Real Function Limits:Sequential Criterion
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> <table class="wiki-content-table"</math> <tr</math> <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> </tr</math> </table</math> <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</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</math>We will now look at the proof of The Sequential Criterion for a Limit of a Function.</p</math> <ul</math> <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> </ul</math> <ul</math> <li</math>Let <math</math>\varepsilon > 0</math</math>. We are given that <math</math>\lim_{x \to c} f(x) = L</math</math> and so for <math</math>\varepsilon > 0</math</math> there exists a <math</math>\delta > 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 > 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> </ul</math> <ul</math> <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> </ul</math> <ul</math> <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> </ul</math> <ul</math> <li</math>Suppose not, in other words, suppose that <math</math>\exists \varepsilon_0 > 0</math</math> such that <math</math>\forall \delta > 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> </ul</math>
