4.6 Limits of functions vs limits of sequences

Limits of functions are closely related to limits of sequences. A precise connection between the two concepts is provided by the following lemma.

Lemma 4.59.

Let $I\subseteq\mathbb{R}$ be an interval and $a\in I$ . Let $f\colon E\to\mathbb{R}$ where either $E=I$ or $E=I\setminus\{a\}$ . Suppose $\displaystyle\lim_{x\to a}f(x)=\ell$ .

If $(a_{n})_{n\in\mathbb{N}}$ satisfies $a_{n}\in I\setminus\{a\}$ for all $n\in\mathbb{N}$ and $a_{n}\to a$ as $n\to\infty$ , then

\lim_{n\to\infty}f(a_{n})=\ell.

Proof.

Let $\varepsilon>0$ . Since $\lim_{x\to a}f(x)=\ell$ , there exists some $\delta>0$ such that

(4.11) (4.11)

|f(x)-\ell|<\varepsilon\qquad\text{for all $x\in I$ with $0<|x-a|<\delta$.}

Since $a_{n}\in I\setminus\{a\}$ , we must have $|a_{n}-a|>0$ for all $n\in\mathbb{N}$ . Moreover, since $a_{n}\to a$ as $n\to\infty$ , there exists some $N\in\mathbb{N}$ such that

(4.12) (4.12)

0<|a_{n}-a|<\delta\qquad\text{for all $n>N$.}

By combining (4.12) with (4.11), we see that for all $n>N$ we have $|f(a_{n})-\ell|<\varepsilon$ . Hence, by the $\varepsilon$ - $N$ definition, $f(a_{n})\to\ell$ as $n\to\infty$ . ∎

As a special case of the above result, we deduce the following important property of continuous functions.

Theorem 4.60 (Sequential continuity).

Let $I\subseteq\mathbb{R}$ be an interval and $f\colon I\to\mathbb{R}$ be continuous at $a\in I$ . If $(a_{n})_{n\in\mathbb{N}}$ satisfies $a_{n}\in I$ for all $n\in\mathbb{N}$ and $a_{n}\to a$ as $n\to\infty$ , then

\lim_{n\to\infty}f(a_{n})=f\big{(}\lim_{n\to\infty}a_{n}\big{)}=f(a).

Proof.

Since $f$ is continuous at $a$ , it follows that $\lim_{x\to a}f(x)=f(a)$ by the characterisation from Lemma 4.42. The result then follows from Lemma 4.59 with $\ell=f(a)$ . ∎

Example 4.61.

Since $\sin\colon\mathbb{R}\to\mathbb{R}$ is continuous at $0$ and $1/n\to 0$ as $n\to\infty$ , it follows from Theorem 4.60 that $\sin(1/n)\to\sin(0)=0$ as $n\to\infty$ .

The following example shows that continuity is needed to guarantee the conclusion of the theorem!

Example 4.62.

Recall from Exercise 4.31 that the function $f\colon\mathbb{R}\to\mathbb{R}$ defined by

f(x):=\begin{cases}0&\text{if $x\leq 0$,}\\ 1&\text{if $x>0$}\end{cases}

is not continuous at $0$ . Let $a_{n}:=\frac{1}{n}$ for all $n\in\mathbb{N}$ . Then $a_{n}\to 0$ as $n\to\infty$ and, since $a_{n}>0$ , we have $f(a_{n})=1$ for all $n\in\mathbb{N}$ . Thus,

\lim_{n\to\infty}f(a_{n})=\lim_{n\to\infty}1=1

while

f\big{(}\lim_{n\to\infty}a_{n}\big{)}=f(0)=0.

Thus, the conclusion of Theorem 4.60 does not necessarily hold when $f$ is discontinuous.

Exercise 4.63.

For each of the following sequences, evaluate the limit as $n\to\infty$ by using the theory of continuous functions. Here $e:=\exp(1)$ .

(i)

$\exp(-1/n^{2})$ ;
(ii)

$\displaystyle\log\Big{(}\frac{2^{n}e+n}{2^{n}+4}\Big{)}$ .