Limit of Image of Sequence

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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 $\displaystyle \lim_{n \to \infty} x_n = \xi$, where $\xi \in S$.


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

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


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


Proof

From Limit of Function by Convergent Sequences, we have:

$\displaystyle \lim_{x \to \xi} f \left({x}\right) = l$ 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 $\displaystyle \lim_{n \to \infty} x_n = \xi$, it is true that $\displaystyle \lim_{n \to \infty} f \left({x_n}\right) = l$.

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

$\blacksquare$


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