Limit of Image of Sequence

Theorem
Let $M_1 = \left({A_1, d_1}\right)$ and $M_2 = \left({A_2, d_2}\right)$ be metric spaces.

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

Let $\left \langle {x_n} \right \rangle$ be a sequence of points in $A_1$ such that:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n = a$

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

Then:
 * $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({a}\right)$

That is:
 * $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({\lim_{n \mathop \to \infty} x_n}\right)$

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:


 * $\displaystyle \lim_{x \mathop \to a} f \left({x}\right) = f \left({a}\right)$

iff
 * for each sequence $\left \langle {x_n} \right \rangle$ of points of $A_1$ such that:
 * $\forall n \in \N_{>0}: x_n \ne a$
 * and:
 * $\displaystyle \lim_{n \mathop \to \infty} x_n = a$
 * it is true that:
 * $\displaystyle \lim_{n \mathop \to \infty} f \left({x_n}\right) = f \left({a}\right)$

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