nyuuzyou/wikis huggingface dataset:
Limit of Increasing Function/Corollary

== Corollary to Limit of Increasing Function ==
Let $f$ be a real function which is increasing on the open interval $\openint a b$.
If $\xi \in \openint a b$, then:
:$\map f {\xi^-}$ and $\map f {\xi^+}$ both exist
and:
:$\map f x \le \map f {\xi^-} \le \map f \xi \le \map f {\xi^+} \le \map f y$
provided that $a < x < \xi < y < b$.
== Proof ==
$f$ is bounded above on $\openint a \xi$ by $\map f \xi$.
By Limit of Increasing Function, the supremum is $\map f {\xi^-}$.
So it follows that:
:$\forall x \in \openint a \xi: \map f x \le \map f {\xi^-} \le \map f \xi$
A similar argument for $\openint \xi b$ holds for the other inequalities.
== Sources ==
* : $\S 12.5$
Category:Limits of Real Functions
Limit of Increasing Sequence of Sets is Union

== Definition ==
Let $\sequence {E_k}_{k \mathop \in \N}$ be an increasing sequence:
:$\forall k \in \N: E_k \subseteq E_{k + 1}$
Then $\sequence {E_k}_{k \mathop \in \N}$ has a limit such that:
:$\ds \lim_{n \mathop \to \infty} E_n = \bigcup_{k \mathop \in \N} {E_k}$
== Proof ==
Let $E = \ds \bigcup_{k \mathop \in \N} {E_k}$.
Let $x \in \ds \limsup_{n \mathop \to \infty} E_n$, where $\ds \limsup_{n \mathop \to \infty} E_n$ denotes the limit superior of $\sequence {E_k}_{k \mathop \in \N}$.
By definition of set union, $x \in E$.
Hence $\ds \limsup_{n \mathop \to \infty} E_n \subseteq E$.
Let $x \in E$.
Then $x \in E_n$ for some $n \in \N$.
Hence, as $\sequence {E_k}_{k \mathop \in \N}$ is an increasing sequence:
:$\forall m \in N: m > n: x \in E_m$
That is, $x \in E_i$ for all but a finite number of $i$.
That is:
:$x \in \ds \liminf_{n \mathop \to \infty} E_n$
So we have shown that:
:$E \subseteq \ds \liminf_{n \mathop \to \infty} E_n$
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:Increasing Sequences of Sets
Limit of Integer to Reciprocal Power

== Theorem ==
Let $\sequence {x_n}$ be the real sequence defined as $x_n = n^{1/n}$, using exponentiation.
Then $\sequence {x_n}$ converges with a limit of $1$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Limits of Sequences
Category:Reciprocals
Category:Limit of Integer to Reciprocal Power
Limit of Integer to Reciprocal Power/Proof 1

== Theorem ==
== Proof ==
From Number to Reciprocal Power is Decreasing we have that the real sequence $\sequence {n^{1/n} }$ is decreasing for $n \ge 3$.
Now, as $n^{1 / n} > 0$ for all positive $n$, it follows that $\sequence {n^{1 / n} }$ is bounded below (by $0$, for a start).
Thus the subsequence of $\sequence {n^{1 / n} }$ consisting of all the terms of $\sequence {n^{1 / n} }$ where $n \ge 3$ is convergent by the Monotone Convergence Theorem (Real Analysis).
Now we need to demonstrate that this limit is in fact $1$.
Let $n^{1 / n} \to l$ as $n \to \infty$.
Having established this, we can investigate the subsequence $\sequence {\paren {2 n}^{1 / {2 n} } }$.
By Limit of Subsequence equals Limit of Real Sequence, this will converge to $l$ also.
From Limit of Root of Positive Real Number, we have that $2^{1 / {2 n} } \to 1$ as $n \to \infty$.
So $n^{1 / {2 n} } \to l$ as $n \to \infty$ by the Combination Theorem for Sequences.
Thus:
:$n^{1 / n} = n^{1 / {2 n} } \cdot n^{1 / {2 n} } \to l \cdot l = l^2$
as $n \to \infty$.
So $l^2 = l$, and as $l \ge 1$ the result follows.
== Sources ==
* : $\S 5$: Subsequences: Exercise $\S 5.7 \ (1)$
Category:Limit of Integer to Reciprocal Power
Limit of Integer to Reciprocal Power/Proof 2

== Theorem ==
== Proof ==
We have the definition of the power to a real number:
:$\ds n^{1/n} = \map \exp {\frac 1 n \ln n}$
From Powers Drown Logarithms, we have that:
:$\ds \lim_{n \mathop \to \infty} \frac 1 n \ln n = 0$
Hence:
:$\ds \lim_{n \mathop \to \infty} n^{1/n} = \exp 0 = 1$
and the result.
== Sources ==
* : $\S 14.7 \ (4)$
Category:Limit of Integer to Reciprocal Power
Limit of Integer to Reciprocal Power/Proof 3

== Theorem ==
== Proof ==
Let $n^{1/n} = 1 + a_n$.
The strategy is to:
:$(1): \quad$ prove that $a_n > 0$ for $n > 1$
:$(2): \quad$ deduce that $n - 1 \ge \dfrac {n \paren {n - 1} } {2!} a_n^2$ for $n > 1$
and hence:
:$(3): \quad$ deduce that $0 \le a_n^2 \le \dfrac 2 n$
Let $n > 1$.
Then:
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 10$
Category:Limit of Integer to Reciprocal Power
Limit of Intersection of Closed Intervals from Zero to Positive Integer Reciprocal

== Theorem ==
For all (strictly) positive integers $n \in \Z_{>0}$, let $A_n$ be the closed real interval:
:$A_n = \closedint 0 {\dfrac 1 n}$
Let $A \subseteq \R$ be the subset of the real numbers defined as:
:$A = \ds \lim_{n \mathop \to \infty} \bigcap A_n$
Then:
:$A = \set 0$
== Proof ==
First it is noted that:
:$\forall x \in \R_{<0}: x \notin A$
and that by definition of closed real interval:
:$\forall n \in \Z_{>0}: 0 \in A_n$
and so by definition of intersection:
:$0 \in A$
It remains to demonstrate that:
:$\forall x \in \R_{>0}: x \notin A$
 $\exists x \in \R_{>0}: x \in A$.
By the Axiom of Archimedes:
:$\exists N \in \Z: N > \dfrac 1 x$
and so from Reciprocal Function is Strictly Decreasing:
:$\exists N \in \Z: \dfrac 1 N < x$
Thus:
:$x \notin A_N$
and so by definition of intersection:
:$x \notin A$
This contradicts our supposition that $x \in A$.
Hence the only element of $A$ is $0$, and so:
:$A = \set 0$
== Sources ==
* : Chapter $1$: Sets and Logic: Exercise $10$
Category:Real Analysis
Limit of Modulo Operation

== Theorem ==
Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then holding $x$ fixed gives:
:$\ds \lim_{y \mathop \to 0} x \bmod y = 0$
:$\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$
== Proof ==
=== Limit 1 ===
=== Limit 2 ===
Category:Limits of Real Functions
Category:Limit of Modulo Operation
Limit of Modulo Operation/Limit 1

== Theorem ==
Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then $\ds \lim_{y \mathop \to 0} x \bmod y = 0$.
== Proof ==
By Range of Modulo Operation for Positive Modulus and Range of Modulo Operation for Negative Modulus we have:
:$-\size y < x \bmod y < \size y$
The result follows from the Squeeze Theorem for Functions.
Category:Limit of Modulo Operation
Limit of Modulo Operation/Limit 2

== Theorem ==
Let $x$ and $y$ be real numbers.
Let $x \bmod y$ denote the modulo operation.
Then $\ds \lim_{y \mathop \to \infty} x \bmod y = x$ if $x \ge 0$.
== Proof ==
As $y \to \infty$:
Therefore by the definition of modulo operation:
Hence the result.
Category:Limit of Modulo Operation
Limit of Monotone Real Function/Decreasing

== Theorem ==
Let $f$ be a real function which is decreasing and bounded below on the open interval $\openint a b$.
Let the infimum of $f$ on $\openint a b$ be $l$.
Then:
:$\ds \lim_{x \mathop \to b^-} \map f x = l$
where $\ds \lim_{x \mathop \to b^-} \map f x$ is the limit of $f$ from the left at $b$.
=== Corollary ===
== Proof ==
Let $\epsilon > 0$.
We have to find a value of $\delta > 0$ such that:
:$\forall x: b - \delta < x < b: \size {\map f x - l} < \epsilon$
That is:
:$l - \epsilon < \map f x < l + \epsilon$
As $l$ is a lower bound for $f$ on $\openint a b$:
:$l - \epsilon < \map f x$
automatically happens
Because $l + \epsilon$ is not a lower bound for $f$ on $\openint a b$:
:$\exists y \in \openint a b: \map f y < l + \epsilon$
But $f$ decreases on $\openint a b$.
So:
:$\forall x: y < x < b: \map f x \le \map f y < l + \epsilon$
We choose $\delta = y - a$ and hence the result.
== Sources ==
* : $\S 12.4 \ \text{(ii)}$
Category:Limits of Real Functions
Limit of Monotone Real Function/Decreasing/Corollary

== Corollary to Limit of Decreasing Function ==
Let $f$ be a real function which is decreasing on the open interval $\openint a b$.
If $\xi \in \openint a b$, then:
:$\map f {\xi^-}$ and $\map f {\xi^+}$ both exist
and:
:$\map f x \ge \map f {\xi^-} \ge \map f \xi \ge \map f {\xi^+} \ge \map f y$
provided that $a < x < \xi < y < b$.
== Proof ==
$f$ is bounded below on $\openint a \xi$ by $\map f \xi$.
By Limit of Decreasing Function, the infimum is $\map f {\xi^-}$.
So it follows that:
:$\forall x \in \openint a \xi: \map f x \ge \map f {\xi^-} \ge \map f \xi$
A similar argument for $\openint \xi b$ holds for the other inequalities.
Category:Limits of Real Functions
Limit of Monotone Real Function/Increasing

== Theorem ==
Let $f$ be a real function which is increasing and bounded above on the open interval $\openint a b$.
Let the supremum of $f$ on $\openint a b$ be $L$.
Then:
:$\ds \lim_{x \mathop \to b^-} \map f x = L$
where $\ds \lim_{x \mathop \to b^-} \map f x$ is the limit of $f$ from the left at $b$.
=== Corollary ===
== Proof ==
Let $\epsilon > 0$.
We have to find a value of $\delta > 0$ such that $\forall x: b - \delta < x < b: \size {\map f x - L} < \epsilon$.
That is, that $L - \epsilon < \map f x < L + \epsilon$.
As $L$ is an upper bound for $f$ on $\openint a b$, $\map f x < L + \epsilon$ automatically happens.
Since $L - \epsilon$ is not an upper bound for $f$ on $\openint a b$, it follows that:
:$\exists y \in \openint a b: \map f y > L - \epsilon$
But $f$ increases on $\openint a b$.
So:
:$\forall x: y < x < b: L - \epsilon < \map f y \le \map f x$
We choose $\delta = b - y$ and hence the result.
== Sources ==
* : $\S 12.4 \ \text{(i)}$
Category:Limits of Real Functions
Limit of Positive Real Sequence is Positive

== Theorem ==
Let $\sequence {x_n}$ be a sequence of positive real numbers.
Let $x_n$ converge to $L$.
Then $L \ge 0$.
== Proof ==
 $L < 0$.
Then for any $n \in \N$:
This contradicts .
Hence we must have $L \ge 0$.
Category:Convergence
Limit of Power of x by Absolute Value of Power of Logarithm of x

== Theorem ==
Let $\alpha$ and $\beta$ be positive real numbers. 
Then: 
:$\ds \lim_{x \mathop \to 0^+} x^\alpha \size {\ln x}^\beta = 0$ 
=== Corollary ===
== Proof ==
From Order of Natural Logarithm Function, we have: 
:$\ln x = \map \OO {x^{-\frac \alpha {2 \beta} } }$ as $x \to 0^+$
That is, by the definition of big-O notation there exists positive real numbers $x_0$ and $C$ such that: 
:$0 \le \size {\ln x} \le C x^{-\frac \alpha {2 \beta} }$
for $0 < x \le x_0$.
So:
:$0 \le \size {\ln x}^\beta \le C^\beta x^{-\alpha/2}$
for $0 < x \le x_0$. 
That is: 
:$0 \le x^\alpha \size {\ln x}^\beta \le C^\beta x^{\alpha/2}$
We have that: 
:$\ds \lim_{x \mathop \to 0^+} C^\beta x^{\alpha/2} = 0$
so by the Squeeze Theorem for Functions:
:$\ds \lim_{x \mathop \to 0^+} x^\alpha \size {\ln x}^\beta = 0$
Category:Logarithms
Category:Limits of Real Functions
Category:Limit of Power of x by Absolute Value of Power of Logarithm of x
Limit of Power of x by Absolute Value of Power of Logarithm of x/Corollary

== Corollary to Limit of Power of x by Absolute Value of Power of Logarithm of x ==
Let $k$ be a positive real number. 
Let $n$ be a positive integer. 
Then: 
:$\ds \lim_{x \mathop \to 0^+} x^k \paren {\ln x}^n = 0$
== Proof ==
From Limit of $x^\alpha \size {\ln x}^\beta$, we have: 
:$\ds \lim_{x \mathop \to 0^+} x^k \size {\ln x}^n = 0$
For $0 < x \le 1$, we have: 
:$\ln x \le 0$
so by the definition of the absolute value, we have: 
:$\size {\ln x} = -\ln x$
so:
:$\ds \lim_{x \mathop \to 0^+} x^k \paren {-\ln x}^n = 0$
That is, from the Multiple Rule for Limits of Real Functions:
:$\ds \paren {-1}^n \lim_{x \mathop \to 0^+} x^k \paren {\ln x}^n = 0$
giving:
:$\ds \lim_{x \mathop \to 0^+} x^k \paren {\ln x}^n = 0$
Category:Limit of Power of x by Absolute Value of Power of Logarithm of x
Limit of Rational Sequence is not necessarily Rational

== Theorem ==
Let $S = \sequence {a_n}$ be a rational sequence.
Let $S$ be convergent to a limit $L$.
Then it is not necessarily the case that $L$ is itself a rational number.
== Proof ==
Proof by Counterexample:
By definition, Euler's number $e$ can be defined as:
:$e = \ds \sum_{n \mathop = 0}^\infty \frac 1 {n!}$
Each of the terms of the sequence of partial sums is rational.
However, from Euler's Number is Irrational, $e$ itself is not.
== Sources ==
* : $\S 1.1$. Number Systems
Category:Rational Sequences
Limit of Real Function/Examples

== Examples of Limits of Real Functions ==
=== Example: Identity Function with $1$ at $0$ ===
=== Example: $\sqrt x$ at $1$ ===
=== Example: $\dfrac 1 x$ at $+\infty$ ===
=== Example: $\dfrac {\sin x} x$ at $+\infty$ ===
=== Example: $e^{-1 / \size x}$ at $0$ ===
=== Example: $\map \sin {\dfrac 1 x}$ at $0$ ===
=== Example: $x \map \sin {\dfrac 1 x}$ at $0$ ===
Category:Examples of Limits of Real Functions
Limit of Real Function/Examples/Identity Function with Discontinuity at 0

== Example of Limit of Real Function ==
Let $f$ be the real function defined as:
:$\map f x = \begin {cases} x & : x \ne 0 \\ 1 & : x = 0 \end {cases}$
Then:
:$\ds \lim_{x \mathop \to 0} \map f x = 0$
== Proof ==
By definition of the limit of a real function:
:$\ds \lim_{x \mathop \to a} \map f x = A$
:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$
In this instance, we have:
:$\map f x = x$ for $x \ne 0$
:$A = 0$
Let $\epsilon \in \R_{>0}$ be arbitrary.
Let $\delta = \epsilon$.
Let $0 < \size {x - 1} < \delta = \epsilon$.
Then we obtain:
Hence the result.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.3$: Limits of functions: Example $1.3.2$
Category:Examples of Limits of Real Functions
Limit of Real Function/Examples/Root of x at 1

== Example of Limit of Real Function ==
:$\ds \lim_{x \mathop \to 1} \sqrt x = 1$
== Proof ==
By definition of the limit of a real function:
:$\ds \lim_{x \mathop \to a} \map f x = A$
:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$
In this instance, we have:
:$\map f x = \sqrt x$
:$A = 1$
So:
for $0 < x < 2$.
Let $\delta = \epsilon$.
Let $0 < \size {x - 1} < \delta = \epsilon$.
Then we obtain:
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity: Example $\text A$
Category:Examples of Limits of Real Functions
Limit of Real Function/Examples/Sine of Reciprocal of x at 0

== Example of Limit of Real Function ==
Let:
:$\map f x = \map \sin {\dfrac 1 x}$
Then:
:$\ds \lim_{x \mathop \to 0} \map f x$
does not exist.
== Proof ==
By definition of the limit of a real function:
:$\ds \lim_{x \mathop \to a} \map f x = A$
:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$
Let $\epsilon = \dfrac 1 2$.
Let $\delta \in \R_{>0}$.
Let $n \in \N$ such that $N = \dfrac 2 {\pi \paren {1 + 4 n} } < \delta$.
Then:
:$\sin N = 1$
and so it is not the case that:
:$0 < \size x < \delta \implies \size {\map f x} < \epsilon$
It follows that $\map \sin {\dfrac 1 x}$ has no limit at $x = 0$.
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.1$ Limits and Continuity: Example $\text B$
* : $1$: Review of some real analysis: $\S 1.3$: Limits of functions: Examples $1.3.4$
* : $1$: Review of some real analysis: Exercise $1.5: 16$
Category:Examples of Limits of Real Functions
Limit of Real Function/Examples/e^-1 over size x at 0

== Example of Limit of Real Function ==
:$\ds \lim_{x \mathop \to 0} e^{-1 / \size x} = 0$
== Proof ==
:
By definition of the limit of a real function:
:$\ds \lim_{x \mathop \to a} \map f x = A$
:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - a} < \delta \implies \size {\map f x - A} < \epsilon$
Let $\epsilon \in \R_{>0}$ be chosen arbitrarily.
Then we have:
So, having been given an arbitrary $\epsilon \in \R_{>0}$, let $\delta = \dfrac 1 {\map \ln {1 / \epsilon} }$.
Then:
:$0 < \size x < \delta \implies \size {e^{-1 / \size x} } < \epsilon$
Hence by definition of limit of a real function:
:$\ds \lim_{x \mathop \to 0} e^{-1 / \size x} = 0$
== Sources ==
* : $1$ Partial Differentiation: $\S 2$. Functions of One Variable: Exercise $1 \, \text {(a)}$
Category:Examples of Limits of Real Functions
Limit of Real Function/Examples/x times Sine of Reciprocal of x at 0

== Example of Limit of Real Function ==
Let:
:$\map f x = x \map \sin {\dfrac 1 x}$
Then:
:$\ds \lim_{x \mathop \to 0} \map f x = 0$
== Proof ==
== Sources ==
* : $1$: Review of some real analysis: $\S 1.3$: Limits of functions: Examples $1.3.4$
Category:Examples of Limits of Real Functions
Limit of Real Function/Examples/x times Sine of Reciprocal of x at 0/Proof

== Example of Limit of Real Function ==
== Proof ==
By definition of the limit of a real function:
:$\ds \lim_{x \mathop \to 0} \map f x = A$
:
:$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size x < \delta \implies \size {\map f x - A} < \epsilon$
Let $\epsilon \in \R_{>0}$ be chosen arbitrarily.
Let $\delta = \epsilon$.
Then we have:
Hence the result.
== Sources ==
* : $1$: Review of some real analysis: Exercise $1.5: 16$
Category:Examples of Limits of Real Functions
Limit of Real Function of 2 Variables/Examples/(x - y) over (x + y) at 0

== Example of Limit of Real Function of 2 Variables ==
Let $f$ be the real function of $2$ variables defined as:
:$\map f {x, y} = \begin {cases} \dfrac {x - y} {x + y} & : x \ne -y \\ 1 & : x = -y \end {cases}$
Then:
:$\ds \lim_{\substack {x \mathop \to 0 \\ y \mathop \to 0} } \map f {x, y}$
does not exist.
However, note that:
=== $\ds \lim_{x \mathop \to 0} \paren {\lim_{y \mathop \to 0} \map f {x, y} }$ equals $1$ ===
=== $\ds \lim_{y \mathop \to 0} \paren {\lim_{x \mathop \to 0} \map f {x, y} }$ equals $-1$ ===
== Proof ==
Let $\epsilon$ be arbitrary.
Consider $\tuple {x, y}$ along the $L$ defined as $x = -y$.
Then selecting points close to $L$ we can make $\dfrac {x - y} {x + y}$ as large as we like by making $x - \paren {-y}$ as small as we like.
Hence any square around $\tuple {0, 0}$ is going to contain $\tuple {x, y}$ for which $\map f {x, y}$ is greater than any $\epsilon$ you pick.
Hence the result.
== Sources ==
* : $1$ Partial Differentiation: $\S 3$. Functions of Several Variables: $3.1$ Limits and Continuity: Example $\text B$
Category:Examples of Limits of Real-Valued Functions
Limit of Real Function of 2 Variables/Examples/x^2 + y^2 at 0

== Example of Limit of Real Function of 2 Variables ==
Let $f$ be the real function of $2$ variables defined as:
:$\map f {x, y} = x^2 + y^2$
Then:
:$\ds \lim_{\substack {x \mathop \to 0 \\ y \mathop \to 0} } \map f {x, y} = 0$
where $\lim$ denotes the limit of $f$.
== Proof ==
Let $\epsilon$ be arbitrary.
Let $\delta = \sqrt {\epsilon / 2}$.
Then we have:
:$\size x < \sqrt {\epsilon / 2}$
and:
:$\size y < \sqrt {\epsilon / 2}$
together imply that:
:$x^2 + y^2 < \epsilon$
Hence the result.
== Sources ==
* : $1$ Partial Differentiation: $\S 3$. Functions of Several Variables: $3.1$ Limits and Continuity: Example $\text A$
Category:Examples of Limits of Real-Valued Functions
Limit of Root of Positive Real Number

== Theorem ==
Let $x \in \R: x > 0$ be a real number.
Let $\sequence {x_n}$ be the sequence in $\R$ defined as:
:$x_n = x^{1 / n}$
Then $x_n \to 1$ as $n \to \infty$.
== Proof 1 ==
== Proof 2 ==
Category:Limits of Sequences
Category:Limit of Root of Positive Real Number
Limit of Root of Positive Real Number/Mistake

== Source Work ==
:
: Chapter $\S 5$: Subsequences
:: $\S 5.4$: Example
This mistake can be seen in the $1977$ edition as published by Cambridge University Press: ISBN 0 521 29167 4
== Mistake ==
:In example $4.4$ we showed that, if $x > 0$, then $x^{1 / n} \to 1$ as $n \to \infty$. We prove this again by a different method. As in example $4.4$ we need only consider the case $x \ge 1$.
== Correction ==
That should actually say:
:In example $4.14$ ... As in example $4.14$ ...
== Sources ==
* : $\S 5$: Subsequences: $\S 5.4$ Example
Category:Mistakes/Mathematical Analysis: A Straightforward Approach
Limit of Root of Positive Real Number/Proof 1

== Theorem ==
== Proof ==
Let us define $a_1 = a_2 = \cdots = a_{n-1} = 1$ and $a_n = x$.
Let $G_n$ be the geometric mean of $a_1, \ldots, a_n$.
Let $A_n$ be the arithmetic mean of $a_1, \ldots, a_n$.
From their definitions:
:$G_n = x^{1/n}$
and:
:$A_n = \dfrac {n - 1 + x} n = 1 + \dfrac{x - 1} n$
From Arithmetic Mean is Never Less than Geometric Mean:
:$x^{1/n} \le 1 + \dfrac{x - 1} n$
That is:
:$x^{1/n} - 1 \le \dfrac{x - 1} n$
There are two cases to consider: $x \ge 1$ and $0 < x < 1$.
Let $x \ge 1$.
From Root of Number Greater than One‎, it follows that:
:$x^{1/n} \ge 1$
Thus:
:$0 \le x^{1/n} - 1 \le \dfrac 1 n \paren {x - 1}$
But from Sequence of Powers of Reciprocals is Null Sequence:
:$\dfrac 1 n \to 0$ as $n \to \infty$
From the Combination Theorem for Sequences:
:$\dfrac 1 n \paren {x - 1} \to 0$ as $n \to \infty$
Thus by the Squeeze Theorem:
:$x^{1/n} - 1 \to 0$ as $n \to \infty$
Hence, again from the Combination Theorem for Sequences:
:$x^{1/n} \to 1$ as $n \to \infty$
Now let $0 < x < 1$.
Then $x = \dfrac 1 y$ where $y > 1$.
But from the above:
:$y^{1/n} \to 1$ as $n \to \infty$
Hence by the Combination Theorem for Sequences:
:$x^{1/n} = \dfrac 1 {y^{1/n} } \to \dfrac 1 1 = 1$ as $n \to \infty$
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.14$: Example
Category:Limit of Root of Positive Real Number
Limit of Root of Positive Real Number/Proof 2

== Theorem ==
== Proof ==
We consider the case where $x \ge 1$; when $0 < x < 1$ the proof can be completed as for proof 1.
From Root of Number Greater than One‎:
:$x^{1/n} \ge 1$
Hence $\sequence {x^{1/n} }$ is bounded below by $1$.
Now consider $x^{1/n} / x^{1 / \paren {n + 1} }$:
So:
:$x^{1/n} > x^{\frac 1 {n + 1} }$
and so $\sequence {x^{1 / n} }$ is strictly decreasing.
Hence from the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {x^{1 / n} }$ converges to a limit $l$ and that $l \ge 1$.
Now, since we know that $\sequence {x^{1 / n} }$ is convergent, we can apply Limit of Subsequence equals Limit of Real Sequence.
That is, any subsequence of $\sequence {x^{1 / n} }$ must also converge to $l$.
So we take the subsequence:
:$\sequence {x^{1 / {2 n} } }$
From what has just been shown:
:$x^{1 / {2 n} } \to l$ as $n \to \infty$
Using the Combination Theorem for Sequences, we have:
:$x^{1 / n} = x^{1 / {2 n} } \cdot x^{1 / {2 n} } \to l \cdot l = l^2$ as $n \to \infty$
But a Convergent Real Sequence has Unique Limit, so $l^2 = l$ and so $l = 0$ or $l = 1$.
But $l \ge 1$ and so $l = 1$.
Hence the result.
== Sources ==
* : $\S 5$: Subsequences: $\S 5.4$ Example
Category:Limit of Root of Positive Real Number
Limit of Sequence in Metric Space in Neighborhood

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\sequence {a_n}$ be a sequence in $A$.
Then $\ds \lim_{n \mathop \to \infty} a_n = a$ for each neighborhood $V$ of $a$:
:$\exists N \in \N: n > N \implies a_n \in V$
== Proof ==
=== Necessary Condition ===
Let $\ds \lim_{n \mathop \to \infty} a_n = a$.
Let $V$ be a neighborhood of $a$.
By definition of neighborhood:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq V$
where $\map {B_\epsilon} a$ denotes the open $\epsilon$-ball of $a$ in $M$.
By definition of limit:
:$\exists N \in \N: n > N \implies \map d {a, a_n} < \epsilon$
Hence $a_n \in V$.
=== Sufficient Condition ===
Let $\sequence {a_n}$ be such that for each neighborhood $V$ of $a$:
:$\exists N \in \N: n > N \implies a_n \in V$
Let $\epsilon \in \R_{>0}$.
Then by Open Ball is Neighborhood of all Points Inside, $\map {B_\epsilon} a$ is a neighborhood of $a$.
Let $N \in \N$ be such that:
:$\forall n > N: a_n \in \map {B_\epsilon} a$
Then:
:$\map d {a, a_n} < \epsilon$
and so by definition of limit:
:$\ds \lim_{n \mathop \to \infty} a_n = a$
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 5$: Limits: Corollary $5.3$
Category:Metric Spaces
Category:Limits of Sequences
Category:Neighborhoods