Difference between revisions of "The Limit Theorems for Functions"

From Department of Mathematics at UTSA
Jump to navigation Jump to search
 
(2 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
<h1 id="toc0">The Uniqueness of Limits of a Function Theorem</h1>
 
<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>
+
<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>
 
<table class="wiki-content-table">
 
<table class="wiki-content-table">
 
<tr>
 
<tr>
Line 15: Line 15:
 
<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>
 
<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>
 
</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>
 
<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>
 
<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>
 
</ul>

Latest revision as of 15:14, 21 October 2021

The Uniqueness of Limits of a Function Theorem

Recall from The Limit of a Function page that for a function where is a cluster point of , then if such that if and then . We have not yet established that the limit is unique, so is it possible that and where ? The following theorem will show that this cannot happen.

Theorem (Uniqueness of Limits): Let be a function and let be a cluster point of . Then if are both limits of at , that is and , then .
  • Proof: Let be a function and let be a cluster point of . Also let and . Suppose that . We will show that this leads to a contradiction. Let be given.
  • Since , then for such that if and then .
  • Similarly, since then for such that if and then . Now let and so we have that:

  • But is arbitrary, so this implies that , that is , a contradiction. So our assumption that was false, and so if then is unique.