Continuous Functions of Convergent Sequences

Passing a Sequence Limit Through a Function

A sequence can approach a number while its terms are transformed by a function. If the function behaves continuously at the number being approached, the transformed terms approach the function’s value there. This gives a useful way to find limits without separately analyzing every transformed term.

The location matters: the function must be continuous at the limiting input. It is not enough that the function is defined at that point, nor is continuity somewhere else relevant. We will make this condition precise, prove the limit rule, and see what can go wrong when continuity fails.

Continuity at a Point

Let $D$ be a subset of the real numbers, let $f:D\to\mathbb{R}$, and let $c\in D$. Continuity at $c$ means that inputs in $D$ sufficiently close to $c$ have function values as close as desired to $f(c)$. The following definition expresses this using two positive tolerances.

Definition: The function $f:D\to\mathbb{R}$ is continuous at $c\in D$ if, for every $\varepsilon>0$, there exists a $\delta>0$ such that for every $x\in D$, $$ |x-c|<\delta \quad\Longrightarrow\quad |f(x)-f(c)|<\varepsilon. $$

Here $\varepsilon$ specifies how close the function values must be, and $\delta$ specifies how close the inputs must be to guarantee it. The choice of $\delta$ may depend on $\varepsilon$ and on the point $c$.

Sequence convergence has a related form: $a_n\to c$ means that for every $\delta>0$, there is an index $N$ such that $n\ge N$ implies $|a_n-c|<\delta$. Continuity lets us choose a suitable input tolerance $\delta$ for a desired output tolerance $\varepsilon$. Sequence convergence then ensures that all sufficiently late terms meet that input tolerance.

The Continuous-Function Limit Theorem

Theorem: Suppose $a_n\in D$ for every $n\in\mathbb{N}$, $a_n\to c$, and $f:D\to\mathbb{R}$ is continuous at $c$. Then $$ f(a_n)\longrightarrow f(c). $$

The conditions are important. The terms must lie in the domain of $f$, their limit $c$ must also be in the domain, and $f$ must be continuous at $c$. No assumption that $f$ is continuous everywhere is needed.

Why the Theorem Works

We prove the theorem directly from the definitions. Let $\varepsilon>0$ be any desired tolerance for the function values. Since $f$ is continuous at $c$, its definition gives a $\delta>0$ such that, for every $x\in D$,

|x-c|<\delta \quad\Longrightarrow\quad |f(x)-f(c)|<\varepsilon.

Since $a_n\to c$, convergence of the input sequence gives an index $N$ such that $n\ge N$ implies $|a_n-c|<\delta$. For each such $n$, $a_n\in D$, so the continuity implication applies with $x=a_n$. Thus $|f(a_n)-f(c)|<\varepsilon$ whenever $n\ge N$. This is exactly the definition of $f(a_n)\to f(c)$, proving the theorem.

The key move is to begin with the output tolerance $\varepsilon$, use continuity to obtain an input tolerance $\delta$, and then use convergence of $a_n$ to meet that $\delta$. The two definitions fit together in that order.

Worked Examples

Worked Example: Squaring a Sequence That Approaches 2

Let $a_n=2+\frac{1}{n}$, and let $f(x)=x^2$. Find the limit of $f(a_n)$.

First, the inputs satisfy

|a_n-2| = \left|2+\frac{1}{n}-2\right| = \frac{1}{n} \longrightarrow 0.

Thus $a_n\to 2$. The function $f(x)=x^2$ is continuous at $2$, so the theorem gives

\lim_{n\to\infty} f(a_n)=f(2)=2^2=4.

To check the result directly, substitute $a_n$:

f(a_n) = \left(2+\frac{1}{n}\right)^2 = 4+\frac{4}{n}+\frac{1}{n^2}.

For $n\ge 1$, $1/n^2\le 1/n$, so

|f(a_n)-4| = \frac{4}{n}+\frac{1}{n^2} \le \frac{5}{n} \longrightarrow 0.

Therefore the transformed sequence does approach $4$, as the theorem predicts.

Worked Example: Taking Square Roots Near 9

Let $a_n=9+\frac{1}{n}$, and let $f(x)=\sqrt{x}$ on $D=[0,\infty)$. Find the limit of $f(a_n)$.

The inputs lie in $D$, and

|a_n-9| = \left|9+\frac{1}{n}-9\right| = \frac{1}{n} \longrightarrow 0.

Hence $a_n\to 9$. The square-root function is continuous at $9$, so

\lim_{n\to\infty}\sqrt{a_n} = \sqrt{9} = 3.

For a direct check, rationalize the difference:

\left|\sqrt{a_n}-3\right| = \left|\sqrt{9+\frac{1}{n}}-3\right| = \frac{\frac{1}{n}}{\sqrt{9+\frac{1}{n}}+3}.

The denominator is greater than $6$, since $\sqrt{9+\frac{1}{n}}>3$. Therefore

0\le \left|\sqrt{a_n}-3\right| < \frac{1}{6n} \longrightarrow 0.

This confirms that the output sequence approaches $3$.

Worked Example: Taking Reciprocals Near 2

Let $a_n=2+\frac{1}{n}$, and let $f(x)=\frac{1}{x}$ on $D=\mathbb{R}\setminus\{0\}$. Find the limit of $f(a_n)$.

Each $a_n$ is positive, so it belongs to $D$, and $|a_n-2|=1/n\to 0$. The reciprocal function is continuous at $2$, which is in its domain. Thus

\lim_{n\to\infty} f(a_n) = f(2) = \frac{1}{2}.

Substitution gives $f(a_n)=\frac{1}{2+1/n}$. To check the limit, calculate the difference:

\left|\frac{1}{2+\frac{1}{n}}-\frac{1}{2}\right| = \left|\frac{2-(2+\frac{1}{n})}{2(2+\frac{1}{n})}\right| = \frac{\frac{1}{n}}{2(2+\frac{1}{n})} < \frac{1}{4n}.

The last bound tends to zero, so the output sequence indeed approaches $1/2$.

Continuity Is the Condition That Protects the Limit

It is tempting to treat the theorem as saying that every function preserves sequence limits. That is false. A function can be defined at the limiting input but fail to be continuous there, and the function values of a sequence approaching that input may then fail to approach the function’s value at the limit.

Worked Example: A Discontinuous Function Can Break the Rule

Define $f:\mathbb{R}\to\mathbb{R}$ by

f(x)= \begin{cases} 0, & x=0,\\ 1, & x\ne 0. \end{cases}

Take $a_n=1/n$. Since $|a_n-0|=1/n\to 0$, we have $a_n\to 0$. But every term $a_n$ is nonzero, so

f(a_n)=1\quad\text{for every }n, \qquad f(0)=0.

Consequently $f(a_n)\to 1$, not $f(0)=0$. The proposed conclusion fails because $f$ is not continuous at $0$. In fact, taking $\varepsilon=1/2$, every $\delta>0$ admits a nonzero $x$ with $|x|<\delta$, while $|f(x)-f(0)|=1$, so the continuity condition fails.

This example also clarifies what “continuity is necessary” means. Continuity at $c$ guarantees that every sequence in the domain converging to $c$ has function values converging to $f(c)$. If continuity fails, the guarantee for all such sequences fails; a particular sequence may still happen to give the expected limit.

Takeaway: If $a_n\to c$ and $f$ is continuous at $c$, then $f(a_n)\to f(c)$. The function’s value at the limiting input is the output limit because continuity controls nearby inputs.

A Practical Way to Apply the Theorem

When using this result, keep the input limit and the function’s continuity condition distinct. The following sequence of checks prevents a common error: evaluating $f(c)$ before confirming that the theorem applies.

Identify the input limit.
Write $a_n\to c$, and verify that the sequence terms approach $c$.

Check the domain.
Confirm that the terms $a_n$ and the limiting input $c$ are in the domain of $f$.

Check continuity at the limit.
Verify that $f$ is continuous at $c$, rather than assuming it is continuous everywhere.

Evaluate the output limit.
Once the conditions hold, conclude $f(a_n)\to f(c)$.

For example, polynomial functions are continuous at every real input, while the reciprocal function is continuous only at inputs in its domain, which excludes zero. Domain restrictions are therefore part of applying the theorem, not an optional detail.

Why This Limit Rule Matters

The theorem connects two ways of studying a function: the behavior of its values near an input and the behavior of a sequence of inputs approaching that point. Rather than expand or simplify every transformed sequence, one can often identify the input limit and evaluate the function at that limit. The proof explains exactly why this shortcut is valid and pinpoints its condition.

The same reasoning is useful whenever a sequence limit is placed inside a familiar continuous function. But the continuity check remains essential: a function’s formula alone does not justify passing a limit through it, and being defined at the limiting input does not guarantee the required behavior.

Check Your Understanding

Answer each question by referring to the theorem’s conditions, not only to its conclusion.

State the continuous-function limit theorem, including its conditions.
In the proof, why do we first use continuity to choose $\delta$, and then use convergence of $a_n$ to choose an index $N$?
If $a_n\to 5$ and $f$ is continuous at $5$, what is the limit of $f(a_n)$?
Why does the reciprocal function require a separate domain check when applying the theorem?
For the function that equals $0$ at $0$ and $1$ elsewhere, what are the limits of $a_n=1/n$ and $f(a_n)$, and why does this not contradict the theorem?