Operations on Functions and Their Limits

This page is intended to be a part of the Real Analysis section of Math Online. Similar topics can also be found in the Calculus section of the site.

# Operations on Functions and Their Limits

Let $A \subseteq \mathbb{R}$ and let $f: A \to \mathbb{R}$ and $g: A \to \mathbb{R}$ be functions. Now $\forall x \in A$ define $f + g$ as $(f + g)(x) = f(x) + g(x)$, $f - g$ as $(f - g)(x) = f(x) - g(x)$, $(fg)(x)$ as $f(x)g(x)$, and $\left ( \frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$ (provided $g(x) \neq 0$).

The following theorem describes the limit laws you are likely familiar with.

 Theorem 1: If $A \subseteq \mathbb{R}$ with $f: A \to \mathbb{R}$ and $g : A \to \mathbb{R}$ and let $c$ be a cluster point of $A$ and $k \in \mathbb{R}$. Then if $\lim_{x \to c} f(x) = L$ and $\lim_{x \to c} g(x) = M$ then: a) $\lim_{x \to c} (f + g)(x) = L + M$. b) $\lim_{x \to c} (f - g)(x) = L - M$. c) $\lim_{x \to c} (fg)(x) = L \cdot M$. d) $\lim_{x \to c} \left ( \frac{f}{g} \right)(x) = \frac{L}{M}$ provided $M \neq 0$. e) $\lim_{x \to c} (kf)(x) = k \cdot L$.

We will prove a), though the rest of these proofs can be obtained in a similar manner to their analogues for sequences.

• Proof of a) Let $A \subseteq \mathbb{R}$ with $f: A \to \mathbb{R}$, and $g: A \to \mathbb{R}$ and let $c$ be a cluster point of $A$. We want to show that $\forall \epsilon > 0$ there exists a $\delta > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta$ then $\mid f(x) + g(x) - (L + M) \mid < \epsilon$.
• First, since $\lim_{x \to c} f(x) = L$, then for $\epsilon_1 = \frac{\epsilon}{2} > 0$ there exists a $\delta_1 > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta_1$ then $\mid f(x) - L \mid < \epsilon_1 = \frac{\epsilon}{2}$.
• Similarly, since $\lim_{x \to c} g(x) = M$, then for $\epsilon_2 = \frac{\epsilon}{2} > 0$ there exists a $\delta_2 > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta_2$ then $\mid g(x) - M \mid < \epsilon_2 = \frac{\epsilon}{2}$.
• Let $\delta = \mathrm{min} \{ \delta_1, \delta_2 \}$. Therefore both of the inequalities above hold. Now notice that:
(1)
\begin{align} \quad \quad \mid f(x) + g(x) - (L + M) \mid = \mid (f(x) - L) + (g(x) - M) \mid ≤ \mid f(x) - L \mid + \mid g(x) - M \mid < \epsilon_1 + \epsilon_2 = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}
• Therefore we have that $\lim_{x \to c} (f + g)(x) = L + M$. $\blacksquare$

The following lemma will be useful in assuring us that our proof of d) will be valid:

 Lemma 1: If $\lim_{x \to c} g(x) = M \neq 0$ then $\exists \delta_0 > 0$ such that for all $x$ such that $x \in A$ and $0 < \mid x - c \mid < \delta_{0}$ we have that $g(x) \neq 0$.
• Proof: First consider the case where $M > 0$. Let $\lim_{x \to c} g(x) = M > 0$, and take $\epsilon_0 = \frac{M}{2} > 0$. Therefore by the definition of a limit, there exists a $\delta_0 > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta_0$ then we have that $\mid g(x) - M \mid < \epsilon_0 = \frac{M}{2}$ which is equivalent to saying $\frac{M}{2} < g(x) < \frac{3M}{2}$, and so all $g(x)$ in this interval are greater than zero since $\frac{M}{2} < g(x)$.
• Now consider the case where $M < 0$. Let $\lim_{x \to c} g(x) = M < 0$, and take $\epsilon_0 = \frac{-M}{2} > 0$. Therefore by the definition of a limit, there exists a $\delta_0 > 0$ such that if $x \in A$ and $0 < \mid x -c \mid < \delta_0$ then we have that $\mid g(x) - M \mid < \epsilon_0 = \frac{-M}{2}$ which is equivalent to $\frac{3M}{2} < g(x) < \frac{M}{2}$, and so all $g(x)$ in this interval are less than zero since $g(x) < \frac{M}{2} < 0$. $\blacksquare$
 Lemma 2: Let $f: A \to \mathbb{R}$ be a function and let $c$ be a cluster point of $A$. If $\lim_{x \to c} f(x)$ exists then $f$ is bounded on some neighbourhood of $c$.
• Proof: Suppose that $\lim_{x \to c} f(x)$ exists, say $\lim_{x \to c} f(x) = L$ where $L \in \mathbb{R}$. Then take $\epsilon_0 = 1$ and so there exists a $\delta_0 > 0$ such that if $x \in A$ and $0 < \mid x - c \mid < \delta_0$ then $\mid f(x) - L \mid < 1$ which is equivalent to $L - 1 < f(x) < L + 1$. Thus $f$ is bounded. $\blacksquare$

The following theorem which you have likely seen in Calculus is known as the squeeze theorem and is yet another analogue from sequence limit theorems. It says that if we have three functions $f, g, h$ such that the values of $g(x)$ are bounded between $f(x)$ and $h(x)$ for all $x \in A$, then if the limits of $f$ and $h$ at some cluster point $c$ of $A$ are equal, then $h$ must also have the same limit at $c$.

 Theorem 2 (Squeeze Theorem): Let $A \subseteq \mathbb{R}$ with $f: A \to \mathbb{R}$, $g: A \to \mathbb{R}$, and $h: A \to \mathbb{R}$ such that $f(x) ≤ g(x) ≤ h(x)$ for all $x \in A$ and let $c$ be a cluster point of $A$. Then if $\lim_{x \to c} f(x) = L = \lim_{x \to c} h(x)$ then $\lim_{x \to c} g(x) = L$.