nyuuzyou/wikis huggingface dataset:
Limit of Sequence in Product of Metric Spaces under Chebyshev Distance

== Theorem ==
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_k = \struct {A_n, d_k}$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^k A_i$ be the cartesian product of $A_1, A_2, \ldots, A_k$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^k \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_k}, y = \tuple {y_1, y_2, \ldots, y_k} \in \AA$.
Let $\sequence {a_n}$ be a sequence of points of $\AA$:
:$a_n = \tuple {a_{n 1}, a_{n 2}, \ldots, a_{n k} } \in \AA$
Let $c = \tuple {c_1, c_2, \ldots, c_k} \in \AA$.
Then:
:$\ds \lim_{n \mathop \to \infty} a_n = c \iff \forall i \in \set {1, 2, \ldots, k}: \lim_{n \mathop \to \infty} a_{n i} = c_i$
where $\ds \lim_{n \mathop \to \infty}$ denotes a limit.
== Proof ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $1$
Category:Chebyshev Distance
Category:Limits of Sequences
Limit of Sequence is Accumulation Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$.
Let $\sequence {x_n}$ be a sequence in $A$.
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then $\alpha$ is also an accumulation point of $\sequence {x_n}$.
== Proof ==
Let $\alpha$ be a limit of $\sequence {x_n}$.
Then by definition of limit, $\sequence {x_n}$ converges to $\alpha$.
By definition of convergence that means:
:for any open set $U \subseteq T$ such that $\alpha \in U$: $\exists N \in \R: n > N \implies x_n \in U$.
As there is an infinite number of values of $n > N$, there are an infinite number of terms of $\sequence {x_n}$.
Hence the result, from definition of accumulation point.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Limit Points
Category:Limit Points
Category:Accumulation Points
Limit of Sequence is Limit of Real Function

== Theorem ==
Let $\sequence {a_n}$ be a real sequence.
Let $f: x \mapsto \map f x$ be a real function.
Suppose the limit:
:$\ds \lim_{x \mathop \to +\infty} \map f x$
exists.
If for every $n$ in the domain of $\sequence {a_n}$:
:$\map f n = a_n$
then:
:$\ds \lim_{n \mathop \to +\infty} \ a_n = \ds \lim_{x \mathop \to +\infty} \map f x$
== Proof ==
This is an instance of Limit of Function by Convergent Sequences, as the reals form a metric space.
== Sources ==
* : $\S 9.1$
Category:Convergence
Category:Sequences
Limit of Sequence to Zero Distance Point

== Theorem ==
Let $S$ be a non-empty subset of $\R$.
Let the distance $\map d {\xi, S} = 0$ for some $\xi \in \R$.
Then there exists a sequence $\sequence {x_n}$ in $S$ such that $\ds \lim_{n \mathop \to \infty} x_n = \xi$.
=== Corollary 1 ===
=== Corollary 2 ===
== Proof ==
First it is shown that:
:$\forall n \in \N_{>0}: \exists x_n \in S: \size {\xi - x_n} < \dfrac 1 n$
 that:
:$\exists n \in \N_{>0}: \not \exists x \in S: \size {\xi - x} < \dfrac 1 n$
Then $\dfrac 1 n$ is a lower bound of the set $T = \set {\size {\xi - x}: x \in S}$.
This contradicts the assertion that $\map d {\xi, S} = 0$.
We have from Sequence of Powers of Reciprocals is Null Sequence that:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$
So as $\size {\xi - x_n} < \dfrac 1 n$ it follows from the Squeeze Theorem for Real Sequences that:
:$\ds \lim_{n \mathop \to \infty} x_n = \xi$
Hence the result.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (6)$
Category:Limits of Sequences
Category:Limit of Sequence to Zero Distance Point
Limit of Sequence to Zero Distance Point/Corollary 1

== Corollary to Limit of Sequence to Zero Distance Point ==
Let $S$ be a non-empty subset of $\R$.
Let the distance $\map d {\xi, S} = 0$ for some $\xi \in \R$.
If $S$ is bounded above, then there exists a sequence $\sequence {x_n}$ in $S$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = \sup S$
== Proof ==
Let $\xi = \sup S$.
Then from Distance from Subset of Real Numbers:
:$\map d {\xi, S} = 0$
The result then follows directly from Limit of Sequence to Zero Distance Point.
Note that the terms of this sequence do not necessarily have to be distinct.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (6)$
Category:Limit of Sequence to Zero Distance Point
Limit of Sequence to Zero Distance Point/Corollary 2

== Corollary to Limit of Sequence to Zero Distance Point ==
Let $S$ be a non-empty subset of $\R$.
Let the distance $\map d {\xi, S} = 0$ for some $\xi \in \R$.
If $S$ is unbounded above, then there exists a sequence $\sequence {x_n}$ in $S$ such that $x_n \to +\infty$ as $n \to \infty$.
== Proof ==
Let $S$ be unbounded above.
Then:
:$\forall n \in \N_{>0}: \exists x_n \in S: x_n > n$
Hence $x_n \to +\infty$ as $n \to \infty$.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (6)$
Category:Limit of Sequence to Zero Distance Point
Limit of Sets Exists iff Limit Inferior contains Limit Superior

== Theorem ==
Let $\Bbb S = \set {E_n : n \in \N}$ be a sequence of sets.
Then $\Bbb S$ converges to a limit :
:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$
== Proof ==
=== Sufficient Condition ===
Let $\Bbb S$ converge to a limit.
Then by definition:
:$\ds \limsup_{n \mathop \to \infty} E_n = \liminf_{n \mathop \to \infty} E_n$
and so by definition of set equality:
:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$
=== Necessary Condition ===
Suppose that:
:$\ds \limsup_{n \mathop \to \infty} E_n \subseteq \liminf_{n \mathop \to \infty}E_n$
From Limit Superior includes Limit Inferior:
:$\ds \liminf_{n \mathop \to \infty} E_n \subseteq \limsup_{n \mathop \to \infty} E_n$
whether or not $\Bbb S$ converges to a limit.
By definition of set equality:
:$\ds \limsup_{n \mathop \to \infty} E_n = \liminf_{n \mathop \to \infty} E_n$
and so by definition $\Bbb S$ converges to a limit.
Category:Limits Superior of Set Sequences
Category:Limits Inferior of Set Sequences
Limit of Sine of X over X at Zero

== Theorem ==
:$\ds \lim_{x \mathop \to 0} \frac {\sin x} x = 1$
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Geometric Proof ==
== Sources ==
* : $\text {II}$. Calculus: Differentiation: Standard Differential Coefficients
* : $\S 16.3 \ (4) \ \text {(i)}$
Category:Differential Calculus
Category:Sine Function
Category:Examples of Limits of Real Functions
Category:Limit of Sine of X over X at Zero
Limit of Sine of X over X at Zero/Corollary

== Theorem ==
:$\ds \lim_{x \mathop \to 0} \frac x {\sin x} = 1$
== Proof ==
We have the inequality:
:$1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$
for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.
Taking the limit of the leftmost term and the rightmost term:
:$\ds \lim_{\theta \mathop \to 0} \ 1 = 1$
:$\ds \lim_{\theta \mathop \to 0} \frac 1 {\cos \theta} = 1$
So by the Squeeze Theorem:
:$\ds \lim_{\theta \mathop \to 0} \frac \theta {\sin \theta} = 1$
Category:Differential Calculus
Category:Sine Function
Category:Limit of Sine of X over X at Zero
Limit of Sine of X over X at Zero/Geometric Proof

== Theorem ==
== Geometric Proof ==
Let $\theta$ be an angle in the unit circle, measured in radians.
Define the following points:
and consider all $\theta$ in the open interval $\openint 0 {\dfrac \pi 2}$.
:
From Area of Triangle in Terms of Side and Altitude, we have that $\triangle OAB$ has an area of $\dfrac 1 2 b h$ where:
:$b = 1$
:$h = \sin \theta$
and so:
:$\Area \triangle OAB = \dfrac 1 2 \sin \theta$
From Area of Sector, the sector formed by arc $AB$ subtending $O$ is $\dfrac \theta 2$.
Clearly this sector cannot be smaller in area than $\triangle OAB$, and so we have the inequality:
:$\dfrac {\sin \theta} 2 \le \dfrac \theta 2$
for all $\theta \in \openint 0 {\dfrac \pi 2}$.
Next, from Area of Triangle in Terms of Side and Altitude, we have that $\triangle OAC$ has an area of $\dfrac 1 2 b h$ where:
:$b = 1$
:$h = \tan \theta$
and so:
:$\Area \triangle OAC = \dfrac 1 2 \tan \theta$ 
$\triangle OAC$ is clearly not smaller than the sector.
We now have the following compound inequality:
:$(\text A) \quad \dfrac 1 2 \sin \theta \le \dfrac 1 2 \theta \le \dfrac 1 2 \tan \theta$
for all $\theta \in \openint 0 {\dfrac \pi 2}$.
Clearly, if any of the plane regions were to be reflected about the $x$-axis, the magnitudes of the signed areas would be the same.
The inequality $(\text A)$, then, will hold in quadrant $\text{IV}$ if the absolute value of all terms is taken, and so:
Now, we have that $\dfrac \theta {\sin\theta} \ge 0$ on $\openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.
Also, we have that $\dfrac 1 {\cos \theta} \ge 0$ on $\openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.
So our absolute value signs are not needed.
Hence we arrive at:
:$1 \le \dfrac \theta {\sin \theta} \le \dfrac 1 {\cos \theta}$
for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.
Inverting all terms and reversing the inequalities:
:$1 \ge \dfrac {\sin\theta} \theta \ge \cos \theta$
for all $\theta \in \openint {-\dfrac \pi 2} 0 \cup \openint 0 {\dfrac \pi 2}$.
Taking to the limit:
:$\ds \lim_{\theta \mathop \to 0} 1 = 1$
:$\ds \lim_{\theta \mathop \to 0} \cos \theta = 1$
So by the Squeeze Theorem:
:$\ds \lim_{\theta \mathop \to 0} \frac {\sin \theta} \theta = 1$
== Sources ==
* : $\S 1.3$
* 
Category:Limit of Sine of X over X at Zero
Limit of Sine of X over X at Zero/Proof 2

== Theorem ==
== Proof ==
From Sine of Zero is Zero:
:$\sin 0 = 0$
From Derivative of Sine Function:
:$\map {D_x} {\sin x} = \cos x$
Then by Cosine of Zero is One:
:$\cos 0 = 1$
From Derivative of Identity Function:
:$\map {D_x} x = 1$
Thus L'Hôpital's Rule applies and so:
:$\ds \lim_{x \mathop \to 0} \frac {\sin x} x = \lim_{x \mathop \to 0} \frac {\map {D_x} {\sin x} } {\map {D_x} x} = \lim_{x \mathop \to 0} \frac {\cos x} 1 = \frac 1 1 = 1$
Category:Limit of Sine of X over X at Zero
Limit of Subsequence equals Limit of Sequence

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\sequence {x_n}$ be a sequence in $T$.
Let $l \in S$ be a limit of $\sequence {x_n}$.
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then $l$ is a limit of $\sequence {x_{n_r} }$
That is, the limit of a convergent sequence in a topological space equals the limit of any subsequence of it.
=== Metric Space ===
=== Normed Division Ring ===
=== Normed Vector Space ===
=== Real Numbers ===
For the real number line under the usual (Euclidean) topology, this translates into the following:
== Proof ==
Let $U \in \tau$ be an open set such that $l \in U$.
By definition of convergence, we have:
:$\exists N \in \N: \forall n > N: x_n \in U$.
When $r > N$, we have $n_r > n_N > N$ by Strictly Increasing Sequence of Natural Numbers.
It follows that:
:$\exists N \in \N: \forall r > N: x_{n_r} \in U$.
Therefore, as $U$ was arbitrary, we have established that $l$ is a limit of $\sequence {x_{n_r} }$, by definition of convergence.
Category:Convergence
Category:Limits of Sequences
Limit of Subsequence equals Limit of Sequence/Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {x_n}$ be a sequence in $A$.
Let $l \in A$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\ds \lim_{r \mathop \to \infty} x_{n_r} = l$
That is, the limit of a convergent sequence equals the limit of a subsequence of it.
== Proof ==
Let $\epsilon > 0$.
Since $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that:
:$\exists N: \forall n > N: \map d {x_n, l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers‎:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\map d {x_n, l} < \epsilon$
The result follows.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Exercise $3$
:::although the question calls for a separate proof for each of $3$ specific metrics on a specific space.
Category:Metric Spaces
Category:Convergence
Category:Limits of Sequences
Limit of Subsequence equals Limit of Sequence/Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring with zero: $0$.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\sequence {x_{n_r} }$ is convergent and $\ds \lim_{r \mathop \to \infty} x_{n_r} = l$
== Proof ==
Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is,
:$d \tuple {x, y} = \norm {x - y}$
By definition of convergence in a normed division ring:
:$\sequence {x_n}$ converges to $l$ in $\struct {R, \norm {\, \cdot \,} }$ $\sequence {x_n}$ converges to $l$ in the metric space $\struct {R, d}$.
We can then apply the result of Limit of Subsequence equals Limit of Sequence for metric spaces to get the result for normed division rings.
Category:Sequences
Category:Normed Division Rings
Limit of Subsequence equals Limit of Sequence/Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\sequence {x_n}$ be a sequence in $X$.
Let $\sequence {x_n}$ be convergent in the norm $\norm {\, \cdot \,}$ to the following limit:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\sequence {x_{n_r} }$ is convergent and $\ds \lim_{r \mathop \to \infty} x_{n_r} = l$
That is, the limit of a convergent sequence equals the limit of a subsequence of it.
== Proof ==
Let $\epsilon > 0$.
Since $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows from the definition of limit that:
:$\exists N \in \N : \forall n \in \N : n > N \implies \norm {x_n - l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers‎:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\norm {x_{n_r} - l} < \epsilon$
The result follows.
Category:Normed Vector Spaces
Category:Convergence
Category:Limits of Sequences
Limit of Subsequence equals Limit of Sequence/Real Numbers

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $l \in \R$ such that:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Let $\sequence {x_{n_r} }$ be a subsequence of $\sequence {x_n}$.
Then:
:$\ds \lim_{r \mathop \to \infty} x_{n_r} = l$
That is, the limit of a convergent sequence equals the limit of a subsequence of it.
== Proof ==
Let $\epsilon > 0$.
As $\ds \lim_{n \mathop \to \infty} x_n = l$, it follows that:
:$\exists N: \forall n > N: \size {x_n - l} < \epsilon$
Now let $R = N$.
Then from Strictly Increasing Sequence of Natural Numbers‎:
:$\forall r > R: n_r \ge r$
Thus $n_r > N$ and so:
:$\size {x_n - l} < \epsilon$
The result follows.
== Sources ==
* : $\S 5$: Subsequences: $\S 5.2$
Category:Convergence
Category:Subsequences
Category:Real Sequences
Limit of Subsequence of Bounded Sequence

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\sequence {x_n}$ be bounded.
Let $b \in \R$ be a real number.
Suppose that $\forall N: \exists n > N: x_n \ge b$.
Then $\sequence {x_n}$ has a subsequence which converges to a limit $l \ge b$.
== Proof ==
Let us pick $N \in \N$.
Then $\exists n_1 > N: x_{n_1} \ge b$.
Again, $\exists n_2 > n_1: x_{n_2} \ge b$.
And so on: for each $n_k$ we find, $\exists n_{k+1} > n_k: x_{n_{k+1}} \ge b$.
In this way we can build a subsequence of $\sequence {x_n}$ each of whose terms are $b$ or bigger.
By the Bolzano-Weierstrass Theorem, this subsequence itself contains a subsequence $\sequence {x_{n_r} }$ which is convergent.
Now, suppose $x_{n_r} \to l$ as $r \to \infty$.
Since $x_{n_r} \ge b$ it follows from Lower and Upper Bounds for Sequences that $l \ge b$.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.15 \ (3)$
Category:Limits of Sequences
Limit of Tail of Decreasing Sequence of Sets

== Theorem ==
Let $X$ be a set.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a decreasing sequence of subsets of $X$ such that: 
:$E_n \downarrow E$ 
where $E_n \downarrow E$ denotes the limit of decreasing sequence of sets.
Then for each $m \in \N$ we have: 
:$E_{n + m} \downarrow E$ 
== Proof ==
Let $m \in \N$.
From Tail of Decreasing Sequence of Sets is Decreasing, we have: 
:$\sequence {E_{n + m} }_{n \mathop \in \N}$ is a decreasing sequence of sets.
Since: 
:$E_n \downarrow E$
we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty E_n = E$
We show that: 
:$\ds \bigcap_{n \mathop = 1}^\infty E_{n + m} = E$
That is: 
:$\ds \bigcap_{n \mathop = m + 1}^\infty E_n = E$
Clearly we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty E_n \subseteq \bigcap_{n \mathop = m + 1}^\infty E_n$
from Intersection is Decreasing.
Now let: 
:$\ds x \in \bigcap_{n \mathop = m + 1}^\infty E_n$
Then: 
:$x \in E_n$ for all $n \in \N$ for all $n \ge m + 1$
and in particular: 
:$x \in E_{m + 1}$ 
Since $\sequence {E_n}_{n \mathop \in \N}$ is a decreasing sequence of sets, we have: 
:$E_{m + 1} \subseteq E_n$ for all $n \in \N$ with $n < m + 1$.
So, we actually have: 
:$x \in E_n$ for all $n \in \N$.
So, from the definition of set intersection, we have: 
:$\ds x \in \bigcap_{n \mathop = 1}^\infty E_n$
So, from the definition of set inclusion, we have: 
:$\ds \bigcap_{n \mathop = m + 1}^\infty E_n \subseteq \bigcap_{n \mathop = 1}^\infty E_n$
So we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty E_{n + m} = E$
Category:Limits of Sets
Category:Decreasing Sequences of Sets
Limit of Tan X over X at Zero

== Theorem ==
:$\ds \lim_{x \mathop \to 0} \frac {\tan x} x = 1$
== Proof 1 ==
 
== Proof 2 ==
 
== Proof 3 ==
 
Category:Tangent Function
Category:Examples of Limits of Real Functions
Category:Limit of Tan X over X at Zero
Limit of Tan X over X at Zero/Proof 3

== Theorem ==
== Proof ==
Let $f$ be the real function defined as:
:$\map f x = \sin x$
Let:
:$c = \pi$
:$h \in \openint 1 {\dfrac \pi 2}$
We have:
Hence:
Hence by the Squeeze Theorem for Functions:
:$\ds \lim_{h \mathop \to 0} \dfrac {\tan h} h = 1$
The result follows on renaming $h$ to $x$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.4$ Law of the Mean: Example $\text H$
Category:Limit of Tan X over X at Zero
Limit of nth Root of Sum of nth Powers equals Maximum

== Theorem ==
Let $a_1, a_2, \ldots, a_r$ be non-negative real numbers.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $a = \max \set {a_1, a_2, \ldots, a_r}$.
Then:
:$\ds \lim_{n \mathop \to \infty} \paren {a_1^n + a_2^n + \cdots + a_r^n}^{1 / n} = a$
== Proof ==
From Sum of $r$ Powers is between Power of Maximum and $r$ times Power of Maximum:
:$a^n \le a_1^n + a_2^n + \cdots + a_r^n \le r a^n$
and so:
:$a \le \paren {a_1^n + a_2^n + \cdots + a_r^n}^{1 / n} \le r^{1/n} a$
From Limit of Integer to Reciprocal Power:
:$n^{1 / n} \to 1$ as $n \to \infty$
Then for $n > r$:
:$1 < r^{1 / n} < n^{1 / n}$
and so:
:$r^{1 / n} = 1$ as $n \to \infty$
Thus as $n \to \infty$:
:$a \le \paren {a_1^n + a_2^n + \cdots + a_r^n}^{1 / n} \le a$
and the result follows by the Squeeze Theorem.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 11$
Category:Examples of Limits of Real Functions
Limit of x to the x

== Theorem ==
Let $f: \R_{>0} \to \R$ be defined as:
:$\forall x \in \R_{>0}: \map f x = x^x$
Then:
:$\ds \lim_{x \mathop \to 0^+} x^x = 1$
== Proof 1 ==
== Proof 2 ==
== Also presented as ==
This can also be presented as:
:$\ds \lim_{x \mathop \to 0^+} \map \exp {x \ln x} = 1$
as is demonstrated during the course of the proof.
Category:Examples of Limits of Real Functions
Category:Limit of x to the x
Limit to 1 of Zeta of s minus Reciprocal of s-1

== Theorem ==
:$\ds \lim_{s \mathop \to 1} \paren {\map \zeta s - \dfrac 1 {s - 1} } = \gamma$
where:
:$\zeta$ denotes the Riemann $\zeta$ (zeta) function
:$\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
== Sources ==
* : $0,57721 56649 \ldots$
Category:Euler-Mascheroni Constant
Category:Riemann Zeta Function
Limit to Infinity of Binomial Coefficient over Power

== Theorem ==
Let $k \in \R \setminus \set {-1, -2, -3, \dotsc}$.
Then:
:$\ds \lim_{r \mathop \to \infty} \dfrac {\dbinom r k} {r^k} = \frac 1 {\map \Gamma {k + 1} }$
=== Integer Case ===
== Proof ==
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $45$
Category:Binomial Coefficients
Category:Limit to Infinity of Binomial Coefficient over Power
Limit to Infinity of Complementary Error Function

== Theorem ==
:$\ds \lim_{x \mathop \to \infty} \map \erfc x = 0$
where $\erfc$ denotes the complementary error function.
== Proof ==
== Sources ==
* : $\S 35$: Complementary Error Function $\ds \map \erfc x = 1 - \map \erf x = \frac 2 {\sqrt \pi} \int_x^\infty e^{-u^2} \rd u$: $35.6$
Category:Complementary Error Function
Limit to Infinity of Error Function

== Theorem ==
:$\ds \lim_{x \mathop \to \infty} \map \erf x = 1$
 
where $\erf$ denotes the error function.
== Proof ==
== Sources ==
* : $\S 35$: Error Function $\ds \map \erf x = \frac 2 {\sqrt \pi} \int_0^x e^{-u^2} \rd u$: $35.3$
Category:Error Function
Category:Examples of Limits of Real Functions
Limit to Infinity of Exponential Integral Function

== Theorem ==
Let $\Ei: \R_{>0} \to \R$ denote the exponential integral function:
:$\map \Ei x = \ds \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {e^{-t} } t \rd t$
Then:
:$\ds \lim_{x \mathop \to \infty} \map \Ei x = 0$
== Proof ==
== Sources ==
* : $\S 35$: Exponential Integral $\ds \map \Ei x = \int_x^\infty \frac {e^{-u} } u \rd u$: $35.10$
Category:Exponential Integral Function
Limit to Infinity of Fresnel Cosine Integral Function

== Theorem ==
:$\ds \lim_{x \mathop \to \infty} \map {\mathrm C} x = \frac 1 2$
where $\mathrm C$ denotes the Fresnel cosine integral function.
== Proof ==
== Sources ==
* : $\S 35$: Fresnel Cosine Integral $\ds \map {\mathrm C} x = \sqrt {\frac 2 \pi} \int_0^x \cos u^2 \rd u$: $35.23$
Category:Fresnel Cosine Integral Function
Limit to Infinity of Fresnel Sine Integral Function

== Theorem ==
:$\ds \lim_{x \mathop \to \infty} \map {\mathrm S} x = \frac 1 2$
where $\mathrm S$ denotes the Fresnel sine integral function.
== Proof ==
== Sources ==
* : $\S 35$: Fresnel Sine Integral $\ds \map {\mathrm S} x = \sqrt {\frac 2 \pi} \int_0^x \sin u^2 \rd u$: $35.20$
Category:Fresnel Sine Integral Function
Limit to Infinity of Laplace Transform

== Theorem ==
Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of the real function $f$.
Then:
:$\ds \lim_{s \mathop \to \infty} \map F s = 0$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $10$. Behavior of $\map f s$ as $s \to 0$: Theorem $1 \text{-} 15$
Category:Laplace Transforms
Limit to Infinity of Number of p-Groups of Order p^m

== Theorem ==
Let $p$ be a prime number.
Let $m \in \N$ be a natural number
Let $\map \nu {p^n}$ denote the $\nu$ function of $p^n$: the number of group types of order $p^m$.
Then:
:$\map \nu {p^m} = p^{A m^3}$
where:
:$\ds \lim_{m \mathop \to \infty} A = \dfrac 2 {27}$
== Proof ==
== Sources ==
* 
* : $0,07407407407 \ldots$
Category:Nu Function
Limit to Infinity of Power of x by Exponential of -a x

== Theorem ==
Let $k$ and $a$ be positive real numbers. 
Then: 
:$\ds \lim_{x \mathop \to \infty} x^k e^{-a x} = 0$
== Proof ==
By Power Series Expansion for Exponential Function, we have: 
:$\ds e^{a x} = \sum_{n \mathop = 0}^\infty \frac {\paren {a x}^n} {n!}$
Since for $x > 0$ each term in this sum is non-negative, we have: 
:$\ds e^{a x} \ge \frac {\paren {a x}^{\floor k + 1} } {\paren {\floor k + 1}!}$
for each $k$.
So, for each $x > 0$ we have: 
:$\ds 0 < e^{-a x} \le \frac {\paren {\floor k + 1}!} {\paren {a x}^{\floor k + 1} }$
So that: 
:$\ds 0 \le x^k e^{-a x} \le \frac 1 {a^{\floor k + 1} } x^{k - \floor k - 1} \paren {\floor k + 1}!$
From the definition of the floor function, we have: 
:$0 \le k - \floor k < 1$
so:
:$k - \floor k - 1 < 0$
Hence by Limit to Infinity of Power:
:$\ds \lim_{x \mathop \to \infty} \frac 1 {a^{\floor k + 1} } x^{k - \floor k - 1} \paren {\floor k + 1}! = 0$
So, by the Squeeze Theorem, we have: 
:$\ds \lim_{x \mathop \to \infty} x^k e^{-a x} = 0$
Category:Exponential Function