nyuuzyou/wikis huggingface dataset:
Limit of Decreasing Sequence of Left Half-Open Intervals with Lower Bound Converging to Upper Bound

== Theorem ==
Let $a, b \in \R$ have $a < b$.
Let $\sequence {a_n}_{n \mathop \in \N}$ be an increasing sequence with $a_n \to b$. 
Then we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b = \set b$ 
That is: 
:$\hointl {a_n} b \downarrow \set b$
where $\downarrow$ denotes the limit of decreasing sequence of sets.
== Proof ==
Clearly:
:$\ds b \in \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b$
so that: 
:$\ds \set b \subseteq \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b$
Now let: 
:$\ds x \in \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b$
Then: 
:$a_n \le x \le b$ 
for each $n \in \N$.
From Limits Preserve Inequalities, we then have: 
:$b \le x \le b$
so that: 
:$x = b$ 
So: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {a_n} b = \set b$
as required.
Category:Decreasing Sequences of Sets
Limit of Decreasing Sequence of Sets is Intersection

== Definition ==
Let $\sequence {E_k}_{k \mathop \in \N}$ be an decreasing sequence:
:$\forall k \in \N: E_{k + 1} \subseteq E_k$
Then $\sequence {E_k}_{k \mathop \in \N}$ has a limit such that:
:$\ds \lim_{n \mathop \to \infty} E_n = \bigcap_{k \mathop \in \N} {E_k}$
== Proof ==
Let $E = \ds \bigcap_{k \mathop \in \N} {E_k}$.
Let $x \in E$.
By definition of set intersection:
:$\forall E_n \in \sequence {E_k}_{k \mathop \in \N}: x \in E_n$
Thus $x \in E_n$ for all but finitely many (that is, zero) terms of $\sequence {E_k}_{k \mathop \in \N}$.
That is:
:$x \in \ds \liminf_{n \mathop \to \infty} E_n$
where $\ds \liminf_{n \mathop \to \infty} E_n$ denotes the limit inferior of $\sequence {E_k}_{k \mathop \in \N}$.
Hence $\ds E \subseteq \liminf_{n \mathop \to \infty} E_n$.
Let $x \in \ds \limsup_{n \mathop \to \infty} E_n$.
 $\exists n \in \N: x \notin E_n$.
Then as $E_{n + 1} \subseteq E_n$ it follows that $n \notin E_{n + 1}$.
Hence:
:$\forall m \in \N: m > n: x \notin E_m$
and so there is only a finite number of $i \in \N$ for which $x \in E_i$, that is, where $m < n$.
That is, by definition of limit superior:
:$x \notin \ds \limsup_{n \mathop \to \infty} E_n$
But this contradicts our assertion that $x \in \ds \limsup_{n \mathop \to \infty} E_n$.
Hence:
:$\forall n \in \N: x \in E_n$
and so by definition of set intersection:
:$x \in E$
So we have shown that:
:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq E$
Hence we have:
:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq E$
and:
:$\ds E \subseteq \liminf_{n \mathop \to \infty} E_n$
Hence by Subset Relation is Transitive:
:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty} E_n$
and it follows from Limit of Sets Exists iff Limit Inferior contains Limit Superior that:
:$\ds \lim_{n \mathop \to \infty} E_n = \bigcup_{k \mathop \in \N} {E_k}$
== Sources ==
* : Chapter $\text {I}$: Sets of Points: $1 \cdot 1$. The algebra of sets
Category:Limits of Sets
Category:Decreasing Sequences of Sets
Limit of Decreasing Sequence of Unbounded Below Closed Intervals

== Theorem ==
Let $x \in \R$. 
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence converging to $x$ such that $x_n \ge x$ for each $n \in \N$. 
Then:
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$
== Proof ==
We first show that: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} \subseteq \hointl {-\infty} x$
Let:
:$\ds t \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Then:
:$t \le x_n$
By the Squeeze Theorem, we then have:
:$t \le x$ 
taking $n \to \infty$. 
So $t \in \hointl {-\infty} x$. 
We now show that:
:$\ds \hointl {-\infty} x \subseteq \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Now suppose that: 
:$t \in \hointl {-\infty} x$ 
Then $t \le x$. 
Since $x \le x_n$ for each $n \in \N$, we have $t \le x_n$ for each $n \in \N$.
That is, $t \in \hointl {-\infty} {x_n}$ for each $n \in \N$.
So:
:$\ds t \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
So, from the definition of set equality, we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \hointl {-\infty} x$
Category:Decreasing Sequences of Sets
Limit of Decreasing Sequence of Unbounded Below Closed Intervals with Endpoint Tending to Negative Infinity

== Theorem ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a decreasing sequence with $x_n \to -\infty$.
Then: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
That is: 
:$\hointl {-\infty} {x_n} \downarrow \O$
where $\downarrow$ denotes the limit of decreasing sequence of sets.
== Proof ==
 suppose that:
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} \ne \O$
Let: 
:$\ds x \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Then: 
:$x \in \hointl {-\infty} {x_n}$ for each $n$.
From the definition of a sequence diverging to $-\infty$:
:there exists $N \in \N$ such that $x_N < x$.
But then: 
:$x \not \in \hointl {-\infty} {x_N}$
contradicting that: 
:$x \in \hointl {-\infty} {x_n}$ for each $n$.
So we have reached a contradiction, and we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
Category:Decreasing Sequences of Sets
Limit of Difference between Consecutive Prime Numbers

== Theorem ==
The Prime Number Theorem indicates that the average value of the difference between two consecutive prime numbers is of the order of $\log p_n$.
Let $E = \ds \liminf_{n \mathop \to \infty} \dfrac {p_{n + 1} - p_n} {\log p_n}$.
If there are infinitely many twin primes, then $E = 0$.
If not, then it is not known what the value of $E$ is.
== Historical Note ==
== Sources ==
* : $0,44634 95408 \ldots$
Category:Prime Numbers
Limit of Difference between Consecutive Prime Numbers/Historical Note

== Historical Note on Limit of Difference between Consecutive Prime Numbers ==
 showed that $E \le \dfrac {15} {16}$.
 and showed in $1966$ that $E \le \dfrac {2 + \sqrt 3} 8 \approx 0 \cdotp 46650 \ldots$
In their $1983$ work , and report that deduced in $1972$ that $E \le \dfrac {2 \sqrt 2 - 1} 4 \approx 0 \cdotp 45706 \ldots$
However, it is not clear who is or was, and this has not been corroborated.
In $1973$ showed that $E \le \dfrac 1 4 + \dfrac \pi {16} \approx 0 \cdotp 44634 \, 95408 \ldots$
== Sources ==
* 
* : $0,44634 95408 \ldots$
Category:Historical Notes
Limit of Distribution Function of Finite Borel Measure at Negative Infinity

== Theorem ==
Let $\mu$ be a finite Borel measure on $\R$.
Let $F_\mu$ be the distribution function of $\mu$.
Then:
:$\ds \lim_{x \mathop \to -\infty} \map {F_\mu} x = 0$
== Proof ==
From Sequential Characterisation of Limit at Minus Infinity of Real Function: Corollary, we aim to show that: 
:for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\ds \lim_{n \mathop \to \infty} \map {F_\mu} {x_n} = 0$.
Since $\sequence {x_n}_{n \mathop \in \N}$ is decreasing, we have: 
:the sequence $\sequence {\hointl {-\infty} {x_n} }_{n \mathop \in \N}$ is decreasing.
From Limit of Decreasing Sequence of Unbounded Below Closed Intervals with Endpoint Tending to Negative Infinity:
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
Further, since $\mu$ is a finite measure, we have: 
:$\ds \map \mu {\hointl {-\infty} {x_1} } < \infty$
So Measure of Limit of Decreasing Sequence of Measurable Sets is admissible, and gives: 
:$\ds \map \mu {\bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} } = \lim_{n \mathop \to \infty} \map {F_\mu} {x_n}$
Since: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
we have: 
:$\ds \map \mu {\bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} } = 0$
from Empty Set is Null Set.
So:
:$\ds \lim_{n \mathop \to \infty} \map {F_\mu} {x_n} = 0$
Since the sequence $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have: 
:$\ds \lim_{x \mathop \to -\infty} \map {F_\mu} x = 0$
by Sequential Characterisation of Limit at Minus Infinity of Real Function: Corollary.
== Sources ==
* : $1.3$: Outer Measures
Category:Distribution Function of Finite Borel Measure
Limit of Error in Stirling's Formula

== Theorem ==
Consider Stirling's Formula:
:$n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
The ratio of $n!$ to its approximation $\sqrt {2 \pi n} \paren {\dfrac n e}^n$ is bounded as follows:
:$e^{1 / \paren {12 n + 1} } < \dfrac {n!} {\sqrt {2 \pi n} n^n e^{-n} } < e^{1 / 12 n}$
== Proof ==
Taking the natural logarithm of the ratio of $n!$ to its approximation $\sqrt {2 \pi n} \paren {\dfrac n e}^n$, we obtain:
From the argument in Stirling's Formula: Proof 2: Lemma 3 we have that $\sequence {d_n - \dfrac 1 {12 n} }$ is an increasing sequence.
Then:
Let:
:$\map f x := \dfrac 1 {2 x} \map \ln {\dfrac {1 + x} {1 - x} } - 1$
for $\size x < 1$.
By Stirling's Formula: Proof 2: Lemma 1:
As $-1 < \dfrac 1 {2 n + 1} < 1$ it can be substituted for $x$ in $(2)$:
Therefore the sequence $\sequence {d_n - \dfrac 1 {12 n + 1} }$ is decreasing.
From Stirling's Formula, we have that:
:$\ds \lim_{n \mathop \to \infty} d_n = \ln \sqrt {2 \pi}$
and so:
:$d_n - \dfrac 1 {12 n} < \ln \sqrt {2 \pi} < d_n - \dfrac 1 {12 n + 1}$
for $n = 1, 2, 3, \ldots$
Therefore:
== Examples ==
== Also see == 
* Stirling's Formula for Gamma Function
== Sources ==
* : $\S 17.4 \ (3)$
* 
Category:Stirling's Formula
Limit of Function by Convergent Sequences

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $S \subseteq A_1$ be an open set of $M_1$.
Let $f: S \to A_2$ be a mapping defined on $S$, except possibly at the point $c \in S$.
Then $\ds \lim_{x \mathop \to c} \map f x = l$ :
:for each sequence $\sequence {x_n}$ of points of $S$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$
it is true that:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
=== Real Number Line ===
On the real number line, this result becomes as follows:
=== Corollary ===
The above result holds for a real function tending to a limit both from the right and from the left:
=== Complex Plane ===
On the complex plane, this result becomes as follows:
== Proof ==
=== Necessary Condition ===
Suppose that:
:$\ds \lim_{x \mathop \to c} \map f x = l$
Let $\epsilon > 0$.
Then by the definition of the limit of a mapping:
:$\exists \delta > 0: \map {d_2} {\map f x, l} < \epsilon$
provided $0 < \map {d_1} {x, c} < \delta$.
Now suppose that $\sequence {x_n}$ is a sequence of points of $S$ such that such that:
:$\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$
Since $\delta > 0$, from the definition of the limit of a sequence:
:$\exists N \in \N_{>0}: \forall n > N: \map {d_1} {x_n, c} < \delta$
But:
:$\forall n \in \N_{>0}: x_n \ne c$
That means:
:$0 < \map {d_1} {x_n, c} < \delta$
by the definition of a metric.
But that implies:
:$\map {d_2} {\map f {x_n}, l} < \epsilon$
That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
:$\forall n > N: \map {d_2} {\map f {x_n}, l} < \epsilon$
Thus:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
=== Sufficient Condition ===
Now suppose that for each sequence $\sequence {x_n}$ of points of $S$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$, it is true that:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
 it is not true that:
:$\ds \lim_{x \mathop \to c} \map f s = l$
So:
:$\exists \epsilon > 0: \forall \delta > 0: \exists x \in S: 0 < \map {d_1} {x, c} < \delta \land \map {d_2} {\map f x, l} \ge \epsilon$
where $\land$ denotes logical and.
For all $n \in \N_{>0}$, define:
:$S_n = \set {x \in S: 0 < \map {d_1} {x, c} < \dfrac 1 n \land \map {d_2} {\map f x, l} \ge \epsilon}$
By hypothesis, $S_n$ is non-empty for all $n \in \N_{>0}$.
Using the axiom of countable choice, there exists a sequence $\sequence {x_n}$ of elements of $S$ such that $x_n \in S_n$ for all $n \in \N_{>0}$.
Then:
:$\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$
but it is not true that:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
So there is our contradiction, and so the result follows.
Category:Metric Spaces
Category:Limits of Sequences
Limit of Function by Convergent Sequences/Complex Plane

== Theorem ==
Let $f$ be a complex function defined on an open subset $S \subseteq \C$, except possibly at the point $c \in S$.
Then $\ds \lim_{x \mathop \to c} \map f z = l$ :
:for each sequence $\sequence {z_n}$ of points of $S$ such that $\forall n \in \N_{>0}: z_n \ne c$ and $\ds \lim_{n \mathop \to \infty} z_n = c$
it is true that:
:$\ds \lim_{n \mathop \to \infty} \map f {z_n} = l$
== Proof ==
=== Necessary Condition ===
Let $\ds \lim_{z \mathop \to c} \map f z = l$.
Let $\epsilon \in \R_{>0}$.
Then by definition of the limit of a complex function:
:$\exists \delta \in \R_{>0}: \cmod {\map f z - l} < \epsilon$
provided $0 < \cmod {z - c} < \delta$.
Now suppose that $\sequence {x_n}$ is a sequence of elements of $S$ such that:
:$\forall n \in \N_{>0}: z_n \ne c$
and:
:$\ds \lim_{n \mathop \to \infty} z_n = c$
Since $\delta > 0$, from the definition of the limit of a complex function:
:$\exists N \in \R_{>0}: \forall n \in \N: n > N \implies \cmod {z_n - c} < \delta$
But:
:$\forall n \in \N_{>0}: z_n \ne c$
That means:
:$0 < \cmod {z_n - c} < \delta$
But that implies:
:$\cmod {\map f {z_n} - l} < \epsilon$
That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
:$\forall n \in \N: n > N \implies \cmod {\map f {z_n} - l} < \epsilon$
Thus:
:$\ds \lim_{n \mathop \to \infty} \map f {z_n} = l$
=== Sufficient Condition ===
Suppose that for each sequence $\sequence {x_n}$ of elements of $S$ such that:
:$\forall n \in \N_{>0}: z_n \ne c$ and $\ds \lim_{n \mathop \to \infty} z_n = c$, it is true that:
::$\ds \lim_{n \mathop \to \infty} \map f {z_n} = l$
 it is not true that:
:$\ds \lim_{z \mathop \to c} \map f z = l$
Thus:
:$\exists \epsilon > 0: \forall \delta > 0: \exists x: 0 < \cmod {z - c} < \delta: \cmod {\map f {z_n} - l} \ge \epsilon$
In particular, if $\delta = \dfrac 1 n$, we can find a $z_n$ where $0 < \cmod {x - c} < \dfrac 1 n$ such that:
:$\cmod {\map f {z_n} - l} \ge \epsilon$
But then $\sequence {z_n}$ is a sequence of elements of $S$ such that:
:$\forall n \in \N_{>0}: z_n \ne c$ and $\ds \lim_{n \mathop \to \infty} z_n = c$
but for which it is not true that:
:$\ds \lim_{n \mathop \to \infty} \map f {z_n} = l$
The result follows by Proof by Contradiction.
Category:Limits of Complex Functions
Category:Limits of Sequences
Limit of Function by Convergent Sequences/Corollary

== Theorem ==
Let $\openint a b$ be an open real interval.
Let $f: \openint a b \to \R$ be a real function.
Let $l \in \R$.
Then:
:$(1): \quad \ds \lim_{x \mathop \to a^+} \map f x = l \iff \forall \sequence {x_n} \subseteq \openint a b: \lim_{n \mathop \to \infty} x_n = a \implies \lim_{n \mathop \to \infty} \map f {x_n} = l$
:$(2): \quad \ds \lim_{x \mathop \to b^-} \map f x = l \iff \forall \sequence {x_n} \subseteq \openint a b: \lim_{n \mathop \to \infty} x_n = b \implies \lim_{n \mathop \to \infty} \map f {x_n} = l$
where:
:$\ds \lim_{x \mathop \to a^+} \map f x$ denotes the limit of $f$ from the right
:$\ds \lim_{x \mathop \to b^-} \map f x$ denotes the limit of $f$ from the left.
== Proof ==
We have that $\struct {\R, d}$ is a metric space, where $d$ is the Euclidean metric on $\R$.
Thus the theorem follows immediately from Limit of Real Function by Convergent Sequences.
== Sources ==
* : $\S 8.9$
Category:Real Analysis
Category:Limits of Sequences
Limit of Function by Convergent Sequences/Real Number Line

== Theorem ==
Let $f$ be a real function defined on an open interval $\openint a b$, except possibly at the point $c \in \openint a b$.
Then $\ds \lim_{x \mathop \to c} \map f x = l$ :
::for each sequence $\sequence {x_n}$ of points of $\openint a b$ such that $\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$
:it is true that:
::$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
== Proof ==
=== Necessary Condition ===
Let $\ds \lim_{x \mathop \to c} \map f x = l$.
Let $\epsilon \in \R_{>0}$.
Then by the definition of the limit of a real function:
:$\exists \delta \in \R_{>0}: \size {\map f x - l} < \epsilon$
provided $0 < \size {x - c} < \delta$.
Now suppose that $\sequence {x_n}$ is a sequence of elements of $\openint a b$ such that:
:$\forall n \in \N_{>0}: x_n \ne c$
and:
:$\ds \lim_{n \mathop \to \infty} x_n = c$
Since $\delta > 0$, from the definition of the limit of a real function:
:$\exists N \in \R_{>0}: \forall n > N: \size {x_n - c} < \delta$
But:
:$\forall n \in \N_{>0}: x_n \ne c$
That means:
:$0 < \size {x_n - c} < \delta$
But that implies:
:$\size {\map f {x_n} - l} < \epsilon$
That is, given a value of $\epsilon > 0$, we have found a value of $N$ such that:
:$\forall n > N: \size {\map f {x_n} - l} < \epsilon$
Thus:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
=== Sufficient Condition ===
Suppose that for each sequence $\sequence {x_n}$ of elements of $\openint a b$ such that:
:$\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$, it is true that:
::$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
 it is not true that:
:$\ds \lim_{x \mathop \to c} \map f x = l$
Thus:
:$\exists \epsilon \in \R_{>0}: \forall \delta \in \R_{>0}: \exists x \in \openint a b: 0 < \size {x - c} < \delta: \size {\map f {x_n} - l} \ge \epsilon$
In particular, if $\delta = \dfrac 1 n$, we can find an $x_n$ where $0 < \size {x - c} < \dfrac 1 n$ such that:
:$\size {\map f {x_n} - l} \ge \epsilon$
But then $\sequence {x_n}$ is a sequence of elements of $\openint a b$ such that:
:$\forall n \in \N_{>0}: x_n \ne c$ and $\ds \lim_{n \mathop \to \infty} x_n = c$
but for which it is not true that:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
The result follows by Proof by Contradiction.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.3$: Limits of functions: Lemma $1.3.5$
* : $\S 8.9$
Category:Limits of Real Functions
Category:Limits of Sequences
Limit of Function in Interval

== Theorem ==
Let $f$ be a real function which is defined on the open interval $\openint a b$.
Let $\xi \in \openint a b$
Suppose that, $\forall x \in \openint a b$, either:
:$\xi \le \map f x \le x$
or:
:$x \le \map f x \le \xi$
Then $\map f x \to \xi$ as $x \to \xi$.
== Proof ==
Note that $\size {\map f x - \xi} \le \size {\xi - x}$.
From Limit of Absolute Value‎ we have that $\size {x - \xi} \to 0$ as $x \to \xi$.
The result follows from the Squeeze Theorem.
== Sources ==
* : $\S 8.15 \ (4)$
Category:Limits of Real Functions
Limit of Functions that Agree

== Theorem ==
Let $f$ and $g$ be real functions.
Let $f$ and $g$ agree for all $x$ in a deleted neighborhood of $c$.
Let the limit:
:$\ds \lim_{x \mathop \to c} \map f x$
exist.
Then the limit:
:$\ds \lim_{x \mathop \to c} \map g x$
also exists, and:
:$\ds \lim_{x \mathop \to c} \map f x = \lim_{x \mathop \to c} \map g x$
== Proof ==
By hypothesis:
:$\map f x = \map g x$
for all $x$ such that:
:$x \in \R: 0 < \size {\alpha - x} < \epsilon$
Putting $c$ for $\alpha$ and $\delta$ for $\epsilon$, this is equivalent to:
:$x \in \R: 0 < \size {x - c} < \delta$
By definition, that the limit of $\map f x$ exists is to say:
:$\exists L: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$
For all $x$ relevant, we can substitute $\map f x$ with $\map g x$ and get:
:$\exists L: \forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: 0 < \size {x - c} < \delta \implies \size {\map g x - L} < \epsilon$
The result follows by the definition of limit.
== Sources ==
* : Appendix $A$: Functions That Agree at All But One Point
Category:Limits of Real Functions
Limit of Hölder Mean as Exponent tends to Infinity

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R_{\ge 0}$ be real numbers.
For $p \in \R_{\ne 0}$, let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.
Then:
:$\ds \lim_{p \mathop \to +\infty} \map {M_p} {x_1, x_2, \ldots, x_n} = \max \set {x_1, x_2, \ldots, x_n}$
== Proof ==
Let $p \in \R$ such that $p \ne 0$.
Let it be assumed (or arranged) that:
:$x_1 \ge x_2 \ge \cdots \ge x_n$
Then:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Generalized Mean: $3.1.16$
Category:Hölder Mean
Limit of Hölder Mean as Exponent tends to Negative Infinity

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R_{\ge 0}$ be real numbers.
For $p \in \R_{\ne 0}$, let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.
Then:
:$\ds \lim_{p \mathop \to -\infty} \map {M_p} {x_1, x_2, \ldots, x_n} = \min \set {x_1, x_2, \ldots, x_n}$
== Proof ==
Let $p \in \R$ such that $p \ne 0$.
Let it be assumed (or arranged) that:
:$x_1 \ge x_2 \ge \cdots \ge x_n$
Then:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Generalized Mean: $3.1.17$
Category:Hölder Mean
Limit of Hölder Mean as Exponent tends to Zero is Geometric Mean

== Theorem ==
Let $x_1, x_2, \ldots, x_n \in \R_{\ge 0}$ be real numbers.
For $p \in \R_{\ne 0}$, let $\map {M_p} {x_1, x_2, \ldots, x_n}$ denote the Hölder mean with exponent $p$ of $x_1, x_2, \ldots, x_n$.
Then:
:$\ds \lim_{p \mathop \to 0} \map {M_p} {x_1, x_2, \ldots, x_n} = \paren {x_1 x_2 \cdots x_n}^{1 / n}$
which is the geometric mean of $x_1, x_2, \ldots, x_n$.
== Proof ==
Let $p \in \R$ such that $p \ne 0$.
With a view to using L'Hôpital's Rule, let us express the argument of the exponential on the in the form:
Then we have:
and:
Hence:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Generalized Mean: $3.1.18$
Category:Hölder Mean
Category:Geometric Mean
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:
== Proof ==
From Limit of Function by Convergent Sequences, we have:
:$\ds \lim_{x \mathop \to a} \map f x = \map f a$
:
: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.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits
Category:Continuous Mappings on Metric Spaces
Category:Limits of Sequences
Limit of Image of Sequence/Real Number Line

== Theorem ==
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:
:$\ds \lim_{n \mathop \to \infty} \map f {x_n} = \map f {\lim_{n \mathop \to \infty} x_n}$
That is, for a continuous function, the limit and function symbols commute.
== Proof ==
From Limit of Real Function by Convergent Sequences, we have:
:$\ds \lim_{x \mathop \to \xi} \map f x = l$
:
:for each sequence $\sequence {x_n}$ of points of $\openint a b$ such that:
::$\forall n \in \N_{>0}: x_n \ne \xi$
:and:
::$\ds \lim_{n \mathop \to \infty} x_n = \xi$
:it is true that:
::$\ds \lim_{n \mathop \to \infty} \map f {x_n} = l$
The result follows directly from this and the definition of continuity on an interval.
== Sources ==
* : $\S 9.6$
Category:Real Analysis
Category:Limits of Sequences