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\chapter*{Appendix A. Properties of Limits} In this appendix we shall establish the fundamental properties of limits stated without proof in Theorem. Before restating the theorem and giving the proof, we recall one of the basic facts about inequalities and absolute values, which we shall use. Called the triangle inequality, it asserts that, for any two real numbers [math]a[/math] and [math]b[/math],

[[math]] |a \pm b| \leq |a| + |b|. [[/math]]

This result is stated and proved for [math]a + b[/math] in. It holds equally well for [math]a - b[/math], since

[[math]] |a - b| = |a + (-b)| \leq |a| + |-b| = |a| + |b|. [[/math]]

The theorem, which we shall prove, is the following:

Theorem

If [math]\lim_{x \rightarrow a} f(x) = b_1[/math], and [math]\lim_{x \rightarrow a} g(x) = b_2[/math], then

\item[(i)] [math]\lim_{x \rightarrow a} [f(x) + g(x)] = b_1 + b_2[/math]. \item[(ii)] [math]\lim_{x \rightarrow a} cf(x) = cb_1[/math]. \item[(iii)] [math]\lim_{x \rightarrow a} f(x)g(x) = b_1b_2[/math]. \item[(iv)] [math]\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{b_1}{b_2} \;provided \; b_2 \neq 0[/math].

According to the definition of limit, the hypotheses tell us that, for any positive number [math]\epsilon[/math], there exist positive numbers [math]\delta_1[/math], and [math]\delta_2[/math] such that if [math]x[/math] is in the domains of both [math]f[/math] and [math]g[/math] and if [math]0 \lt |x - a| \lt \delta_1[/math] and [math]0 \lt |x - a| \lt \delta_2[/math], then [math]|f(x) - b_1| \lt \epsilon[/math] and [math]|g(x) - b_2| \lt \epsilon[/math]. Where it is relevant in the proofs which follow, we shall assume without explicitly stating it the condition that [math]x[/math] lies in the appropriate domain of [math]f[/math] or [math]g[/math] (or both). \medskip Proof of (i). Let [math]\epsilon[/math] be an arbitrary positive number. Then there exist positive numbers [math]\delta_1[/math] and [math]\delta_2[/math] such that [math]|f(x) - b_1| \lt \frac{\epsilon}{2}[/math] whenever [math]0 \lt |x - a| \lt \delta_1[/math], and [math]|g(x) - b_2| \lt \frac{\epsilon}{2}[/math], whenever [math]0 \lt |x - a| \lt \delta_2[/math]. (It is legitimate to write [math]\frac{\epsilon}{2}[/math] in these inequalities, since the definition specifies the existence of [math]\delta'[/math]s for any positive number [math]\epsilon[/math]. Given a choice of [math]\epsilon[/math], we can then take [math]\frac{\epsilon}{2}[/math] to be the number which implies the existence of [math]\delta_1[/math] and [math]\delta_2[/math].) We set

[[math]] \delta = \mbox{minimum}\; \{ \delta_1, \delta_2 \}. [[/math]]

Let us now suppose that [math]0 \lt |x - a| \lt \delta[/math]. It follows that [math]0 \lt |x - a| \lt \delta_1[/math] and [math]0 \lt |x - a| \lt \delta_2[/math], and thence that [math]|f(x) - b_1| \lt \frac{\epsilon}{2}[/math] and [math]|g(x) - b_2| \lt \frac{\epsilon}{2}[/math]. Clearly,

[[math]] |[f(x) + g(x)] - [b_1 + b_2]| = |[f(x) - b_1|] + [g(x) - b_2]|. [[/math]]

Hence, using the triangle inequality, we obtain

[[math]] |[f(x) + g(x)] - [b_1 + b_2]| \lt |f(x) - b_1| + |g(x) - b_2| \lt \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon . [[/math]]

Thus we have shown that, for any [math]\epsilon \gt 0[/math], there exists a [math] \delta \gt 0[/math] such that, whenever [math]0 \lt |x - a| \lt \delta[/math], then [math]|[f(x) + g(x)] - [b_1 + b_2]| \lt \epsilon[/math]. By the definition of limit we have therefore proved that

[[math]] \lim_{x \rightarrow a} [f(x) + g(x)] = b_1 + b_2, [[/math]]

which is the result (i). \medskip Proof of (ii). Suppose first that [math]c = 0[/math]. Then [math]cf[/math] is the constant function with value 0, and [math]cb_1 = 0[/math]. Hence

[[math]] |cf(x) - cb_1| = |0 - 0| = 0 , [[/math]]

for every [math]x[/math] in the domain of [math]f[/math]. Thus, for any two positive numbers [math]\epsilon[/math] and [math]\delta[/math], it is trivially true that

[[math]] |cf(x) - cb_1| \lt \epsilon, \;\;\;\mbox{whenever}\; 0 \lt |x - a| \lt \delta, [[/math]]

and (ii) is therefore proved in this special case. We next assume that [math]c \neq 0[/math], and choose an arbitrary positive number [math]\epsilon[/math]. There then exists a positive number [math]\delta[/math] such that

[[math]] |f(x) - cb_1| \lt \frac{\epsilon}{|c|} , \;\;\;\mbox{whenever}\; 0 \lt |x - a| \lt \delta. [[/math]]

It follows immediately that

[[math]] |cf(x) - cb_1| = |c[ f(x) - b_1]| = |c| |f(x) - b_1| \lt |c| \frac{\epsilon}{|c|} = \epsilon [[/math]]

whenever [math]0 \lt |x - a| \lt \delta[/math]. This completes the proof of (ii). \medskip Proof of (iii). Let [math]\epsilon[/math] be an arbitrary positive number. Select a positive number [math]M[/math] such that [math]|b_1| \lt M[/math] and [math]|b_2| \lt M[/math]. Then there exist positive numbers [math]\delta_1[/math], [math]\delta_2[/math], and [math]\delta_3[/math] such that

[[math]] \begin{array}{ll} |f(x) - b_1| \lt \frac{\epsilon}{2M}, \;\mbox{provided}\; &0 \lt |x - a| \lt \delta_1, \\ |g(x) - b_2| \lt \frac{\epsilon}{2M}, \;\mbox{provided}\; &0 \lt |x - a| \lt \delta_2,\\ |g(x) - b_2| \lt M - |b_2|, \;\mbox{provided}\; &0 \lt |x - a| \lt \delta_3. \end{array} [[/math]]

We set

[[math]] \delta = \mbox{minimum}\; \{ \delta_1, \delta_2, \delta_3 \}, [[/math]]

and in the remainder of the argument we assume that [math]0 \lt |x - a| \lt \delta[/math]. It then follows that all three of the above inequalities hold. Using the last one together with the triangle inequality, we first observe that

[[math]] |g(x)| = |(g(x) - b_2) + b_2| \leq |g(x) - b_2| + |b_2| \lt (M - |b_2|) + |b_2| = M. [[/math]]

Next we obtain

[[math]] \begin{eqnarray*} |f(x)g(x) - b_ib_2| &=& | f(x)g(x) - b_1g(x) + b_1g(x) - b_1b_2 | \\ &=& | g(x)[f(x) - b_1] + b_1[g(x) - b_2] | \\ &\leq& | g(x) [f(x) - b_1] | + |b_1[g(x) - b_2] | \\ &=& | g(x)| |f(x) - b_1| + | b_1| |g(x) - b_2 |. \end{eqnarray*} [[/math]]

Finally, therefore,

[[math]] |f(x)g(x) - b_1b_2| \lt M \frac{\epsilon}{2M} + M \frac{\epsilon}{2M} = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon , [[/math]]

and the proof of (iii) is finished. \medskip Proof of (iv). We shall prove the simpler statement:

[[math]] \lim_{x \rightarrow a} \frac{1}{g(x)} = \frac{1}{b_2}, \;\;\;\mbox{provided}\; b_2 \neq 0. ( 1 ) [[/math]]

This fact, together with (iii), then implies

[[math]] \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} f(x) \frac{1}{g(x)} = b_1 \frac{1}{b_2} = \frac{b_1}{b_2} , [[/math]]

which is the result (iv). Since it is assumed that [math]b_2 \neq 0[/math], there exists a number m such that [math]0 \lt m \lt |b_2|[/math]. Hence there exists a positive number [math]\delta_1[/math] such that

[[math]] |g(x) - b_2| \lt |b_2| - m, [[/math]]

whenever [math]0 \lt |x - a| \lt \delta_1[/math]. But

[[math]] \begin{eqnarray*} |b_2| = |-b_2| &=& |(g(x) - b_2) - g(x)| \\ &\leq& |g(x) - b_2| + |g(x)| . \end{eqnarray*} [[/math]]

Hence, if [math]0 \lt |x - a| \lt \delta_1[/math], we have

[[math]] |g(x)| \gt |b_2| - |g(x) - b_2| \gt |b_2| - (|b_2| - m) = m . [[/math]]

Taking reciprocals, we therefore obtain

[[math]] \frac{1}{|g(x)|} \lt \frac{1}{m}, \;\mbox{whenever}\; 0 \lt |x - a| \lt \delta_1. [[/math]]

Now let [math]\epsilon[/math] be an arbitrary positive number. There exists a positive number [math]\delta_2[/math] such that

[[math]] |g(x) - b_2| \lt m |b_2| \epsilon, \;\mbox{whenever}\; 0 \lt |x - a| \lt \delta_2. [[/math]]

We set

[[math]] \delta = \mbox{minimum}\; \{ \delta_1, \delta_2 \}. [[/math]]

It follows that, if [math]0 \lt |x - a| \lt \delta[/math], then

[[math]] \begin{eqnarray*} \Big|\frac{1}{g(x)} - \frac{1}{b_2}\Big| &=& \Big|\frac{b_2 - g(x)}{b_2 g(x)} \Big| \\ &=& \frac{1}{|g(x)|} \frac{1}{|b_2|} |g(x) - b_2| \\ & \lt & \frac{1}{m} \frac{1}{|b_2|} |g(x) - b_2| \\ & \lt & \frac{1}{m |b_2|} m |b_2| \epsilon = \epsilon. \end{eqnarray*} [[/math]]

Thus (1) is proved, and, as we have seen, (1) and (iii) imply (iv). This completes the proof of the theorem.==General references== Doyle, Peter G. (2008). "Crowell and Slesnick's Calculus with Analytic Geometry" (PDF). Retrieved Oct 29, 2024.