Real Function Limits:Sequential Criterion - Revision history

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Lila at 15:45, 20 October 2021

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-The Sequential Criterion for a Limit of a Function
-<p>We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function <math>f</math> at a cluster point <math>c</math> from <math>A</math> with regards to sequences <math>(a_n)</math> from <math>A</math> that converge to <math>c</math>.</p>
+<p>We will now look at a very important theorem known as '''The Sequential Criterion for a Limit''' which merges the concept of the limit of a function <math>f</math> at a cluster point <math>c</math> from <math>A</math> with regards to sequences <math>(a_n)</math> from <math>A</math> that converge to <math>c</math>.</p>
 <tr>

@@ Line 27: / Line 27: @@
 </ul>
+== Licensing ==
+Content obtained and/or adapted from:
-==Resources==
+* [http://mathonline.wikidot.com/the-sequential-criterion-for-a-limit-of-a-function The Sequential Criterion for a Limit of a Function, mathonline.wikidot.com] under a CC BY-SA license

@@ Line 1: / Line 1: @@
 The Sequential Criterion for a Limit of a Function
 <p>We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function <math>f</math> at a cluster point <math>c</math> from <math>A</math> with regards to sequences <math>(a_n)</math> from <math>A</math> that converge to <math>c</math>.</p>
-<table class="wiki-content-table">
 <tr>
 <td><strong>Theorem 1 (The Sequential Criterion for a Limit of a Function):</strong> Let <math>f : A \to \mathbb{R}</math> be a function and let <math>c</math> be a cluster point of <math>A</math>. Then <math>\lim_{x \to c} f(x) = L</math> if and only if for all sequences <math>(a_n)</math> from the domain <math>A</math> where <math>a_n \neq c</math> <math>\forall n \in \mathbb{N}</math> and <math>\lim_{n \to \infty} a_n = c</math> then <math>\lim_{n \to \infty} f(a_n) = L</math>.</td>
 </tr>
-</table>
 <p>Consider a function <math>f</math> that has a limit <math>L</math> when <math>x</math> is close to <math>c</math>. Now consider all sequences <math>(a_n)</math> from the domain <math>A</math> where these sequences converge to <math>c</math>, that is <math>\lim_{n \to \infty} a_n = c</math>. The Sequential Criterion for a Limit of a Function says that then that as <math>n</math> goes to infinity, the function <math>f</math> evaluated at these <math>a_n</math> will have its limit go to <math>L</math>.</p>
 <p>For example, consider the function <math>f: \mathbb{R} \to \mathbb{R}</math> defined by the equation <math>f(x) = x</math>, and suppose we wanted to compute <math>\lim_{x \to 0} x</math>. We should already know that this limit is zero, that is <math>\lim_{x \to 0} x = 0</math>. Now consider the sequence <math>(a_n) = \left ( \frac{1}{n} \right)</math>. This sequence <math>(a_n)</math> is clearly contained in the domain of <math>f</math>. Furthermore, this sequence converges to 0, that is <math>\lim_{n \to \infty} \frac{1}{n} = 0</math>. If all such sequences <math>(a_n)</math> that converge to <math>0</math> have the property that <math>(f(a_n))</math> converges to <math>f(0) = 0</math>, then we can say that <math>\lim_{n \to 0} f(x) = 0</math>.</p>

@@ Line 13: / Line 13: @@
 </ul>
 <ul>
-<li>Let <math>\varepsilon > 0</math>. We are given that <math>\lim_{x \to c} f(x) = L</math> and so for <math>\varepsilon > 0</math> there exists a <math>\delta > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta</math> then we have that <math>\mid f(x) - L \mid < \varepsilon</math>. Now since <math>\delta > 0</math>, since we have that <math>\lim_{n \to \infty} a_n = c</math> then there exists an <math>N \in \mathbb{N}</math> such that if <math>n ≥ N</math> then <math>\mid a_n - c \mid < \delta</math>. Therefore <math>a_n \in V_{\delta} (c) \cap A</math>.</li>
+<li>Let <math>\varepsilon > 0</math>. We are given that <math>\lim_{x \to c} f(x) = L</math> and so for <math>\varepsilon > 0</math> there exists a <math>\delta > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta</math> then we have that <math>\mid f(x) - L \mid < \varepsilon</math>. Now since <math>\delta > 0</math>, since we have that <math>\lim_{n \to \infty} a_n = c</math> then there exists an <math>N \in \mathbb{N}</math> such that if <math>n \geq N</math> then <math>\mid a_n - c \mid < \delta</math>. Therefore <math>a_n \in V_{\delta} (c) \cap A</math>.</li>
 </ul>
 <ul>
-<li>Therefore it must be that <math>\mid f(a_n) - L \mid < \varepsilon</math>, in other words, <math>\forall n ≥ N</math> we have that <math>\mid f(a_n) - L \mid < \varepsilon</math> and so <math>\lim_{n \to \infty} f(a_n) = L</math>.</li>
+<li>Therefore it must be that <math>\mid f(a_n) - L \mid < \varepsilon</math>, in other words, <math>\forall n \geq N</math> we have that <math>\mid f(a_n) - L \mid < \varepsilon</math> and so <math>\lim_{n \to \infty} f(a_n) = L</math>.</li>
 </ul>
 <ul>
@@ Line 22: / Line 22: @@
 </ul>
 <ul>
-<li>Suppose not, in other words, suppose that <math>\exists \varepsilon_0 > 0</math> such that <math>\forall \delta > 0</math> then <math>\exists x_{\delta} \in A \cap V_{\delta} (c) \setminus \{ c \}</math> such that <math>\mid f(x_{\delta}) - L \mid ≥ \varepsilon_0</math>. Let <math>\delta_n = \frac{1}{n}</math>. Then there exists <math>x_{\delta_n} = a_n \in A \cap V_{\delta_n} (c) \setminus \{ c \}</math>, in other words, <math>0 < \mid a_n - c \mid < \frac{1}{n}</math> and <math>\lim_{n \to \infty} a_n = c</math>. However, <math>\mid f(a_n) - L \mid ≥ \varepsilon_0</math> so <math>\lim_{n \to \infty} f(a_n) \neq L</math>, a contradiction. Therefore <math>\lim_{x \to c} f(x) = L</math>.
+<li>Suppose not, in other words, suppose that <math>\exists \varepsilon_0 > 0</math> such that <math>\forall \delta > 0</math> then <math>\exists x_{\delta} \in A \cap V_{\delta} (c) \setminus \{ c \}</math> such that <math>\mid f(x_{\delta}) - L \mid \geq \varepsilon_0</math>. Let <math>\delta_n = \frac{1}{n}</math>. Then there exists <math>x_{\delta_n} = a_n \in A \cap V_{\delta_n} (c) \setminus \{ c \}</math>, in other words, <math>0 < \mid a_n - c \mid < \frac{1}{n}</math> and <math>\lim_{n \to \infty} a_n = c</math>. However, <math>\mid f(a_n) - L \mid \geq \varepsilon_0</math> so <math>\lim_{n \to \infty} f(a_n) \neq L</math>, a contradiction. Therefore <math>\lim_{x \to c} f(x) = L</math>.
 </ul>

@@ Line 1: / Line 1: @@
 The Sequential Criterion for a Limit of a Function
-<p</math>We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function <math</math>f</math</math> at a cluster point <math</math>c</math</math> from <math</math>A</math</math> with regards to sequences <math</math>(a_n)</math</math> from <math</math>A</math</math> that converge to <math</math>c</math</math>.</p</math>
+<p>We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function <math>f</math> at a cluster point <math>c</math> from <math>A</math> with regards to sequences <math>(a_n)</math> from <math>A</math> that converge to <math>c</math>.</p>
-<table class="wiki-content-table"</math>
+<table class="wiki-content-table">
-<tr</math>
+<tr>
-<td</math><strong</math>Theorem 1 (The Sequential Criterion for a Limit of a Function):</strong</math> Let <math</math>f : A \to \mathbb{R}</math</math> be a function and let <math</math>c</math</math> be a cluster point of <math</math>A</math</math>. Then <math</math>\lim_{x \to c} f(x) = L</math</math> if and only if for all sequences <math</math>(a_n)</math</math> from the domain <math</math>A</math</math> where <math</math>a_n \neq c</math</math> <math</math>\forall n \in \mathbb{N}</math</math> and <math</math>\lim_{n \to \infty} a_n = c</math</math> then <math</math>\lim_{n \to \infty} f(a_n) = L</math</math>.</td</math>
+<td><strong>Theorem 1 (The Sequential Criterion for a Limit of a Function):</strong> Let <math>f : A \to \mathbb{R}</math> be a function and let <math>c</math> be a cluster point of <math>A</math>. Then <math>\lim_{x \to c} f(x) = L</math> if and only if for all sequences <math>(a_n)</math> from the domain <math>A</math> where <math>a_n \neq c</math> <math>\forall n \in \mathbb{N}</math> and <math>\lim_{n \to \infty} a_n = c</math> then <math>\lim_{n \to \infty} f(a_n) = L</math>.</td>
-</tr</math>
+</tr>
-</table</math>
+</table>
-<p</math>Consider a function <math</math>f</math</math> that has a limit <math</math>L</math</math> when <math</math>x</math</math> is close to <math</math>c</math</math>. Now consider all sequences <math</math>(a_n)</math</math> from the domain <math</math>A</math</math> where these sequences converge to <math</math>c</math</math>, that is <math</math>\lim_{n \to \infty} a_n = c</math</math>. The Sequential Criterion for a Limit of a Function says that then that as <math</math>n</math</math> goes to infinity, the function <math</math>f</math</math> evaluated at these <math</math>a_n</math</math> will have its limit go to <math</math>L</math</math>.</p</math>
+<p>Consider a function <math>f</math> that has a limit <math>L</math> when <math>x</math> is close to <math>c</math>. Now consider all sequences <math>(a_n)</math> from the domain <math>A</math> where these sequences converge to <math>c</math>, that is <math>\lim_{n \to \infty} a_n = c</math>. The Sequential Criterion for a Limit of a Function says that then that as <math>n</math> goes to infinity, the function <math>f</math> evaluated at these <math>a_n</math> will have its limit go to <math>L</math>.</p>
-<p</math>For example, consider the function <math</math>f: \mathbb{R} \to \mathbb{R}</math</math> defined by the equation <math</math>f(x) = x</math</math>, and suppose we wanted to compute <math</math>\lim_{x \to 0} x</math</math>. We should already know that this limit is zero, that is <math</math>\lim_{x \to 0} x = 0</math</math>. Now consider the sequence <math</math>(a_n) = \left ( \frac{1}{n} \right)</math</math>. This sequence <math</math>(a_n)</math</math> is clearly contained in the domain of <math</math>f</math</math>. Furthermore, this sequence converges to 0, that is <math</math>\lim_{n \to \infty} \frac{1}{n} = 0</math</math>. If all such sequences <math</math>(a_n)</math</math> that converge to <math</math>0</math</math> have the property that <math</math>(f(a_n))</math</math> converges to <math</math>f(0) = 0</math</math>, then we can say that <math</math>\lim_{n \to 0} f(x) = 0</math</math>.</p</math>
+<p>For example, consider the function <math>f: \mathbb{R} \to \mathbb{R}</math> defined by the equation <math>f(x) = x</math>, and suppose we wanted to compute <math>\lim_{x \to 0} x</math>. We should already know that this limit is zero, that is <math>\lim_{x \to 0} x = 0</math>. Now consider the sequence <math>(a_n) = \left ( \frac{1}{n} \right)</math>. This sequence <math>(a_n)</math> is clearly contained in the domain of <math>f</math>. Furthermore, this sequence converges to 0, that is <math>\lim_{n \to \infty} \frac{1}{n} = 0</math>. If all such sequences <math>(a_n)</math> that converge to <math>0</math> have the property that <math>(f(a_n))</math> converges to <math>f(0) = 0</math>, then we can say that <math>\lim_{n \to 0} f(x) = 0</math>.</p>
-<p</math>We will now look at the proof of The Sequential Criterion for a Limit of a Function.</p</math>
+<p>We will now look at the proof of The Sequential Criterion for a Limit of a Function.</p>
-<ul</math>
+<ul>
-<li</math><strong</math>Proof:</strong</math> <math</math>\Rightarrow</math</math> Suppose that <math</math>\lim_{x \to c} f(x) = L</math</math>, and let <math</math>(a_n)</math</math> be a sequence in <math</math>A</math</math> such that <math</math>a_n \neq c</math</math> <math</math>\forall n \in \mathbb{N}</math</math> such that <math</math>\lim_{n \to \infty} a_n = c</math</math>. We thus want to show that <math</math>\lim_{n \to \infty} f(a_n) = L</math</math>.</li</math>
+<li><strong>Proof:</strong> <math>\Rightarrow</math> Suppose that <math>\lim_{x \to c} f(x) = L</math>, and let <math>(a_n)</math> be a sequence in <math>A</math> such that <math>a_n \neq c</math> <math>\forall n \in \mathbb{N}</math> such that <math>\lim_{n \to \infty} a_n = c</math>. We thus want to show that <math>\lim_{n \to \infty} f(a_n) = L</math>.</li>
-</ul</math>
+</ul>
-<ul</math>
+<ul>
-<li</math>Let <math</math>\varepsilon &gt; 0</math</math>. We are given that <math</math>\lim_{x \to c} f(x) = L</math</math> and so for <math</math>\varepsilon &gt; 0</math</math> there exists a <math</math>\delta &gt; 0</math</math> such that if <math</math>x \in A</math</math> and <math</math>0 < \mid x - c \mid < \delta</math</math> then we have that <math</math>\mid f(x) - L \mid < \varepsilon</math</math>. Now since <math</math>\delta &gt; 0</math</math>, since we have that <math</math>\lim_{n \to \infty} a_n = c</math</math> then there exists an <math</math>N \in \mathbb{N}</math</math> such that if <math</math>n ≥ N</math</math> then <math</math>\mid a_n - c \mid < \delta</math</math>. Therefore <math</math>a_n \in V_{\delta} (c) \cap A</math</math>.</li</math>
+<li>Let <math>\varepsilon > 0</math>. We are given that <math>\lim_{x \to c} f(x) = L</math> and so for <math>\varepsilon > 0</math> there exists a <math>\delta > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta</math> then we have that <math>\mid f(x) - L \mid < \varepsilon</math>. Now since <math>\delta > 0</math>, since we have that <math>\lim_{n \to \infty} a_n = c</math> then there exists an <math>N \in \mathbb{N}</math> such that if <math>n ≥ N</math> then <math>\mid a_n - c \mid < \delta</math>. Therefore <math>a_n \in V_{\delta} (c) \cap A</math>.</li>
-</ul</math>
+</ul>
-<ul</math>
+<ul>
-<li</math>Therefore it must be that <math</math>\mid f(a_n) - L \mid < \varepsilon</math</math>, in other words, <math</math>\forall n ≥ N</math</math> we have that <math</math>\mid f(a_n) - L \mid < \varepsilon</math</math> and so <math</math>\lim_{n \to \infty} f(a_n) = L</math</math>.</li</math>
+<li>Therefore it must be that <math>\mid f(a_n) - L \mid < \varepsilon</math>, in other words, <math>\forall n ≥ N</math> we have that <math>\mid f(a_n) - L \mid < \varepsilon</math> and so <math>\lim_{n \to \infty} f(a_n) = L</math>.</li>
-</ul</math>
+</ul>
-<ul</math>
+<ul>
-<li</math><math</math>\Leftarrow</math</math> Suppose that for all <math</math>(a_n)</math</math> in <math</math>A</math</math> such that <math</math>a_n \neq c</math</math> <math</math>\forall n \in \mathbb{N}</math</math> and <math</math>\lim_{n \to \infty} a_n = c</math</math>, we have that <math</math>\lim_{n \to \infty} f(a_n) = L</math</math>. We want to show that <math</math>\lim_{x \to c} f(x) = L</math</math>.</li</math>
+<li><math>\Leftarrow</math> Suppose that for all <math>(a_n)</math> in <math>A</math> such that <math>a_n \neq c</math> <math>\forall n \in \mathbb{N}</math> and <math>\lim_{n \to \infty} a_n = c</math>, we have that <math>\lim_{n \to \infty} f(a_n) = L</math>. We want to show that <math>\lim_{x \to c} f(x) = L</math>.</li>
-</ul</math>
+</ul>
-<ul</math>
+<ul>
-<li</math>Suppose not, in other words, suppose that <math</math>\exists \varepsilon_0 &gt; 0</math</math> such that <math</math>\forall \delta &gt; 0</math</math> then <math</math>\exists x_{\delta} \in A \cap V_{\delta} (c) \setminus \{ c \}</math</math> such that <math</math>\mid f(x_{\delta}) - L \mid ≥ \varepsilon_0</math</math>. Let <math</math>\delta_n = \frac{1}{n}</math</math>. Then there exists <math</math>x_{\delta_n} = a_n \in A \cap V_{\delta_n} (c) \setminus \{ c \}</math</math>, in other words, <math</math>0 < \mid a_n - c \mid < \frac{1}{n}</math</math> and <math</math>\lim_{n \to \infty} a_n = c</math</math>. However, <math</math>\mid f(a_n) - L \mid ≥ \varepsilon_0</math</math> so <math</math>\lim_{n \to \infty} f(a_n) \neq L</math</math>, a contradiction. Therefore <math</math>\lim_{x \to c} f(x) = L</math</math>. <math</math>\blacksquare</math</math></li</math>
+<li>Suppose not, in other words, suppose that <math>\exists \varepsilon_0 > 0</math> such that <math>\forall \delta > 0</math> then <math>\exists x_{\delta} \in A \cap V_{\delta} (c) \setminus \{ c \}</math> such that <math>\mid f(x_{\delta}) - L \mid ≥ \varepsilon_0</math>. Let <math>\delta_n = \frac{1}{n}</math>. Then there exists <math>x_{\delta_n} = a_n \in A \cap V_{\delta_n} (c) \setminus \{ c \}</math>, in other words, <math>0 < \mid a_n - c \mid < \frac{1}{n}</math> and <math>\lim_{n \to \infty} a_n = c</math>. However, <math>\mid f(a_n) - L \mid ≥ \varepsilon_0</math> so <math>\lim_{n \to \infty} f(a_n) \neq L</math>, a contradiction. Therefore <math>\lim_{x \to c} f(x) = L</math>.
-</ul</math>
+</ul>
 ==Resources==
 * [http://mathonline.wikidot.com/the-sequential-criterion-for-a-limit-of-a-function The Sequential Criterion for a Limit of a Function]

@@ Line 1: / Line 1: @@
-<h1 id="toc0"><span>The Sequential Criterion for a Limit of a Function</span></h1>
+The Sequential Criterion for a Limit of a Function
-<p>We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function <math>f</math> at a cluster point <math>c</math> from <math>A</math> with regards to sequences <math>(a_n)</math> from <math>A</math> that converge to <math>c</math>.</p>
+<p>We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function $f$ at a cluster point $c$ from $A$ with regards to sequences $(a_n)$ from $A$ that converge to $c$.</p>
 <table class="wiki-content-table">
 <tr>
-<td><strong>Theorem 1 (The Sequential Criterion for a Limit of a Function):</strong> Let <span class="math-inline"><math>f : A \to \mathbb{R}</math></span> be a function and let <span class="math-inline"><math>c</math></span> be a cluster point of <span class="math-inline"><math>c</math></span>. Then <span class="math-inline"><math>\lim_{x \to c} f(x) = L</math></span> if and only if for all sequences <span class="math-inline"><math>(a_n)</math></span> from the domain <span class="math-inline"><math>A</math></span> where <span class="math-inline"><math>a_n \neq c</math></span> <span class="math-inline"><math>\forall n \in \mathbb{N}</math></span> and <span class="math-inline"><math>\lim_{n \to \infty} a_n = c</math></span> then <span class="math-inline"><math>\lim_{n \to \infty} f(a_n) = L</math></span>.</td>
+<td><strong>Theorem 1 (The Sequential Criterion for a Limit of a Function):</strong> Let $f : A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. Then $\lim_{x \to c} f(x) = L$ if and only if for all sequences $(a_n)$ from the domain $A$ where $a_n \neq c$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = c$ then $\lim_{n \to \infty} f(a_n) = L$.</td>
 </tr>
 </table>
-<p>Consider a function <span class="math-inline">$f$</span> that has a limit <span class="math-inline">$L$</span> when <span class="math-inline">$x$</span> is close to <span class="math-inline">$c$</span>. Now consider all sequences <span class="math-inline">$(a_n)$</span> from the domain <span class="math-inline">$A$</span> where these sequences converge to <span class="math-inline">$c$</span>, that is <span class="math-inline">$\lim_{n \to \infty} a_n = c$</span>. The Sequential Criterion for a Limit of a Function says that then that as <span class="math-inline">$n$</span> goes to infinity, the function <span class="math-inline">$f$</span> evaluated at these <span class="math-inline">$a_n$</span> will have its limit go to <span class="math-inline">$L$</span>.</p>
+<p>Consider a function $f$ that has a limit $L$ when $x$ is close to $c$. Now consider all sequences $(a_n)$ from the domain $A$ where these sequences converge to $c$, that is $\lim_{n \to \infty} a_n = c$. The Sequential Criterion for a Limit of a Function says that then that as $n$ goes to infinity, the function $f$ evaluated at these $a_n$ will have its limit go to $L$.</p>
-<p>For example, consider the function <span class="math-inline">$f: \mathbb{R} \to \mathbb{R}$</span> defined by the equation <span class="math-inline">$f(x) = x$</span>, and suppose we wanted to compute <span class="math-inline">$\lim_{x \to 0} x$</span>. We should already know that this limit is zero, that is <span class="math-inline">$\lim_{x \to 0} x = 0$</span>. Now consider the sequence <span class="math-inline">$(a_n) = \left ( \frac{1}{n} \right)$</span>. This sequence <span class="math-inline">$(a_n)$</span> is clearly contained in the domain of <span class="math-inline">$f$</span>. Furthermore, this sequence converges to 0, that is <span class="math-inline">$\lim_{n \to \infty} \frac{1}{n} = 0$</span>. If all such sequences <span class="math-inline">$(a_n)$</span> that converge to <span class="math-inline">$0$</span> have the property that <span class="math-inline">$(f(a_n))$</span> converges to <span class="math-inline">$f(0) = 0$</span>, then we can say that <span class="math-inline">$\lim_{n \to 0} f(x) = 0$</span>.</p>
+<p>For example, consider the function $f: \mathbb{R} \to \mathbb{R}$ defined by the equation $f(x) = x$, and suppose we wanted to compute $\lim_{x \to 0} x$. We should already know that this limit is zero, that is $\lim_{x \to 0} x = 0$. Now consider the sequence $(a_n) = \left ( \frac{1}{n} \right)$. This sequence $(a_n)$ is clearly contained in the domain of $f$. Furthermore, this sequence converges to 0, that is $\lim_{n \to \infty} \frac{1}{n} = 0$. If all such sequences $(a_n)$ that converge to $0$ have the property that $(f(a_n))$ converges to $f(0) = 0$, then we can say that $\lim_{n \to 0} f(x) = 0$.</p>
 <p>We will now look at the proof of The Sequential Criterion for a Limit of a Function.</p>
 <ul>
-<li><strong>Proof:</strong> <span class="math-inline">$\Rightarrow$</span> Suppose that <span class="math-inline">$\lim_{x \to c} f(x) = L$</span>, and let <span class="math-inline">$(a_n)$</span> be a sequence in <span class="math-inline">$A$</span> such that <span class="math-inline">$a_n \neq c$</span> <span class="math-inline">$\forall n \in \mathbb{N}$</span> such that <span class="math-inline">$\lim_{n \to \infty} a_n = c$</span>. We thus want to show that <span class="math-inline">$\lim_{n \to \infty} f(a_n) = L$</span>.</li>
+<li><strong>Proof:</strong> $\Rightarrow$ Suppose that $\lim_{x \to c} f(x) = L$, and let $(a_n)$ be a sequence in $A$ such that $a_n \neq c$ $\forall n \in \mathbb{N}$ such that $\lim_{n \to \infty} a_n = c$. We thus want to show that $\lim_{n \to \infty} f(a_n) = L$.</li>
 </ul>
 <ul>
-<li>Let <span class="math-inline">$\epsilon &gt; 0$</span>. We are given that <span class="math-inline">$\lim_{x \to c} f(x) = L$</span> and so for <span class="math-inline">$\epsilon &gt; 0$</span> there exists a <span class="math-inline">$\delta &gt; 0$</span> such that if <span class="math-inline">$x \in A$</span> and <span class="math-inline">$0 &lt; \mid x - c \mid &lt; \delta$</span> then we have that <span class="math-inline">$\mid f(x) - L \mid &lt; \epsilon$</span>. Now since <span class="math-inline">$\delta &gt; 0$</span>, since we have that <span class="math-inline">$\lim_{n \to \infty} a_n = c$</span> then there exists an <span class="math-inline">$N \in \mathbb{N}$</span> such that if <span class="math-inline">$n ≥ N$</span> then <span class="math-inline">$\mid a_n - c \mid &lt; \delta$</span>. Therefore <span class="math-inline">$a_n \in V_{\delta} (c) \cap A$</span>.</li>
+<li>Let $\epsilon &gt; 0$. We are given that $\lim_{x \to c} f(x) = L$ and so for $\epsilon &gt; 0$ there exists a $\delta &gt; 0$ such that if $x \in A$ and $0 &lt; \mid x - c \mid &lt; \delta$ then we have that $\mid f(x) - L \mid &lt; \epsilon$. Now since $\delta &gt; 0$, since we have that $\lim_{n \to \infty} a_n = c$ then there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - c \mid &lt; \delta$. Therefore $a_n \in V_{\delta} (c) \cap A$.</li>
 </ul>
 <ul>
-<li>Therefore it must be that <span class="math-inline">$\mid f(a_n) - L \mid &lt; \epsilon$</span>, in other words, <span class="math-inline">$\forall n ≥ N$</span> we have that <span class="math-inline">$\mid f(a_n) - L \mid &lt; \epsilon$</span> and so <span class="math-inline">$\lim_{n \to \infty} f(a_n) = L$</span>.</li>
+<li>Therefore it must be that $\mid f(a_n) - L \mid &lt; \epsilon$, in other words, $\forall n ≥ N$ we have that $\mid f(a_n) - L \mid &lt; \epsilon$ and so $\lim_{n \to \infty} f(a_n) = L$.</li>
 </ul>
 <ul>
-<li><span class="math-inline">$\Leftarrow$</span> Suppose that for all <span class="math-inline">$(a_n)$</span> in <span class="math-inline">$A$</span> such that <span class="math-inline">$a_n \neq c$</span> <span class="math-inline">$\forall n \in \mathbb{N}$</span> and <span class="math-inline">$\lim_{n \to \infty} a_n = c$</span>, we have that <span class="math-inline">$\lim_{n \to \infty} f(a_n) = L$</span>. We want to show that <span class="math-inline">$\lim_{x \to c} f(x) = L$</span>.</li>
+<li>$\Leftarrow$ Suppose that for all $(a_n)$ in $A$ such that $a_n \neq c$ $\forall n \in \mathbb{N}$ and $\lim_{n \to \infty} a_n = c$, we have that $\lim_{n \to \infty} f(a_n) = L$. We want to show that $\lim_{x \to c} f(x) = L$.</li>
 </ul>
 <ul>
-<li>Suppose not, in other words, suppose that <span class="math-inline">$\exists \epsilon_0 &gt; 0$</span> such that <span class="math-inline">$\forall \delta &gt; 0$</span> then <span class="math-inline">$\exists x_{\delta} \in A \cap V_{\delta} (c) \setminus \{ c \}$</span> such that <span class="math-inline">$\mid f(x_{\delta}) - L \mid ≥ \epsilon_0$</span>. Let <span class="math-inline">$\delta_n = \frac{1}{n}$</span>. Then there exists <span class="math-inline">$x_{\delta_n} = a_n \in A \cap V_{\delta_n} (c) \setminus \{ c \}$</span>, in other words, <span class="math-inline">$0 &lt; \mid a_n - c \mid &lt; \frac{1}{n}$</span> and <span class="math-inline">$\lim_{n \to \infty} a_n = c$</span>. However, <span class="math-inline">$\mid f(a_n) - L \mid ≥ \epsilon_0$</span> so <span class="math-inline">$\lim_{n \to \infty} f(a_n) \neq L$</span>, a contradiction. Therefore <span class="math-inline">$\lim_{x \to c} f(x) = L$</span>. <span class="math-inline">$\blacksquare$</span></li>
+<li>Suppose not, in other words, suppose that $\exists \epsilon_0 &gt; 0$ such that $\forall \delta &gt; 0$ then $\exists x_{\delta} \in A \cap V_{\delta} (c) \setminus \{ c \}$ such that $\mid f(x_{\delta}) - L \mid ≥ \epsilon_0$. Let $\delta_n = \frac{1}{n}$. Then there exists $x_{\delta_n} = a_n \in A \cap V_{\delta_n} (c) \setminus \{ c \}$, in other words, $0 &lt; \mid a_n - c \mid &lt; \frac{1}{n}$ and $\lim_{n \to \infty} a_n = c$. However, $\mid f(a_n) - L \mid ≥ \epsilon_0$ so $\lim_{n \to \infty} f(a_n) \neq L$, a contradiction. Therefore $\lim_{x \to c} f(x) = L$. $\blacksquare$</li>
 </ul>
 ==Resources==
 * [http://mathonline.wikidot.com/the-sequential-criterion-for-a-limit-of-a-function The Sequential Criterion for a Limit of a Function]