2.1 Limits of functions

Definition 2.1.1. Let \(X\) and \(Y\) be subsets of \(\mathbb {R}\). Suppose \(E\subset X\), \(\, f\) maps \(E\) into \(Y\) and \(p\) is a limit point of \(E\). We say that the limit of \(f(x)\) as \(x\) approaches \(p\) in \(E\) is \(q\) and write \[\lim _{x\longrightarrow p} f(x) = q\] If there is a point \(q\in Y\) such that for every \(\varepsilon >0\, \exists \,\) a \(\, \delta (\varepsilon ) > 0 \ni \begin {vmatrix} f(x) - q\\ \end {vmatrix} < \varepsilon \,\, \forall \, x \in E\,\) for which \(\, 0 < \begin {vmatrix} x - p\\ \end {vmatrix} < \delta \)

This definition can be recast in terms of limits of sequences as follows:

Theorem 2.1.2. Let \(X\) and \(Y\) be subsets of \(\mathbb {R}\). Suppose \(E\subset X, \, f\) maps \(E\) into \(Y\) and \(p\) is a limit point of \(E\). Then \(\, \displaystyle {\lim _{x\rightarrow p} \, f(x) = q }\hspace {0.3cm}\cdots \cdots \hspace {0.3cm} (i)\hspace {0.2cm}\) if and only if \(\, \displaystyle {\lim _{n\rightarrow \infty } \, f(P_n) = q}\hspace {0.3cm}\cdots \cdots \hspace {0.3cm} (ii)\hspace {0.3cm}\) for every sequence \(\big \{P_n\big \}^{\infty }_{n = 1}\hspace {0.2cm}\) in \(E\) such that \(\, P_n \neq p\,\hspace {0.2cm}, \hspace {0.2cm} \displaystyle {\lim _{n\rightarrow \infty } \, P_n = p}\hspace {0.3cm}\cdots \cdots \hspace {0.3cm} (iii)\)

Proof. Suppose \((i)\) holds choose a sequence \(\, \big \{P_n\big \}^{\infty }_{n= 1}\,\) in \(E\) which satisfies \((iii)\). Let \(\varepsilon > 0\) be given. Then \(\, \exists \, \delta > 0 \, \ni \, \begin {vmatrix} f(x) - q\\ \end {vmatrix} < \varepsilon \). If \(x \in E\) and \(\, 0 < \begin {vmatrix} x - P\\ \end {vmatrix} < \delta \). Also \(\, \exists \, N \in \mathbb {N}\ni \, 0 <\begin {vmatrix} P_n - P\\ \end {vmatrix} < \delta \). Thus, we have \(\begin {vmatrix} f(P_n) - q\\ \end {vmatrix} < \varepsilon \, \, \forall n > N\), which shows that \((ii)\) holds.
Conversely, suppose \((i)\) is false. Then \(\exists \,\) some \(\varepsilon > 0\, \ni \, \) every \(\delta > 0\, \exists \, \) a point \(x \in E\) (depends on \(\delta )\), for which \(\begin {vmatrix} f(x) - q\\ \end {vmatrix}\geq \varepsilon \) but \(0 < \begin {vmatrix} x - P\\ \end {vmatrix} < \delta \). Taking \(\delta _n = \dfrac {1}{n}\, , \hspace {0.2cm} (n = 1, 2, \cdots \cdots , )\), we thus find a sequence in \(E\) satisfying \((iii)\) for which \((ii)\) is false. Hence, a contradiction. Therefore \((i)\) holds when \((ii)\) and \((iii)\) are satisfied.

□

Corollary 2.1.3. If \(f\) has a limit at \(p\) thus limit is unique.

Proof. The proof follows from the fact that the limit of sequence is unique.

□

Theorem 2.1.4. If \(\displaystyle {\lim _{x\rightarrow p} \, f(x) = q \neq 0\, , \, \exists K \in \mathbb {R}^+\,}\) and \(\, \delta > 0 \, \ni \, 0<\begin {vmatrix} x - p\\ \end {vmatrix} < \delta \, \implies \, \begin {vmatrix} f(x)\\ \end {vmatrix} > K\).

Proof. Since \(\, \displaystyle {\lim _{x\rightarrow p} f(x)= q \neq 0}\), for \(\varepsilon = \dfrac {1}{2}\,\begin {vmatrix} q\\ \end {vmatrix}\, \exists \, \delta > 0, \ni \, 0 < \begin {vmatrix} x - p\\ \end {vmatrix}< \delta \, \implies \begin {vmatrix} f(x) - q\\ \end {vmatrix} < \dfrac {1}{2}\, \begin {vmatrix} q\\ \end {vmatrix}\,\\ \implies \begin {vmatrix} q\\ \end {vmatrix} - \begin {vmatrix} f(x)\\ \end {vmatrix} < \dfrac {1}{2}\, \begin {vmatrix} q\\ \end {vmatrix}\) i.e \(\begin {vmatrix} f(x)\\ \end {vmatrix} > \dfrac {1}{2}\,\begin {vmatrix} q\\ \end {vmatrix}\). Hence, the number \(K = \dfrac {1}{2}\, \begin {vmatrix} q\\ \end {vmatrix}\in \mathbb {R}^+\) as required.

□

Theorem 2.1.5. Let \(X\) and \(Y\) be subsets of \(\mathbb {R}\). Suppose \(E\subset X,\, p\) is a limit point of \(E\) and \(f\) and \(g\) are real valued functions on \(E\), and \(\,\displaystyle {\lim _{x\rightarrow p} f(x) = A}\, \hspace {0.3cm} , \hspace {0.3cm}\displaystyle {\lim _{x\rightarrow p} g(x) = B}\,\), then

(a).: \(\displaystyle {\lim _{x\rightarrow p} (f + g)(x) = A + B}\)
(b).: \(\displaystyle {\lim _{x\rightarrow p} (fg)(x) = AB}\)
(c).: \(\displaystyle {\lim _{x\rightarrow p} \Big (\dfrac {f}{g}\Big )(x) = \dfrac {A}{B}\hspace {0.3cm}}\) if \(B\neq 0\)

Proof. These assertions follow immediately from analogous properties of sequences.

□

[next] [front] [up]

Questions on this section

Stuck on something here? Ask below and it stays attached to this topic.