# Sequential Criterion for the Limit of a Function on Metric Spaces

Recall from the Limits of Functions on Metric Spaces page that if $(S, d_S)$ and $(T, d_T)$ are both metric spaces, $A \subseteq S$, $f : A \to T$, and $p \in S$ is an accumulation point of $A$. Then the following notation:

(1)means that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in D(f) \setminus \{ p \}$ and $d_S(x, p) < \delta$ then $d_T(f(x), b) < \epsilon$.

Equivalently we said that $\lim_{x \to p} f(x) = b$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $x \in (D(f) \setminus \{ p \}) \cap B_S(p, \delta)$ then $f(x) \in B_T(b, \epsilon)$.

We will now look at an important theorem which says that $\lim_{x \to p} f(x) = b$ if and only if for every sequence $(x_n)_{n=1}^{\infty}$ in $D(f) \setminus \{ p \} = A \setminus \{ p \}$ that converge to $p$ we have that the sequences $(f(x_n))_{n=1}^{\infty}$ converge to $b$.

Theorem 1: Let $(S, d_S)$ and $(T, d_T)$ be metric spaces, $A \subseteq S$, $f : A \to T$, $p \in S$ be an accumulation point of $A$, and $b \in T$. Then $\lim_{x \to p} f(x) = b$ if and only if for all sequences $(x_n)_{n=1}^{\infty}$ in $A \setminus \{ p \}$ (sequences of elements from $A$ not containing $p$) that converge to $p$ in $S$ we have that the sequences $(f(x_n))_{n=1}^{\infty}$ converge to $b$ in $T$. |

**Proof:**$\Rightarrow$ Suppose that $\lim_{x \to p} f(x) = b$ and let $(x_n)_{n=1}^{\infty}$ be a sequence of elements from $A \setminus \{ p \}$ such that $(x_n)_{n=1}^{\infty}$ converges to $p$, i.e., $\lim_{n \to \infty} x_n = p$.

- Now since $\lim_{x \to p} f(x) = b$ then for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x \in A \setminus \{ p \}$ such that $d_S(x, p) < \delta$ $(*)$ we have that $d_T(f(x), b) < \epsilon$.

- Since $\lim_{n \to \infty} x_n = p$ we have that for $\delta > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $d_S(x_n, p) < \delta$. So whenever $n \geq N$ we have by $(*)$ that then $d_T(f(x_n), b) < \epsilon$. Therefore $\lim_{n \to \infty} f(x_n) = b$.

- $\Leftarrow$ Now suppose that for all sequences $(x_n)_{n=1}^{\infty}$ contained in $A \setminus \{ p \}$ that converge to $p$ we have that $\lim_{n \to \infty} f(x_n) = b$.

- Assume instead that $\lim_{x \to p} f(x) \neq b$. Then there exists an $\epsilon_0 > 0$ such that there exists an $x_n \in D(f) \cap B \left (p, \frac{1}{n} \right ) \setminus \{ p \}$ where $d_T(f(x_n),b) \geq \epsilon_0$. Consider the sequence $(x_n)_{n=1}^{\infty}$. Then $d_S(x_n, p) < \frac{1}{n}$ for all $n \in \mathbb{N}$ and so $\lim_{n \to \infty} d_S(x_n, p) \leq \lim_{n \to \infty} \frac{1}{n} = 0$, so $\lim_{n \to \infty} x_n = p$, i.e., $(x_n)_{n=1}^{\infty}$ converges to $p$. However $d_T(f(x_n), b) \geq \epsilon_0 > 0$ for all $n \in \mathbb{N}$ so $\lim_{n \to \infty} f(x_n) \neq b$ which is a contradiction.

- Hence, the assumption that $\lim_{x \to p} f(x) \neq b$ was false. Therefore $\lim_{x \to p} f(x) = b$. $\blacksquare$