*This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.*

# The Sequential Criterion for a Limit of a Function

We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function $f$ at a cluster point $c$ from $A$ with regards to sequences $(a_n)$ from $A$ that converge to $c$.

Theorem 1 (The Sequential Criterion for a Limit of a Function): Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then $\lim_{x \to c} f(x) = L$ if and only if for all sequences $(a_n)$ from the domain $A$ where $a_n \neq c$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = c$ then $\lim_{n \to \infty} f(a_n) = L$. |

Consider a function $f$ that has a limit $L$ when $x$ is close to $c$. Now consider all sequences $(a_n)$ from the domain $A$ where these sequences converge to $c$, that is $\lim_{n \to \infty} a_n = c$. The Sequential Criterion for a Limit of a Function says that then that as $n$ goes to infinity, the function $f$ evaluated at these $a_n$ will have its limit go to $L$.

For example, consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by the equation $f(x) = x$, and suppose we wanted to compute $\lim_{x \to 0} x$. We should already know that this limit is zero, that is $\lim_{x \to 0} x = 0$. Now consider the sequence $(a_n) = \left ( \frac{1}{n} \right)$. This sequence $(a_n)$ is clearly contained in the domain of $f$. Furthermore, this sequence converges to 0, that is $\lim_{n \to \infty} \frac{1}{n} = 0$. If all such sequences $(a_n)$ that converge to $0$ have the property that $(f(a_n))$ converges to $f(0) = 0$, then we can say that $\lim_{n \to 0} f(x) = 0$.

We will now look at the proof of The Sequential Criterion for a Limit of a Function.

**Proof:**$\Rightarrow$ Suppose that $\lim_{x \to c} f(x) = L$, and let $(a_n)$ be a sequence in $A$ such that $a_n \neq c$ $\forall n \in \mathbb{N}$ such that $\lim_{n \to \infty} a_n = c$. We thus want to show that $\lim_{n \to \infty} f(a_n) = L$.

- Let $\epsilon > 0$. We are given that $\lim_{x \to c} f(x) = L$ and so for $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then we have that $\mid f(x) - L \mid < \epsilon$. Now since $\delta > 0$, since we have that $\lim_{n \to \infty} a_n = c$ then there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - c \mid < \delta$. Therefore $a_n \in V_{\delta} (c) \cap A$.

- Therefore it must be that $\mid f(a_n) - L \mid < \epsilon$, in other words, $\forall n ≥ N$ we have that $\mid f(a_n) - L \mid < \epsilon$ and so $\lim_{n \to \infty} f(a_n) = L$.

- $\Leftarrow$ Suppose that for all $(a_n)$ in $A$ such that $a_n \neq c$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = c$, we have that $\lim_{n \to \infty} f(a_n) = L$. We want to show that $\lim_{x \to c} f(x) = L$.

- Suppose not, in other words, suppose that $\exists \epsilon_0 > 0$ such that $\forall \delta > 0$ then $\exists x_{\delta} \in A \cap V_{\delta} (c) \setminus \{ c \}$ such that $\mid f(x_{\delta}) - L \mid ≥ \epsilon_0$. Let $\delta_n = \frac{1}{n}$. Then there exists $x_{\delta_n} = a_n \in A \cap V_{\delta_n} (c) \setminus \{ c \}$, in other words, $0 < \mid a_n - c \mid < \frac{1}{n}$ and $\lim_{n \to \infty} a_n = c$. However, $\mid f(a_n) - L \mid ≥ \epsilon_0$ so $\lim_{n \to \infty} f(a_n) \neq L$, a contradiction. Therefore $\lim_{x \to c} f(x) = L$. $\blacksquare$