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:

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

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

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

where:

$(1): \quad \xi \in \Bbb I$

$(2): \quad \ds \lim_{n \mathop \to \infty} x_n$ denotes the limit of $x_n$.

Then:

$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f \xi$

That is:

$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f {\lim_{n \mathop \to \infty} x_n}$

Proof

From Limit of Function by Convergent Sequences, we have:

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

if and only if:

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:

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

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

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $2$: Metric Spaces: $\S 5$: Limits