A.2: Proofs of Theorems
This appendix provides proofs of selected properties and theorems stated in the main text. In many cases, the proofs are deferred because they are lengthy or require concepts introduced later in the relevant chapter. In this appendix we prove the following topics:
Limit Laws
PROOF OF LIMITS WITH CONSTANTS The limit \(\lim_{x \to a} c\) means for any given \(\varepsilon \gt 0,\) a number \(\delta \gt 0\) exists such that \[\abs{c - c} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \delta \pd\] But \(\abs{c - c} = 0,\) so the first inequality reduces to \(0 \lt \varepsilon,\) which is always true, regardless of the value of \(x.\) The condition \(0 \lt \abs{x - a} \lt \delta\) therefore becomes irrelevant, so we are free to select any positive number \(\delta\)—for example, \(\delta = 1.\) Hence, \(\lim_{x \to a} c = c\) is proved. Likewise, the limit \(\lim_{x \to a} x = a\) means for any given \(\varepsilon \gt 0,\) there is a number \(\delta \gt 0\) such that \[\abs{x - a} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \delta \pd\] Comparing the inequalities, it is simplest to select \(\delta = \varepsilon.\) Then \[\abs{x - a} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \varepsilon \cma\] proving \(\lim_{x \to a} x = a.\) \[\qedproof\]
PROOF OF THE SUM AND DIFFERENCE LAWS Let \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M.\) Then for any number \(\varepsilon_1 \gt 0,\) a corresponding number \(\delta_1 \gt 0\) exists such that \[\abs{f(x) - L} \lt \varepsilon_1 \if 0 \lt \abs{x - a} \lt \delta_1 \pd\] Likewise, for any number \(\varepsilon_2 \gt 0,\) there exists a number \(\delta_2 \gt 0\) for which \[\abs{g(x) - M} \lt \varepsilon_2 \if 0 \lt \abs{x - a} \lt \delta_2 \pd\] By the Triangle Inequality (Section 0.2), \[ \ba \abs{[f(x) + g(x)] - (L + M)} &\leq \abs{f(x) - L} + \abs{g(x) - M} \nl &\lt \varepsilon_1 + \varepsilon_2 \pd \ea \] Now let \(\varepsilon \gt 0\) be given. We can choose \[\varepsilon_1 = \varepsilon_2 = \frac{\varepsilon}{2} \pd\] Taking \(\delta\) to be the smaller of \(\delta_1\) and \(\delta_2,\) we have \[\abs{[f(x) + g(x)] - (L + M)} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \delta,\] thus proving that \[\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x) \pd\] The proof of the Difference Law is nearly identical. \[\qedproof\]
PROOF OF THE CONSTANT MULTIPLE LAW Let \(\lim_{x \to a} f(x) = L.\) Then, for a given \(\varepsilon_1 \gt 0,\) there exists a number \(\delta_1 \gt 0\) such that \[\abs{f(x) - L} \lt \varepsilon_1 \if 0 \lt \abs{x - a} \lt \delta_1 \pd\] Likewise, \(\lim_{x \to a} [cf(x)] = cL\) means for any number \(\varepsilon \gt 0,\) a number \(\delta \gt 0\) exists such that \[\abs{c f(x) - cL} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \delta \pd\] The first inequality becomes \[ \abs{c f(x) - cL} = \abs{c} \abs{f(x) - L} \pd \] If \(c = 0,\) then the limit becomes \(\lim_{x \to a} 0 = 0,\) which holds trivially. Otherwise, assuming \(c \ne 0,\) we can choose \(\varepsilon_1 = \varepsilon/\abs c.\) Then \[\abs{c} \abs{f(x) - L} \lt \abs{c} \varepsilon_1 = \abs{c} \par{\frac{\varepsilon}{\abs c}} = \varepsilon \pd\] In addition, we can set \(\delta = \delta_1.\) Therefore, there exists a number \(\delta \gt 0\) such that for any given \(\varepsilon \gt 0,\) \[\abs{c f(x) - cL} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \delta \pd\] This proves that \(\lim_{x \to a} [cf(x)] = c \lim_{x \to a} f(x).\) \[\qedproof\]
PROOF OF THE PRODUCT LAW If \(\lim_{x \to a} f(x) = L\) and \(\lim_{x \to a} g(x) = M,\) then \(\lim_{x \to a} [f(x) g(x)] = LM\) means for any given \(\varepsilon \gt 0,\) a number \(\delta \gt 0\) exists such that \[\abs{f(x) g(x) - LM} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \delta \pd\] To attain terms with \(\abs{f(x) - L}\) and \(\abs{g(x) - M},\) we use algebraic manipulation: \[ \baat{2} \abs{f(x) g(x) - LM} &= \abs{f(x) g(x) - L g(x) + L g(x) - LM} \nl &= \abs{g(x) \parbr{f(x) - L} + L \parbr{g(x) - M}} \nl &\leqslant \abs{g(x)} \abs{f(x) - L} + \abs{L} \abs{g(x) - M} \pd &&\comment{\text{by the Triangle Inequality}} \eaat \] Because \(\lim_{x \to a} g(x) = M,\) there exists a \(\delta_1 \gt 0\) such that when \(0 \lt \abs{x - a} \lt \delta_1,\) \[\abs{g(x) - M} \lt 1 \implies \abs{g(x)} \leqslant 1 + \abs M \pd\] In addition, there is a number \(\delta_2 \gt 0\) such that \[\abs{g(x) - M} \lt \frac{\varepsilon}{2 \par{1 + \abs L}} \if 0 \lt \abs{x - a} \lt \delta_2 \pd\] Similarly, a number \(\delta_3\) exists for which \[\abs{f(x) - L} \lt \frac{\varepsilon}{2 \par{1 + \abs M}} \if 0 \lt \abs{x - a} \lt \delta_3 \pd\] Let \(\delta\) be the minimum of \(\delta_1,\) \(\delta_2,\) and \(\delta_3.\) Then when \(0 \lt \abs{x - a} \lt \delta,\) the three inequalities \(0 \lt \abs{x - a} \lt \delta_1,\) \(0 \lt \abs{x - a} \lt \delta_2,\) \(0 \lt \abs{x - a} \lt \delta_3\) are simultaneously true. Thus, \[ \ba \abs{f(x) g(x) - LM} &\leq \abs{g(x)} \abs{f(x) - L} + \abs{L} \abs{g(x) - M} \nl &\lt \par{1 + \abs M} \frac{\varepsilon}{2 \par{1 + \abs M}} + \abs L \frac{\varepsilon}{2 \par{1 + \abs L}} \nl &\lt \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \nl &= \varepsilon \pd \ea \] This proves that \(\lim_{x \to a} [f(x) g(x)] = LM.\) \[\qedproof\]
PROOF OF THE QUOTIENT LAW
First let's show that
\[\lim_{x \to a} \frac{1}{g(x)} = \frac{1}{M} \cmaa M \ne 0 \pd\]
This limit statement means, for any given \(\varepsilon \gt 0,\)
a number \(\delta \gt 0\) exists such that
\[\abs{\frac{1}{g(x)} - \frac{1}{M}} \lt \varepsilon \if 0 \lt \abs{x - a} \lt \delta \pd\]
Rewriting shows
\[\abs{\frac{1}{g(x)} - \frac{1}{M}} = \abs{\frac{M - g(x)}{M g(x)}} \pd\]
Because \(\lim_{x \to a} g(x) = M,\)
there exists some \(\delta_1 \gt 0\) such that
\[\abs{g(x) - M} \lt \frac{\abs M}{2} \if 0 \lt \abs{x - a} \lt \delta_1 \pd\]
(The choice of \(\abs{M}/2\) is highly convenient, as we will see later in the proof.)
At this point, we aim to attain a bound for \(\abs{g(x)}.\)
First note that
\[
\ba
\abs M &= \abs{g(x) + \parbr{M - g(x)}} \nl
&\leq \abs{g(x)} + \abs{M - g(x)} \nl
&\lt \abs{g(x)} + \frac{\abs M}{2} \pd
\ea
\]
Solving for \(\abs{g(x)}\) then shows, for \(0 \lt \abs{x - a} \lt \delta_1,\)
\[\abs{g(x)} \gt \frac{\abs M}{2} \or \frac{1}{\abs{g(x)}} \lt \frac{2}{\abs M} \pd\]
Likewise, some number \(\delta_2 \gt 0\) exists such that
\[\abs{g(x) - M} \lt \frac{M^2}{2} \varepsilon \if 0 \lt \abs{x - a} \lt \delta_2 \pd\]
By taking \(\delta\) to be the minimum of \(\delta_1\) and \(\delta_2,\)
the inequalities
\[\frac{1}{\abs{g(x)}} \lt \frac{2}{\abs M} \and \abs{g(x) - M} \lt \frac{M^2}{2} \varepsilon\]
are simultaneously true for \(0 \lt \abs{x - a} \lt \delta.\)
For these
Squeeze Theorem
PROOF
Let \(\varepsilon \gt 0\) be given, and assume that \(g(x) \leq f(x) \leq h(x)\)
on some interval containing \(a\) (with
Power Rule
We split the proof into four cases of \(n \col\) positive integers, negative integers, rational numbers, and irrational numbers.
PROOF FOR POSITIVE INTEGERS Let \(f(x) = x^n,\) where \(n\) is a positive integer. You may verify that the formula \[x^n - a^n = (x - a) \par{x^{n - 1} + x^{n - 2} a + \cdots + xa^{n - 2} + a^{n - 1}}\] holds by multiplying out the terms on the right-hand side. By the definition of a derivative at a point, \[ \ba f'(a) &= \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a} \frac{x^n - a^n}{x - a} \nl &= \lim_{x \to a} \frac{(x - a) \par{x^{n - 1} + x^{n - 2} a + \cdots + xa^{n - 2} + a^{n - 1}}}{x - a} \nl &= \lim_{x \to a} \par{x^{n - 1} + x^{n - 2} a + \cdots + xa^{n - 2} + a^{n - 1}} \nl &= \underbrace{a^{n - 1} + a^{n - 2} a + \cdots + aa^{n - 2} + a^{n - 1}}_{n \; \text{terms}} \nl &= na^{n - 1} \pd \ea \] Thus, \(f'(x) = n x^{n - 1}.\)
PROOF FOR NEGATIVE INTEGERS If \(n\) is a negative integer, then \(p = -n\) is a positive integer. We begin by writing \(y = x^n\) as \(y = x^{-p} =\) \(1/x^p,\) whose derivative is given by the Quotient Rule (Section 2.3) to be \[ \ba \deriv{y}{x} &= \frac{(0) x^p - 1 \deriv{}{x} \par{x^p}}{x^{2p}} \nl &= -\frac{px^{p - 1}}{x^{2p}} \nl &= -p x^{-p - 1} \nl &= n x^{n - 1} \pd \ea \]
PROOF FOR RATIONAL NUMBERS If \(n\) is rational, then it can be written in the form \(n = p/q\) where \(p\) and \(q\) are integers. Let \(y= x^n\) \(= x^{p/q}.\) Raising both sides to the power of \(q\) gives \[y^q = x^p \pd\] Using Implicit Differentiation (Section 2.5) to differentiate both sides with respect to \(x,\) we attain \[ q y^{q - 1} \deriv{y}{x} = p x^{p - 1} \pd \] This step is true because \(p\) and \(q\) are integers; we proved the Power Rule for integers in cases (a) and (b). Then solving for \(\textderiv{y}{x}\) gives \[ \deriv{y}{x} = \frac{p x^{p - 1}}{q y^{q - 1}} \pd \] Simplifying shows \[ \ba \deriv{y}{x} &= \frac{p x^{p - 1}}{q \par{x^{p/q}}^{q - 1}} \nl &= \frac{p}{q} \frac{x^{p - 1}}{x^{p - p/q}} \nl &= \frac{p}{q} x^{p/q - 1} \nl &= n x^{n - 1} \pd \ea \]
PROOF FOR IRRATIONAL NUMBERS Let \(y = x^n,\) where \(x \gt 0.\) We perform Logarithmic Differentiation (Section 2.6) by first taking the natural logarithm of both sides: \[ \ba \ln y &= \ln \par{x^n} \nl \ln y &= n \ln x \pd \ea \] Differentiating both sides by Implicit Differentiation shows \[ \ba \frac{1}{y} \deriv{y}{x} &= \frac{n}{x} \nl \deriv{y}{x} &= \frac{n}{x} y \nl \deriv{y}{x} &= \frac{n}{x} x^n \nl \deriv{y}{x} &= n x^{n - 1} \pd \ea \] Hence, the Power Rule is applicable for any real \(n.\) \[\qedproof\]
L'Hôpital's Rule
The proof of L'Hôpital's Rule relies on one intermediate theorem, the Cauchy Mean Value Theorem: If \(f\) and \(g\) are continuous on \([a, b]\) and differentiable on \((a, b)\) with \(g'(x) \ne 0\) for any \(x \in (a, b),\) then there exists a number \(c\) in \((a, b)\) such that \[\frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)} \pd\] We begin by proving the Cauchy Mean Value Theorem, from which a proof of L'Hôpital's Rule follows.
PROOF OF CAUCHY MEAN VALUE THEOREM Let \(f\) and \(g\) be continuous on \([a, b]\) and differentiable on \((a, b)\) with \(g'(x) \ne 0\) for any \(x \in (a, b).\) We first define a new function \(h\) such that \[h(x) = [f(b) - f(a)] g(x) - [g(b) - g(a)] f(x) \pd\] Evaluating \(h\) at \(a\) and \(b\) shows \[ \ba h(a) &= [f(b) - f(a)] g(a) - [g(b) - g(a)] f(a) \cma \nl h(b) &= [f(b) - f(a)] g(b) - [g(b) - g(a)] f(b) \pd \ea \] Note that \[ \ba h(b) - h(a) &= [f(b) - f(a)] [g(b) - g(a)] - [g(b) - g(a)] [f(b) - f(a)] \nl &= [g(b) - g(a)] \par{[f(b) - f(a)] - [f(b) - f(a)]} \nl &= 0 \cma \ea \] so \(h(a) = h(b).\) Because \(f\) and \(g\) are continuous on \([a, b],\) \(h\) is also continuous on \([a, b].\) Likewise, \(f\) and \(g\) are differentiable on \((a, b),\) so \(h\) is also differentiable on \((a, b).\) Thus, by Rolle's Theorem, a number \(c\) exists in \((a, b)\) such that \(h'(c) = 0 \col\) \[h'(c) = [f(b) - f(a)] g'(c) - [g(b) - g(a)] f'(c) = 0 \pd\] Rearranging shows \[ \frac{f(b) - f(a)}{g(b) - g(a)} = \frac{f'(c)}{g'(c)} \cma \] which is Cauchy's Mean Value Theorem. We safely divided by \(g(b) - g(a)\) because \(g'(x) \ne 0\) on \((a, b),\) so \(g(a) \ne g(b).\) Because \(g' \ne 0\) on \((a, b)\) and \(a \lt c \lt b,\) it is justifiable to divide by \(g'(c).\) \[\qedproof\]
PROOF OF L'HÔPITAL'S RULE
Let \(f\) and \(g\) be differentiable on some open interval \(I\)
containing \(a\) (except possibly at \(a\)),
with \(g'(x) \ne 0\) in some neighborhood around \(a\) (except possibly at