nyuuzyou/wikis huggingface dataset:
Limit Superior/Examples/(-1)^n (1 + n^-1)

== Example of Limit Superior ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{>0}: a_n = \paren {-1}^n \paren {1 + \dfrac 1 n}$
The limit superior of $\sequence {a_n}$ is given by:
:$\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = 1$
This is not the same as:
:$\ds \sup_{n \mathop \ge 1} \paren {-1}^n \paren {1 + \dfrac 1 n}$
== Proof ==
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
We have that:
and:
:$\sequence {a_n}$ contains no subsequence which converges to a limit different from $-1$ or $1$.
Hence:
:$L = \set {-1, 1}$
and it follows that:
:$\ds \map {\limsup_{n \mathop \to \infty} } {\paren {-1}^n \paren {1 + \dfrac 1 n} } = 1$
But note that:
:$\ds \sup_{n \mathop \ge 1} \paren {-1}^n \paren {1 + \dfrac 1 n}$
occurs when $n = 2$, at which point:
:$\paren {-1}^n \paren {1 + \dfrac 1 n} = \paren {-1}^2 \paren {1 + \dfrac 1 2} = \dfrac 3 2$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: Exercise $\S 5.15 \ (1) \ \text {(i)}$
Category:Examples of Limits Superior
Limit Superior/Examples/Farey Sequence

== Example of Limit Superior ==
Consider the Farey sequence:
:$\sequence {a_n} = \dfrac 1 2, \dfrac 1 3, \dfrac 2 3, \dfrac 1 4, \dfrac 2 4, \dfrac 3 4, \dfrac 1 5, \dfrac 2 5, \dfrac 3 5, \dfrac 4 5, \dfrac 1 6, \ldots$
The limit superior of $\sequence {a_n}$ is given by:
:$\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = 1$
== Proof ==
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
From the definition of $F$ we have that:
:$\forall n \in \N_{>0}: 0 < a_n < 1$
From Lower and Upper Bounds for Sequences we have that:
:$L \subseteq \closedint 0 1$
Consider the subsequences:
:$(1): \quad \sequence {a_{n_r} } = \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \dfrac 1 5 \to 0$ as $n \to \infty$
:$(2): \quad \sequence {a_{n_s} } = \dfrac 1 2, \dfrac 2 3, \dfrac 3 4, \dfrac 4 5 \to 1$ as $n \to \infty$
Hence by definition of limit superior:
:$\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = 1$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: $\S 5.14$: Example $\text {(iii)}$
Category:Examples of Limits Superior
Limit Superior/Examples/Sequence of Reciprocals

== Example of Limit Superior ==
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{>0}: a_n = \dfrac 1 n$
The limit superior of $\sequence {a_n}$ is given by:
:$\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = 0$
== Proof ==
From Sequence of Reciprocals is Null Sequence, $\sequence {a_n}$ is convergent:
:$\ds \lim_{n \mathop \to \infty} \dfrac 1 n = 0$
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
By Limit of Subsequence equals Limit of Real Sequence, all such subsequences have limit $0$.
 
Hence by definition of limit superior:
:$\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = \sup \set 0 = 0$
== Sources ==
* : $\S 5$: Subsequences: Lim sup and lim inf: $\S 5.14$: Example $\text {(i)}$
Category:Examples of Limits Superior
Limit Superior includes Limit Inferior

== Theorem ==
Let $\set {E_n : n \in \N}$ be a sequence of sets.
Then:
:$\ds \liminf_{n \mathop \to \infty} E_n \subseteq \limsup_{n \mathop \to \infty} E_n$
== Proof ==
Select any $\ds x \in \liminf_{n \mathop \to \infty} E_n$.
Then by definition 2 of Limit Inferior of Sequence of Sets, $x$ belongs to $E_i$ for all but finitely many values of $i$.
Therefore $x$ belongs to $E_i$ for infinitely many values of $i$.
So by the definition 2 of Limit Superior of Sequence of Sets:
:$\ds x \in \limsup_{n \mathop \to \infty} E_n$
== Sources ==
* : Chapter $\text {I}$: Sets of Points: $1 \cdot 1$. The algebra of sets
Category:Limits Superior of Set Sequences
Category:Limits Inferior of Set Sequences
Limit at Infinity of Polynomial over Complex Exponential

== Theorem ==
Let $n \in \N$.
Let $\map {P_n} x$ be a real polynomial, of degree $n$.
Let $e^z$ be the complex exponential function, where $z = x + i y$.
Let $a \in \R_{>0}$.
Then:
:$\ds \lim_{x \mathop \to +\infty} \frac {\map {P_n} x} {e^{a z} } = 0$
== Proof ==
Let $\epsilon > 0$.
By the definition of limits at infinity, we need to show that there is some $M \in \R$ such that:
:$\ds x > M \implies \size {\frac {\map {P_n} x} {e^{a z} } - 0} < \epsilon$
But:
This means it is sufficient to find an $M \in \R$ such that:
Let $\ds \map {P_n} x = \sum_{j \mathop = 0}^n \paren {a_j x^j}$ where $a_j \in \R$ for every $j$ in $\set {0, \ldots, n}$.
Let $\ds M' = \frac 1 \epsilon \sum_{j \mathop = 0}^n \size {a_j}$.
We observe that $M' \ge 0$.
Also, for every $x > M'$:
:$x > 0$ as $M' \ge 0$
We have for every $x > M'$:
We have that for every $k \in \N$ there exists $N_k \in \N$ such that:
:$x^k < e^{a x}$ for all $x > N_k$ by Exponential Dominates Polynomial
Let $N = \map \max {N_1, \ldots, N_{n + 1} }$.
We have for every $k \in \set {1, \ldots, n + 1}$:
Let $M = \map \max {M', N}$.
We get for every $x > M$:
Category:Limits of Mappings
Category:Polynomial Theory
Category:Exponential Function
Limit at Infinity of Real Identity Function

== Theorem ==
Let $I_\R: \R \to \R$ be the identity function on $\R$.
Then:
:$(1): \quad \ds \lim_{x \mathop \to +\infty} \map {I_\R} x = +\infty$
:$(2): \quad \ds \lim_{x \mathop \to -\infty} \map {I_\R} x = -\infty$
== Proof ==
We have that the Derivative of Identity Function is $1$.
Hence, by Derivative of Monotone Function, $x$ is strictly increasing.
Now, by the definition of infinite limit at infinity, the first assertion is:
:$\forall M \in \R_{>0}: \exists N \in \R_{>0}: x > N \implies \map f x > M$
For every $M$, choose $N = M$.
The second assertion is proved similarly.
Category:Unbounded Mappings
Category:Identity Mappings
Limit at Infinity of Sine Integral Function

== Theorem ==
Let $\Si: \R \to \R$ denote the sine integral function.
Then $\Si$ has a (finite) limit at infinity:
:$\ds \lim_{x \mathop \to +\infty} \map \Si x = \frac \pi 2$
=== Corollary ===
== Proof ==
The limit:
:$\ds \lim_{x \mathop \to +\infty} \map \Si x = \lim_{x \mathop \to +\infty} \int_{t \mathop \to 0}^{t \mathop = x} \frac {\sin t} t \rd t$
is the Dirichlet Integral.
== Sources ==
* : $\S 35$: Sine Integral $\ds \map \Si x = \int_0^x \frac {\sin u} u \rd u$: $35.13$
Category:Sine Integral Function
Limit at Infinity of x^n

== Theorem ==
Let $x \mapsto x^n$, $n \in \R$ be a real function which is continuous on the open interval $\openint 1 {+\infty}$.
Let $n > 0$.
Then $x^n \to +\infty$ as $x \to +\infty$.
== Proof ==
From Upper Bound of Natural Logarithm:
:$\forall n > 0: n \ln x < x^n$
which, by Multiple Rule for Continuous Real Functions, implies:
From Logarithm Tends to Infinity:
:$n \ln x \to +\infty$ as $x \to +\infty$
The result follows from the Push Theorem.
== Also see ==
* Sequence of Powers of Reciprocals is Null Sequence
* Limit at Infinity of Identity Function
Category:Examples of Limits of Real Functions
Limit iff Limits from Left and Right

== Theorem ==
Let $f$ be a real function defined on an open interval $\openint a b$ except possibly at a point $c \in \openint a b$.
Then:
:$\map f x \to l$ as $x \to c$
:
:$\map f x \to l$ as $x \to c^-$
and
:$\map f x \to l$ as $x \to c^+$
== Proof ==
=== Necessary Condition ===
Let $\map f x \to l$ as $x \to c$.
Then from the definition of the limit of a function:
:$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$
So for any given $\epsilon$, there exists a $\delta$ such that:
:$0 < \size {x - c} < \delta$
implies that:
:$l - \epsilon < \map f x < l + \epsilon$
Now:
That is: $\forall \epsilon > 0: \exists \delta > 0$:
:$(1): \quad c - \delta < x < c \implies \norm {\map f x - l} < \epsilon$
:$(2): \quad c < x < c + \delta \implies \norm {\map f x - l} < \epsilon$
So given that particular value of $\epsilon$, we can find a value of $\delta$ such that the conditions for both:
:$(1): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the left
and :
:$(2): \quad f$ tending to the limit $l$ as $x$ tends to $c$ from the right.
Thus:
:$\ds \lim_{x \mathop \to c} \map f x = l$
implies that:
:$\ds \lim_{x \mathop \to c^-} \map f x = l$
and:
:$\ds \lim_{x \mathop \to c^+} \map f x = l$
=== Sufficient Condition ===
Let $\map f x \to l$ as $x \to c^-$ and $\map f x \to l$ as $x \to c^+$.
This means that:
:$(1): \quad \forall \epsilon > 0: \exists \delta > 0: c - \delta < x < c \implies \size {\map f x - l} < \epsilon$
and :
:$(2): \quad \forall \epsilon > 0: \exists \delta > 0: c < x < c + \delta \implies \size {\map f x - l} < \epsilon$
In the same manner as above, the conditions on $\delta$ give us that:
So:
:$\forall \epsilon > 0: \exists \delta > 0: 0 < \size {x - c} < \delta \implies \size {\map f x - l} < \epsilon$
Thus:
:$\ds \lim_{x \mathop \to c^-} \map f x = l$
and:
:$\ds \lim_{x \mathop \to c^+} \map f x = l$
together imply that:
:$\ds \lim_{x \mathop \to c} \map f x = l$
Hence the result.
== Sources ==
* : $\S 8.4$
* : Appendix: $\S 18.6$: Limits of functions
Category:Limits of Real Functions
Limit of (Cosine (X) - 1) over X at Zero

== Theorem ==
:$\ds \lim_{x \mathop \to 0} \frac {\cos x - 1} x = 0$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
Category:Differential Calculus
Category:Cosine Function
Category:Examples of Limits of Real Functions
Category:Limit of (Cosine (X) - 1) over X at Zero
Limit of (Cosine (X) - 1) over X at Zero/Proof 1

== Theorem ==
== Proof ==
This proof works directly from the definition of the cosine function:
Now let:
:$\map {f_n} x = \paren {-1}^n \dfrac {x^{2 n - 1} } {\paren {2 n - 1}!}$
Then for every $n \in \N_{> 0}$, and for all $x \in \closedint {\dfrac 1 2} {\dfrac 1 2}$:
But from Sum of Infinite Geometric Sequence:
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {2^n} = 2 < \infty$
By the Weierstrass M-Test, $\ds \sum_{n \mathop = 1}^\infty \map {f_n} x$ converges uniformly to some function $f$ on $\closedint {\dfrac 1 2} {\dfrac 1 2}$.
But from Real Polynomial Function is Continuous, and the Uniform Limit Theorem $f$ is continuous on $\closedint {\dfrac 1 2} {\dfrac 1 2}$.
So:
:$\ds \lim_{x \mathop \to 0} \map f x = \map f 0 = \sum_{n \mathop = 1}^\infty \paren {-1} \frac {0^{2 n - 1} } {\paren {2 n - 1}!} = 0$
Category:Limit of (Cosine (X) - 1) over X at Zero
Limit of (Cosine (X) - 1) over X at Zero/Proof 2

== Theorem ==
== Proof ==
This proof assumes the truth of the Derivative of Cosine Function:
From Cosine of Zero is One:
:$\cos 0 = 1$
From Derivative of Cosine Function:
:$\map {D_x} {\cos x} = -\sin x$
and by Derivative of Constant:
:$\map {D_x} {-1} = 0$
So by Sum Rule for Derivatives:
:$\map {D_x} {\cos x - 1} = -\sin x$
By Sine of Zero is Zero:
:$\sin 0 = 0$
From Derivative of Identity Function:
:$\map {D_x} x = 1$
Thus L'Hôpital's Rule applies and so:
:$\ds \lim_{x \mathop \to 0} \frac {\cos x - 1} x = \lim_{x \mathop \to 0} \frac {-\sin x} 1 = \frac {-0} 1 = 0$
Category:Limit of (Cosine (X) - 1) over X at Zero
Limit of Absolute Value

== Theorem ==
Let $x, \xi \in \R$ be real numbers.
Then:
:$\size {x - \xi} \to 0$ as $x \to \xi$
where $\size {x - \xi}$ denotes the Absolute Value.
== Proof ==
Let $\epsilon > 0$.
Let $\delta = \epsilon$.
From the definition of a limit of a function, we need to show that $\size {\map f x - 0} < \epsilon$ provided that $0 < \size {x - \xi} < \delta$, where $\map f x = \size {x - \xi}$.
Thus, provided $0 < \size {x - \xi} < \delta$, we have:
Hence the result.
== Sources ==
* : $\S 8.15 \ (4)$
Category:Examples of Limits of Real Functions
Category:Absolute Value Function
Limit of Bounded Convergent Sequence is Bounded

== Theorem ==
Let $\sequence {x_n}$, $\sequence {a_n}$, and $\sequence {b_n}$ be convergent sequences in $\R$.
Let $\sequence {x_n}$, $\sequence {a_n}$, and $\sequence {b_n}$ converge to $x, a, b \in \R$, respectively.
Suppose that:
:$\exists N \in \N: n \ge N \implies a_n \le x_n \le b_n$
Then:
:$a \le x \le b$
== Proof ==
 that $x < a$.
Let $\epsilon = \dfrac {a - x} 2 > 0$
From the convergence of $\sequence {x_n}$:
:$\exists M_1 \in \N : n \ge M \implies x - \epsilon < x_n < x + \epsilon$
Or, equivalently:
:$\exists M_1 \in \N : n \ge M \implies \dfrac {3 x - a} 2 < x_n < \dfrac {x + a} 2$
From the convergence of $\sequence {a_n}$:
:$\exists M_2 \in \N : n \ge M \implies a - \epsilon < a_n < a + \epsilon$
Or, equivalently:
:$\exists M_2 \in \N : n \ge M \implies \dfrac {x + a} 2 < a_n < \dfrac {3 a - x} 2$
Let $M = \max \set {N, M_1, M_2}$
Then, for any $n \ge M$:
This contradicts the hypothesis that:
:$\forall n \ge N : a_n \le x_n$
The same argument, mutatis mutandis, brings us to a contradiction if we suppose $x > b$.
Hence the result, by Proof by Contradiction.
Category:Limits of Sequences
Category:Real Analysis
Limit of Complex Function is Unique

== Theorem ==
Let $f: S \to \C$ be a complex function.
Let $z_0$ be a limit point of $S$.
Suppose that $\ds \lim_{z \mathop \to z_0} \map f z = L$.
Then that limit $L$ is unique.
== Proof ==
 $L' \ne L$ is another limit point of $\map f z$ at $z_0$.
Let us take $\epsilon = \dfrac {\cmod {L - L'} } 2$.
Then we can find $\delta_1 > 0, \delta_2 > 0$ such that:
:$z \in S, 0 < \cmod {z - z_0} < \delta_1 \implies \cmod {\map f z - L} < \epsilon$
:$z \in S, 0 < \cmod {z - z_0} < \delta_2 \implies \cmod {\map f z - L'} < \epsilon$
Because $z_0$ is a limit point:
:$\exists z^* \in S: 0 < \cmod {z - z_0} < \min \set {\delta_1, \delta_2}$
Then:
This contradicts the choice we made of $\epsilon$.
Hence the result by Proof by Contradiction.
Category:Limits of Complex Functions
Limit of Composite Function

== Theorem ==
Let $f$ and $g$ be real functions.
Let:
Then, if either:
:Hypothesis 1: $f$ is continuous at $\eta$ (that is $l = \map f \eta$)
or:
:Hypothesis 2: for some open interval $I$ containing $\xi$, it is true that $\map g x \ne \eta$ for any $x \in I$ except possibly $x = \xi$
then:
:$\ds \lim_{x \mathop \to \xi} \map f {\map g x} = \lim_{y \mathop \to \eta} \map f y$
=== Corollary ===
== Proof ==
Let $\epsilon > 0$.
Since $\ds \lim_{y \mathop \to \eta} \map f y = l$, we can find $\Delta > 0$ such that:
:$\size {\map f y - l} < \epsilon$ provided $0 < \size {y - \eta} < \Delta$
Let $y = \map g x$.
Then, provided that $0 < \size {\map g x - \eta} < \Delta$, we have:
:$\size {\map f {\map g x} - l} < \epsilon$
But $\ds \lim_{x \mathop \to \xi} \map g x = \eta$ and $\Delta > 0$.
Hence:
:$\exists \delta > 0: \size {\map g x - \eta} < \Delta$ provided that $0 < \size {x - \xi} < \delta$
We now need to establish the reason for the conditions under which $0 < \size {x - \xi} < \delta \implies \size {\map f {\map g x} - l} < \epsilon$.
=== Counterexample ===
As it stands, this is not generally the case, as follows.
Now, if Hypothesis 1: $f$ is continuous at $\eta$, then $l = \map f \eta$ and so $\size {\map f y - l} < \epsilon$ even when $y = \eta$.
So we can write: provided that $\size {\map g x - \eta} < \Delta$, we have:
:$\size {\map f {\map g x} - l} < \epsilon$
and the argument holds.
Otherwise, let us assume Hypothesis 2: For some open interval $I$ containing $\xi$, it is true that $\map g x \ne \eta$ for any $x \in I$ except possibly $x = \xi$.
Then we can be sure that $\map g x \ne \eta$ provided that $0 < \size {x - \xi} < \delta$ for sufficiently small $\delta > 0$.
So we can write: $0 < \size {\map g x - \eta} < \Delta$ provided that $0 < \size {x - \xi} < \delta$, and the argument holds.
== Sources ==
* : $\S 8.17$
Category:Limits of Real Functions
Limit of Composite Function/Corollary

== Corollary to Limit of Composite Function ==
Let $I$ and $J$ be real intervals.
Let:
:$(1): \quad g: I \to J$ be a real function which is continuous on $I$
:$(2): \quad f: J \to \R$ be a real function which is continuous on $J$.
Then the composite function $f \circ g$ is continuous on $I$.
== Proof ==
This follows directly and trivially from:
:the definition of continuity at a point
:the definition of continuity on an interval
:Limit of Composite Function.
Category:Limits of Real Functions
Limit of Composite Function/Counterexample

== Counterexample to Limit of Composite Function is Image of Limit ==
Let $f$ and $g$ be the real functions defined as:
for $\eta, y_1, y_2 \in \R$.
Then:
:$\ds \lim_{y \mathop \to \eta} \map f y = y_2$
and:
:$\ds \lim_{x \mathop \to \xi} \map g x = \eta$
But we have:
:$\forall x \in \R: \map f {\map g x} = y_1$
and so:
:$\ds \lim_{x \mathop \to \xi} \map f {\map g x} = y_1$
Hence it is not true that:
:$\ds \lim_{x \mathop \to \xi} \map f {\map g x} = \lim_{y \mathop \to \eta} \map f y$
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 15$
* : $\S 8.17$
Category:Limits of Real Functions
Limit of Constant Function/Limit at Infinity

== Theorem ==
Let $a, b \in \R$.
Define $f : \R \to \R$ by: 
:$\map f x = a$ for each $x \in \R$. 
Then: 
:$\ds \lim_{x \mathop \to \infty} \map f x = a$
where $\ds \lim_{x \mathop \to \infty}$ denotes the limit at $+\infty$.
== Proof ==
We have: 
:$\size {\map f x - a} = 0$ for all $x \in \R$. 
So for any $\epsilon > 0$ and $M \in \R$, we have:
:$\size {\map f x - a} < \epsilon$ for all $x \in \R$ with $x \ge M$.
So from the definition of the limit at $+\infty$, we have: 
:$\ds \lim_{x \mathop \to \infty} \map f x = a$
Category:Limit of Constant Function
Category:Limit at Infinity
Limit of Constant Function/Limit at Minus Infinity

== Theorem ==
Let $a, b \in \R$.
Define $f : \R \to \R$ by: 
:$\map f x = a$ for each $x \in \R$. 
Then: 
:$\ds \lim_{x \mathop \to -\infty} \map f x = a$
where $\ds \lim_{x \mathop \to -\infty}$ denotes the limit at $-\infty$.
== Proof ==
We have: 
:$\size {\map f x - a} = 0$ for all $x \in \R$. 
So for any $\epsilon > 0$ and $M \in \R$, we have:
:$\size {\map f x - a} < \epsilon$ for all $x \in \R$ with $x \le M$.
So from the definition of the limit at $-\infty$, we have: 
:$\ds \lim_{x \mathop \to -\infty} \map f x = a$
Category:Limit of Constant Function
Category:Limit at Minus Infinity
Limit of Constant Function/Two-Sided Limit at Real Number

== Theorem ==
Let $a, b \in \R$.
Define $f : \R \to \R$ by: 
:$\map f x = a$ for each $x \in \R$. 
Then: 
:$\ds \lim_{x \mathop \to b} \map f x = a$
where $\ds \lim_{x \mathop \to b}$ denotes the limit as $x \to b$.
== Proof ==
We have: 
:$\size {\map f x - a} = 0$ for all $x \in \R$. 
So for any $\epsilon > 0$ and $\delta > 0$, we have: 
:$\size {\map f x - a} < \epsilon$ for all $x \in \R$ with $\size {x - b} < \delta$.
So by the definition of the limit as $x \to b$, we have: 
:$\ds \lim_{x \mathop \to b} \map f x = a$
Category:Limit of Constant Function
Limit of Countable Union 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 $x_n < x$ for each $n \in \N$.
Then we have:
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \openint {-\infty} x$
== Proof ==
First suppose that:
:$\ds t \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Then we have:
:$t \le x_n$ for some $n \in \N$.
We have $x_n < x$ for each $n \in \N$.
Hence $t < x$ and $t \in \openint {-\infty} x$.
Now suppose that $t \in \openint {-\infty} x$.
Then $t < x$. 
Hence we can take $\epsilon > 0$ such that:
:$t + \epsilon < x$
so that:
:$t < x - \epsilon$ 
Since $x_n \to x$, there exists $N \in \N$ such that:
:$x - \epsilon \le x_n \le x + \epsilon$ for $n \ge N$. 
Hence we have:
:$t \le x_N$ 
In particular, $t \in \openinl {-\infty} {x_N}$, so:
:$\ds t \in \bigcup_{n \mathop = 1}^\infty \openinl {-\infty} {x_n}$ 
So we have:
:$\ds \openint {-\infty} x \subseteq \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Category:Limits of Sequences of Intervals
Limit of Cumulative Distribution Function at Negative Infinity

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $F_X$ be the cumulative distribution function.
Then:
:$\ds \lim_{x \mathop \to -\infty} \map {F_X} x = 0$
where $\ds \lim_{x \mathop \to -\infty} \map {F_X} x$ denotes the limit at $-\infty$ of $F_X$. 
== 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 $\map {F_X} {x_n} \to 0$
at which point we will obtain: 
:$\ds \lim_{x \mathop \to -\infty} \map {F_X} x = 0$
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, we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
Let $P_X$ be the probability distribution of $X$. 
From Probability Distribution is Probability Measure, we have: 
:$P_X$ is a finite measure.
From Measure of Limit of Decreasing Sequence of Measurable Sets, we have: 
:$\ds \map {P_X} \O = \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }$ 
So, we obtain:
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have: 
:for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map {F_X} {x_n} \to 0$.
So, from Sequential Characterisation of Limit at Minus Infinity of Real Function: Corollary, we have: 
:$\ds \lim_{x \mathop \to \infty} \map {F_X} x = 0$
== Sources ==
* 
* 
Category:Limit of Cumulative Distribution Function at Negative Infinity
Category:Limit at Minus Infinity
Category:Cumulative Distribution Functions
Limit of Cumulative Distribution Function at Positive Infinity

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $F_X$ be the cumulative distribution function.
Then:
:$\ds \lim_{x \mathop \to \infty} \map {F_X} x = 1$
where $\ds \lim_{x \mathop \to \infty} \map {F_X} x$ denotes the limit at $+\infty$ of $F_X$. 
== Proof ==
From Sequential Characterization of Limit at Positive Infinity of Real Function: Corollary, we aim to show that: 
:for all increasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map {F_X} {x_n} \to 1$
at which point we will obtain: 
:$\ds \lim_{x \mathop \to \infty} \map {F_X} x = 1$
Since $\sequence {x_n}_{n \mathop \in \N}$ is increasing:
:the sequence $\sequence {\hointl {-\infty} {x_n} }_{n \mathop \in \N}$ is increasing.
We show that: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \R$
=== Lemma ===
Let $P_X$ be the probability distribution of $X$. 
So, from Measure of Limit of Increasing Sequence of Measurable Sets, we have: 
:$\ds \map {P_X} \R = \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }$
So, we obtain:
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have: 
:for all increasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map {F_X} {x_n} \to 1$.
So, from Sequential Characterization of Limit at Positive Infinity of Real Function: Corollary, we have: 
:$\ds \lim_{x \mathop \to \infty} \map {F_X} x = 1$
== Sources ==
* 
* 
Category:Limit of Cumulative Distribution Function at Positive Infinity
Category:Cumulative Distribution Functions
Category:Limit at Infinity
Limit of Cumulative Distribution Function at Positive Infinity/Lemma

== Lemma ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be an increasing with $x_n \to \infty$. 
Then:
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \R$
== Proof ==
Clearly we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} \subseteq \R$
So we only need to show that: 
:$\ds \R \subseteq \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$ 
Let $x \in \R$. 
From the definition of a sequence diverging to $\infty$:
:there exists $N \in \N$ such that $x_N > x$.
So:
:$x \in \hointl {-\infty} {x_N}$
giving:
:$\ds x \in \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
So:
:$\ds \R \subseteq \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$ 
from the definition of subset. 
So, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \R$
from the definition of set equality.
Category:Limit of Cumulative Distribution Function at Positive Infinity