View source for Real Function Limits:Sequential Criterion


<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>

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<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>
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<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>
<p>We will now look at the proof of The Sequential Criterion for a Limit of a Function.</p>
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<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>
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<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>
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<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>
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<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>
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<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>. 
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