Limit of Image of Sequence

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

Let $$S \subseteq M$$ be an open set in $$M$$.

Let $$f$$ be a mapping which is continuous on $$S$$.

Let $$\left \langle {x_n} \right \rangle$$ be a sequence of points in $$S$$ such that $$\lim_{n \to \infty} x_n = \xi$$, where $$\xi \in S$$.

Then $$\lim_{n \to \infty} f \left({x_n}\right) = f \left({\xi}\right)$$.

Alternatively, this can be put as $$\lim_{n \to \infty} f \left({x_n}\right) = f \left({\lim_{n \to \infty} x_n}\right)$$.

That is, for a continuous function, the limit and function symbols commute.

Proof
From Limit of Function by Convergent Sequences, we have:

$$\lim_{x \to \xi} f \left({x}\right) = l$$ iff, for each sequence $$\left \langle {x_n} \right \rangle$$ of points of $$S$$ such that $$\forall n \in \N^*: x_n \ne \xi$$ and $$\lim_{n \to \infty} x_n = \xi$$, it is true that $$\lim_{n \to \infty} f \left({x_n}\right) = l$$.

The result follows directly from this and the definition of continuity.