## Limit in a metric space

### Set

todo: clean up this definition
 context $\langle X,d\rangle$ … metric space context $x$ … infinite seqeunce in $X$ definiendum $\mathrm{lim}_{n\to\infty}\ x_n$ range $\varepsilon\in\mathbb R$ range $\varepsilon>0$ range $m\in\mathbb N$ range $m\ge 0$ range $y\equiv\mathrm{lim}_{n\to\infty}\ x_n$ postulate $\forall\varepsilon.\,\exists m.\,\forall (n\ge m).\,d(x_n,y)<\varepsilon$

#### Examples

$\sum_{k=1}^\infty a_k = y$

means

$\forall (\varepsilon\in{\mathbb R}_{>0}).\,\exists (m\in{\mathbb N}).\,\forall (n\ge_{\mathbb N} m).\,| \sum_{k=1}^n a_k - y \, |<\varepsilon$

##### Infinite sum

$\sum_{n=0}^\text{Classical} f(n):=\lim_{m\to\infty}\sum_{n=0}^m f(n)$

##### Riemann integral

$h_m=\frac{(b-a)}{m}$

$\int_a^b f(x)\,{\mathrm d}x := \lim_{m\to\infty}\sum_{n=0}^{m-1} f\left(a+h_mx\right)\cdot h_m$

h = (b - a)/m;
int[f_] = Sum[f[a + h*n]*h, {n, 0, m - 1}];

(* Example f(x)=5+7x^2 *)
int[5 + 7 #^2 &] // Expand
Limit[%, m -> Infinity]

With $L_mx:=a+h_mx$, that reads

$\int_a^b f(x)\,{\mathrm d}x := (b-a)\cdot\lim_{m\to\infty} \dfrac{1}{m}\sum_{n=0}^{m-1} f\left(L_mn\right)$

Note that

$\int_a^b f(x)\,{\mathrm d}x = (b-a)\int_0^1 f(L_1(x))\,{\mathrm d}x$

so any limit of the form

$\lim_{m\to\infty} \frac{1}{m} \sum_{n=0}^{m-1} g\left(\frac{n}{m}\right)$

can be rewritten as $\int_0^1 f_g(x)\,{\mathrm d}x$, where $f_g$ is a reverse engineered function from $g$.

##### Infinite upper bound

$\int_a^\infty f(x)\,{\mathrm d}x := \lim_{b\to \infty} \int_a^b f(x)\,{\mathrm d}x$

#### sci

7944543

I take one for the notebook:

In calculus/analysis, infinity isn't used as an entity (like a number), but instead

limit n to infinity

means

for whatever m you choose (arbitry), there is such and such, that such and such

For example you may consider a sequence $s_n$ given by 1/2, 1/4, 1/8, 1/16, … then the “limit of n to infinity” is the number y=0. Why? Because for all real numbers $\varepsilon$ bigger than zero, you can find a natural number m, so that for all numbers n after that, you have that $s_n$ became smaller than $\varepsilon$. For example, choose the small number $\varepsilon = 0.0003$. The sequence actualy becomes forever smaller than that after, say, m = 4000. Indeed, for any n after 4000, the number $s_n$ is something smaller than 1/4000, which is 0.00025.

So the limit to infinity is formalized as something to do with

for arbitrary high values of the index, the thing itself is still restricted.

Coming back to the definition: The limit of n to infinity of a seqeunce $s_n$ is y, if for all real numbers $\varepsilon$ bigger than zero, you can find a natural number m, so that for all numbers n after that, you have that the difference (here given by the distance on the real number line) between the value $s_n$ and this y became smaller than $\varepsilon$.

In formulas

$\lim_{n \to \infty} s_n = y$

iff

$\forall ( \varepsilon \in {\mathbb R}_{>0} ) . \, \exists ( m \in {\mathbb N} ) . \, \forall ( n \ge_{\mathbb N} m) . \, | s_n - y \, |<\varepsilon$

Another example: Consider again the sequence $s_n$ given by 1/2, 1/4, 1/8, 1/16, … and create a new sequence

$S_m = \sum_{k=1}^n a_n$

which has members 1/2, 1/2+1/4, 1/2+1/4+1/8, …

You can prove that with y=1 the above formula regarding $\forall ( \varepsilon \in {\mathbb R}_{>0} )$ holds.

So we say

$\sum_{k=0}^\infty s_n := \lim_{n \to \infty} \sum_{k=0}^n s_k = 1$

#### Reference

Wikipedia: Limit of a seqence

#### Context 