Limit of Image of Sequence

Theorem
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.

Let $f: A_1 \to A_2$ be a mapping which is continuous at $a \in A_1$.

Let $\sequence {x_n}$ be a sequence of points in $A_1$ such that:
 * $\ds \lim_{n \mathop \to \infty} x_n = a$

where $\ds \lim_{n \mathop \to \infty} x_n$ is the limit of $x_n$.

Then:
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$

That is:
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f {\lim_{n \mathop \to \infty} x_n}$

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

Real Number Line
On the real number line, this result becomes as follows:

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


 * $\ds \lim_{x \mathop \to a} \map f x = \map f a$


 * for each sequence $\sequence {x_n}$ of points of $A_1$ such that:
 * $\forall n \in \N_{>0}: x_n \ne a$
 * and:
 * $\ds \lim_{n \mathop \to \infty} x_n = a$
 * it is true that:
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$
 * $\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f a$

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