Limit of Image of Sequence

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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:


Let $f$ be a real function which is continuous on the interval $\Bbb I$.

Let $\left \langle {x_n} \right \rangle$ be a sequence of points in $\Bbb I$ such that:

$\displaystyle \lim_{n \mathop \to \infty} x_n = \xi$

where:

$(1): \quad \xi \in \Bbb I$
$(2): \quad \displaystyle \lim_{n \mathop \to \infty} x_n$ denotes the limit of $x_n$.


Then:

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

That is:

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


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.

$\blacksquare$


Sources