(L\). Without continuity, the implication \(x_n\to L\Rightarrow f(x_n)\to f(L)\) may fail.
The discontinuous example makes the error concrete. Its terms satisfy \(x_n=1/n\), so \(x_n\to0\), and the recurrence gives \(x_{n+1}=f(x_n)\). But \(f(x_n)=1/(n+1)\to0\), while \(f(0)=1\). The false step would be to replace the limit of \(f(x_n)\) by \(f(0)\). The correction is to check continuity before making that replacement: here the two values differ precisely because continuity fails.
There is a second point to watch: the theorem gives a necessary condition, not a convergence test. Solving \(L=f(L)\) finds possible limits, but the equation does not establish that the terms converge. Convergence must be known or proved by a separate argument, for example by applying the monotone convergence theorem after verifying its hypotheses. Even once convergence is established, an equation with multiple solutions may leave more than one candidate.
Why the Fixed-Point Equation Is Useful
For a convergent recursive sequence, the fixed-point equation turns a limit problem into an algebra problem. One first proves or establishes convergence, then uses continuity to derive \(L=f(L)\), and finally solves that equation to obtain necessary candidates. This can make the limit easier to identify than evaluating the terms directly.
The order matters. Numerical terms may suggest a limit, but they do not prove it. A fixed-point calculation may produce plausible values, but it does not prove convergence or decide among multiple candidates. The theorem links a known limit to the recurrence under precise conditions; it does not replace the work needed to establish convergence.
Check Your Understanding
Use the theorem's assumptions and conclusion to answer these questions.
- Why does the shifted sequence \(x_{n+1}\) have the same limit as \(x_n\)?
- Which continuity condition is required to conclude that \(f(x_n)\to f(L)\)?
- If a recursive sequence converges to \(L\), what equation must \(L\) satisfy under the theorem's assumptions?
- Why does solving \(L=f(L)\) not by itself prove that the sequence converges?
- In the discontinuous example, what are \(\lim_{n\to\infty} f(x_n)\) and \(f(\lim_{n\to\infty}x_n)\), and why are they different?