Difference between revisions of "The Limit Theorems for Functions"

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<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 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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<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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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.