The Limit Theorems for Functions - Revision history

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The Uniqueness of Limits of a Function Theorem

Recall from The Limit of a Function page that for a function $..."$

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Created page with "<h1 id="toc0">The Uniqueness of Limits of a Function Theorem</h1> <p>Recall from <a href="/the-limit-of-a-function">The Limit of a Function</a> page that for a function <math>..."

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<h1 id="toc0">The Uniqueness of Limits of a Function Theorem</h1>
<p>Recall from <a href="/the-limit-of-a-function">The Limit of a Function</a> page that for a function <math>f : A \to \mathbb{R}</math> where <math>c</math> is a cluster point of <math>A</math>, then <math>\lim_{x \to c} f(x) = L</math> if <math>\forall \epsilon > 0</math> <math>\exists \delta > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta</math> then <math>\mid f(x) - L \mid < \epsilon</math>. We have not yet established that the limit <math>L</math> is unique, so is it possible that <math>\lim_{x \to c} f(x) = L</math> and <math>\lim_{x \to c} f(x) = M</math> where <math>L \neq M</math>? The following theorem will show that this cannot happen.</p>
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<td><strong>Theorem (Uniqueness of Limits):</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 if <math>L, M \in \mathbb{R}</math> are both limits of <math>f</math> at <math>c</math>, that is <math>\lim_{x \to c} f(x) = L</math> and <math>\lim_{x \to c} f(x) = M</math>, then <math>L = M</math>.</td>
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<ul>
<li><strong>Proof:</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>. Also let <math>\lim_{x \to c} f(x) = L</math> and <math>\lim_{x \to c} f(x) = M</math>. Suppose that <math>L \neq M</math>. We will show that this leads to a contradiction. Let <math>\epsilon > 0</math> be given.</li>
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<ul>
<li>Since <math>\lim_{x \to c} f(x) = L</math>, then for <math>\epsilon_1 = \frac{\epsilon}{2}</math> <math>\exists \delta_1 > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta_1</math> then <math>\mid f(x) - L \mid < \epsilon_1 = \frac{\epsilon}{2}</math>.</li>
</ul>
<ul>
<li>Similarly, since <math>\lim_{x \to c} f(x) = M</math> then for <math>\epsilon_2 = \frac{\epsilon}{2}</math> <math>\exists \delta_2 > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta_2</math> then <math>\mid f(x) - M \mid < \epsilon_2 = \frac{\epsilon}{2}</math>. Now let <math>\delta = \mathrm{min} \{ \delta_1, \delta_2 \}</math> and so we have that:</li>
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<span class="equation-number">(1)
<math>\begin{align} \quad \quad \mid L - M \mid = \mid L - f(x) + f(x) - M \mid ≤ \mid L - f(x) \mid + \mid f(x) - M \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}</math>
<ul>
<li>But <math>\epsilon > 0</math> is arbitrary, so this implies that <math>\mid L - M \mid = 0</math>, that is <math>L = M</math>, a contradiction. So our assumption that <math>L \neq M</math> was false, and so if <math>\lim_{x \to c} f(x) = L</math> then <math>L</math> is unique. <math>\blacksquare</math></li>
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 <h1 id="toc0">The Uniqueness of Limits of a Function Theorem</h1>
-<p>Recall from The Limit of a Function</a> page that for a function <math>f : A \to \mathbb{R}</math> where <math>c</math> is a cluster point of <math>A</math>, then <math>\lim_{x \to c} f(x) = L</math> if <math>\forall \epsilon > 0</math> <math>\exists \delta > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta</math> then <math>\mid f(x) - L \mid < \epsilon</math>. We have not yet established that the limit <math>L</math> is unique, so is it possible that <math>\lim_{x \to c} f(x) = L</math> and <math>\lim_{x \to c} f(x) = M</math> where <math>L \neq M</math>? The following theorem will show that this cannot happen.</p>
+<p>Recall from The Limit of a Function page that for a function <math>f : A \to \mathbb{R}</math> where <math>c</math> is a cluster point of <math>A</math>, then <math>\lim_{x \to c} f(x) = L</math> if <math>\forall \epsilon > 0</math> <math>\exists \delta > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta</math> then <math>\mid f(x) - L \mid < \epsilon</math>. We have not yet established that the limit <math>L</math> is unique, so is it possible that <math>\lim_{x \to c} f(x) = L</math> and <math>\lim_{x \to c} f(x) = M</math> where <math>L \neq M</math>? The following theorem will show that this cannot happen.</p>
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 <li>Similarly, since <math>\lim_{x \to c} f(x) = M</math> then for <math>\epsilon_2 = \frac{\epsilon}{2}</math> <math>\exists \delta_2 > 0</math> such that if <math>x \in A</math> and <math>0 < \mid x - c \mid < \delta_2</math> then <math>\mid f(x) - M \mid < \epsilon_2 = \frac{\epsilon}{2}</math>. Now let <math>\delta = \mathrm{min} \{ \delta_1, \delta_2 \}</math> and so we have that:</li>
 </ul>
-<math>\begin{align} \quad \quad \mid L - M \mid = \mid L - f(x) + f(x) - M \mid ≤ \mid L - f(x) \mid + \mid f(x) - M \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}</math>
+<math>\begin{align} \quad \quad \mid L - M \mid = \mid L - f(x) + f(x) - M \mid \leq \mid L - f(x) \mid + \mid f(x) - M \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}</math>
 <ul>
 <li>But <math>\epsilon > 0</math> is arbitrary, so this implies that <math>\mid L - M \mid = 0</math>, that is <math>L = M</math>, a contradiction. So our assumption that <math>L \neq M</math> was false, and so if <math>\lim_{x \to c} f(x) = L</math> then <math>L</math> is unique. <math>\blacksquare</math></li>
 </ul>