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1\defineenumeration
2  [theorem]
3  [alternative=serried,
4   width=fit,
5   distance=\emwidth,
6   text=Theorem,
7   style=italic,
8   title=yes,
9   titlestyle=normal,
10   prefix=yes,
11   headcommand=\groupedcommand{}{.}]
12
13\defineenumeration
14  [lemma]
15  [theorem]
16  [text=Lemma]
17
18\defineenumeration
19  [proof]
20  [alternative=serried,
21   width=fit,
22   distance=\emwidth,
23   text=Proof,
24   number=no,
25   headstyle=italic,
26   headcommand=\groupedcommand{}{.},
27   title=yes,
28   titlestyle=normal]
29
30\starttext
31
32\startalignment[flushleft,tight]
33\bfb\setupinterlinespace  Proving the l'Hospital rule directly from the Lagrange mean value theorem
34\stopalignment
35
36\blank[big]
37
38\startlines
39Anders Holst
40Mikael P. Sundqvist
41\stoplines
42
43\blank[big]
44
45\startnarrower[2*middle]
46  \bold{Abstract.} At our first-year calculus course for engineers we discuss Lagrange's mean value theorem but not Cauchy's mean value theorem, and for this reason we usually give a weak form of l'Hospital's rule on limits. In this note we give a simple proof of the stronger version of l'Hospital's rule, using only Lagrange's mean value theorem and elementary results on limits and derivatives.
47\stopnarrower
48
49\blank[big]
50
51We formulate and prove the l'Hospitals rule for one-sided limits. This in fact
52strengthen the usual formulation slightly.
53
54\starttheorem
55  [title={l'Hospital's rule},
56   reference={thm:lHospital}]
57  Assume that the functions \m{f} and \m{g} are continuous in \m{\rightopeninterval{a,b}} and differentiable in \m{\openinterval{a,b}}. Assume further that \m{f(a) = g(a) = 0} and that \m{g'(x) \neq 0} in \m{\openinterval{a,b}}. If \m{f'(x)/g'(x)\tendsto A} as \m{x \tendsto a^+}, then \m{f(x)/g(x)\tendsto A} as \m{x\tendsto a^+}.
58\stoptheorem
59
60A geometric interpretation of the l'Hospital rule goes as follow. In the \m{uv}-plane, draw the curve parametrized by \m{u=g(x)} and \m{v=f(x)}. Then the direction coefficient \m{f(x)/g(x)} of the secant (dotted in \in{Figure}[fig:lHospital]) connecting \m{(g(x),f(x))} with \m{(g(a),f(a)) = (0,0)} should approach the same value as the direction coefficient \m{f'(x)/g'(x)} of the tangent to the curve at \m{(g(x),f(x))} (dashed in \in{Figure}[fig:lHospital]) as \m{x} approaches \m{a}. Our proof of the theorem uses that we can parametrize this curve locally around the origin as a function graph \m{u=t} and \m{v = f(g^{-1}(t))}.
61
62\startplacefloat
63  [figure]
64  [location=top,
65   reference=fig:lHospital]
66  \startMPcode[offset=1TS]
67  numeric u ; u:=7.5ts ;
68  path p,tangent,sekant ;
69
70  p:=(0,0){dir 10}..(1.5,1){dir 50}..(3,2) ;
71  z0 = point 1 of p ;
72  tangent:=(((-1,0)--(1,0)) rotated 50) shifted z0 ;
73  sekant:=origin--z0 ;
74
75  drawarrow ((-0.25,0)--(3,0)) scaled u ;
76  drawarrow ((0,-0.25)--(0,2)) scaled u ;
77
78  pickup pencircle scaled 1 ;
79  draw p scaled u ;
80  draw tangent scaled u dashed evenly ;
81  draw sekant scaled u dashed withdots ;
82
83  dotlabel.ulft("\m{(g(x),f(x))}", z0 scaled u) ;
84  dotlabel.lrt ("\m{(g(a),f(a))}", origin) ;
85  label.bot("\m{u}", (2.9u,0)) ;
86  label.lft("\m{v}", (0,1.9u)) ;
87  \stopMPcode
88\stopplacefloat
89
90The only place in our proof where Lagrange's mean value theorem occurs is
91in this useful property of right-hand side derivatives.
92
93\startlemma
94  [reference=lemma:rightderivative]
95  Let \m{c > 0}. Assume that \m{\phi\maps \rightopeninterval{0,c} \to \reals} is continuous in \m{\rightopeninterval{0,c}} and differentiable in \m{\openinterval{0,c}}, and that \m{\lim_{t \tendsto 0^+} \phi'(t)} exists and equals \m{A}. Then
96
97  \startformula
98    \lim_{h \tendsto 0^+} \frac{\phi(0 + h) - \phi(0)}{h} = A.
99  \stopformula
100\stoplemma
101
102\startproof
103  For \m{h \in \openinterval{0,c}} the differential quotient \m{(\phi(0 + h) - \phi(0))/h} equals \m{\phi'(\xi_h)} for some \m{\xi_h \in \openinterval{0,h}}, by Lagrange's mean value theorem. As \m{h\tendsto 0^+} we have \m{\xi_h \tendsto 0^+}, and so
104
105  \startformula
106    \lim_{h\tendsto 0^+}\frac{\phi(0+h)-\phi(0)}{h}=\lim_{h\tendsto 0^+}\phi'(\xi_h)=A.
107  \stopformula
108\stopproof
109
110\startproof
111  [title={of \in{Theorem}[thm:lHospital]}]
112  Since \m{g'} is a Darboux function it will not change sign in \m{\openinterval{a,b}}, and for simplicity we assume that \m{g' > 0} in this interval. Lagrange's mean value theorem assures that \m{g} is strictly monotone in the interval \m{\rightopeninterval{a,b}} and thus that it has an inverse \m{g^{-1}\maps \rightopeninterval{0,g(b)} \to \rightopeninterval{a,b}}.
113
114  The composite function \m{\phi \mapsas t\mapsto f(g^{-1}(t))}, \m{t \in \rightopeninterval{0,g(b)}} is continuous at \m{t = 0} and differentiable for \m{t \in \openinterval{0, g(b)}}. By the substitution \m{t = g(x)} in the given limit, together with the chain rule and the rule of derivatives of inverse functions, we get
115
116  \startformula
117    A = \lim_{x\tendsto a^+} \frac{f'(x)}{g'(x)}
118      = \lim_{t\tendsto 0^+} \frac{f'(g^{-1}(t))}{g'(g^{-1}(t))}
119      = \lim_{t\tendsto 0^+} \frac{\dd}{\dd t}f(g^{-1}(t))
120      = \lim_{t\tendsto 0^+} \phi'(t).
121  \stopformula
122
123  By \in{Lemma}[lemma:rightderivative], and by substitution \m{t = g(x)} again, we conclude that
124
125  \startformula
126    A = \lim_{t\tendsto 0^+} \frac{\phi(0+t)-\phi(0)}{t}
127      = \lim_{t\tendsto 0^+} \frac{f(g^{-1}(t))}{t}
128      = \lim_{x\tendsto a^+} \frac{f(x)}{g(x)}.
129  \stopformula
130
131  This completes the proof.
132\stopproof
133
134\stoptext
135