Chapter 1. Real Limits, Continuity and Differentiation
Introduction
Real analysis is similar to calculus with a strong emphasis placed on rigorous math- ematical proofs. In this first chapter, we shall prove some of the theorems, about limits, continuity and differentiation, which are normally omitted in calculus courses. We shall also introduce a few additional concepts. In the second chapter, we give a careful definition of the Riemann integral, which is usually described informally in calculus courses, and we prove many of the basic properties of the integral (which cannot be proven using only an informal definition of the integral). In the third chapter, we discuss sequences of functions and power series. We introduce the concept of uniform convergence (which is not usually mentioned in calculus courses) and prove various properties which allow us to show, for example, that power series can be differentiated and integrated term by term (a fact which is used often in calculus, physics and engineering, but which is not usually proven).
In the fourth chapter, we shall discuss inner product spaces and normed linear spaces (which students meet in linear algebra courses) and metric spaces. These include the Euclidean space Rn, but they also include some infinite dimensional spaces such as spaces of sequences and spaces of functions. This is of interest because a differential or integral equation can often be interpreted as being given by a linear operator (a differential or integral operator) acting on an (infinite dimensional) vector space of functions, with the solutions forming a subspace. This can be studied using a combination of analysis and linear algebra.
In the next chapter, we study limits and continuity in metric spaces, and in the following chapter we introduce completeness and compactness.
We finish the course with some applications of our study of metric spaces. We shall prove an existence and uniqueness theorem for differential equations, which is often stated and used without proof in applied differential equations courses and in physics and engi- neering. We shall also prove results about polynomial approximations and trigonometric approximations and explore their relevance to Fourier series, which are a useful tool used in the solution of a number of differential equations.
Let us give a more detailed introduction to this first chapter. We shall prove the Monotone Convergence Theorem for sequences, the Intermediate Value Theorem and the Extreme Value Theorem for continuous functions, and the Chain Rule, the Inverse Function Theorem, and l’Hˆopital’s Rule for differentiable functions. These proofs are usually omitted in calculus courses. We shall also introduce a few new concepts including Cauchy sequences and uniform continuity.
What makes most of these theorems difficult to prove is that they involve a subtle property of R (involving the order relation ≤) which is not shared by Q. This property can be vaguely described by saying that intervals in R do not have any holes while intervals in Q do have holes (the holes being the irrational numbers). The precise statement of this property is given in Theorem 1.5, which we shall accept, axiomatically, without proof. In mathematics, we must accept some properties, called axioms, without proof and then use these properties to prove other results.
1
Order Properties in R
1.1 Theorem: (Discreteness Property of Z) For all k, n ∈ Z we have k ≤ n if and only if
k
(2) For every x ∈ R there exists m ∈ Z with m < x.
Proof: Letx∈R. Letn=⌊x⌋+1andm=⌊x⌋−1. Sincex−1<⌊x⌋wehave
x < ⌊x⌋ + 1 = n and since ⌊x⌋ ≤ x we have m = ⌊x⌋ − 1 ≤ x − 1 < x.
1.11Theorem: (DensityofQinR)Foralla,b∈Rwitha
na
Similarly we say that (xk)k≥p diverges to −∞, or that the limit of (xk)k≥p is equal to
negative infinity, and we write xk → −∞ (as k → ∞), or we write lim xk = −∞ when k→∞
∀r∈R∃m∈Z∀k∈Z≥p k≥m=⇒xk
4
1.17 Theorem: (Monotonic Convergence Theorem) Let (xk) be a sequence in R.
(1) Suppose (xk) is increasing. If (xk) is bounded above then xk → sup{xk}, and if (xk) is not bounded above then xk → ∞.
(2) Suppose (xk) is decreasing. If (xk) is bounded below then xk → inf{xk}, and if (xk) is not bounded below then xk → −∞.
Proof: We prove Part 1 in the case that (xk)k≥p is increasing and bounded above, say by b∈R. LetA={xk|k≥p}(soAistherangeofthesequence(xk)). NotethatAis nonempty and bounded above (indeed b is an upper bound for A). By the Completeness PropertyofR,AhasasupremuminR. Leta=sup{xk|k≥p}. Notethata≥xk forall k ≥ p and a ≤ b, by the definition of the supremum. Let ε > 0. By the Approximation Property of the supremum, we can choose an index m ≥ p so that the element xm ∈ A satisfiesa−ε
choose k2 > k1 with xk2 ∈ I2, and so on. In this way, we obtain a subsequence (xkj )j≥0
of(xk)withxkj ∈Ij forallj≥0. Weclaimthatxkj →casj→∞. Letε>0Choose
m∈Zsothat 1 (b−a)<ε. Forj≥m,sincec∈[a,b]⊆[a,b]andx ∈[a,b],it 2m jj kjjj
follows that
|x −c|=max{x ,c}−min{x ,c}≤b −a = 1 (b−a)≤ 1 (b−a)<ε.
kj kj kj jj2j 2m Thusxkj →casj→∞,asclaimed.
6
1.24 Definition: Let (xk)k≥p be a sequence in R. We say that (xk) is Cauchy when ∀ε>0∃m∈Z≥p ∀k,l∈Z≥p k,l≥m=⇒|xk −xl|<ε.
1.25 Theorem: (Cauchy Criterion for Convergence) For a sequence (xk) in R, the se- quence (xk) converges if and only if it is Cauchy.
Proof: Let (xk) be a sequence in R. Suppose that (xk) converges, say xk → a. Let ε > 0 andchoosem∈Zsothatk≥m=⇒|xk −a|< ε. Thenfork,l≥mwehave
2
|xk −xl|=|xk −a+a−xl|≤|xk −a|+|a−xl|< ε + ε =ε. 22
Thus (xk) is Cauchy.
Now suppose that (xk ) is Cauchy. We claim that (xk ) is bounded. Since (xk ) is Cauchy, we can choose m ∈ Z so that k,l ≥ m =⇒ |xk −xl| < 1. In particular, for all k ≥ m we have |xk −xm| < 1 and so |xk| = |xk −xm +xm| ≤ |xk −xm|+|xm| < 1+|xm|. It follows that (xk) is bounded by b = max |xp|, |xp+1|, · · · , |xm−1| , 1 + |xm|.
Because (xk) is bounded, it has a convergent subsequence, by the Bolzano Weierstrass Theorem. Let (xkj ) be a convergent subsequence of (xk) and let a = lim xkj . We claim
j→∞
that the original sequence (xk) converges with lim xk = a. Let ε > 0. Since (xk) is
k→∞ Cauchy,wecanchoosem∈Zsothatk,l≥m=⇒|xk−xl|<ε.Sincexkj →awecan
2
choosem0 ∈Zsothatj≥m0 =⇒|xkj −a|< ε. Chooseanindexj≥m0 sothatkj ≥m.
2
Then for all k ≥ m we have
|xk −a|=|xk −xkj +xkj −a|≤|xk −xkj|+|xkj −a|< ε + ε =ε.
22
Thus xk → a, as claimed.
1.26 Definition: Let (xk)k≥p be a sequence in R. The series xk is defined to
k≥p
be the sequence (Sn)n≥k where Sn = xk = xp +xp+1 +···+xn. The term Sn is
n k=p
called the nth partial sum of the series xk. The sum of the series, denoted by k≥p
∞
S = xk = xp +xp+1 +xp+2 +··· , is the limit of the sequence of partial sums, if it k=p
exists, and we say the series converges when the sum exists and is finite.
1.27 Note: As with sequences, we assume students are familiar with series and various
tests for convergence.
1.28 Theorem: (Cauchy Criterion for Series) Let (xk)k≥p be a sequence. Then the series
xk converges if and only if
∀ε>0∃l∈Z≥p ∀m,n∈Z≥p m>n≥l=⇒
m
xk<ε .
Proof: This follows from the Cauchy Criterion for the convergence of the sequence of
partial sums. Indeed (Sn) converges if and only if for all ε > 0 there exists l ≥ p such that mnm
m > n ≥ l =⇒ |Sm − Sn| < ε, and we have |Sm − Sn| = xk − k=p
k=p
xk = xk. k=n+1
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k=n+1
Limits of Functions and Continuity in R
1.29 Definition: Let A ⊆ R. For a ∈ R, we say that a is a limit point of A when ∀δ > 0 ∃x ∈ A 0 < |x − a| < δ.
When a ∈ A and a is not a limit point of A we say that a is an isolated point of A. 1.30Definition: Letf:A⊆R→R. WhenaisalimitpointofA,wemakethe
following definitions.
(1) For b ∈ R, we say that the limit of f(x) as x tends to a is equal to b, and we write
lim f(x) = b or we write f(x) → b as x → a, when x→a
∀ ε > 0 ∃ δ > 0 ∀ x ∈ A 0 < | x − a | < δ =⇒ | f ( x ) − b | < ε .
(2) We say the limit of f(x) as x tends to a is equal to infinity, and we write lim f(x) = ∞,
x→a
or we write f(x) → ∞ as x → a, when
∀ r ∈ R ∃ δ > 0 ∀ x ∈ A 0 < | x − a | < δ =⇒ f ( x ) > r .
(3) We say that the limit of f(x) as x tends to a is equal to negative infinity, and we write limf(x)=−∞,orwewritef(x)→−∞asx→a,when
x→a
∀ r ∈ R ∃ δ > 0 ∀ x ∈ A 0 < | x − a | < δ =⇒ f ( x ) < r .
1.31 Note: We assume that students are familiar with limits of functions and are able to
calculate limits using various limit rules (such as Operations on Limits, and the Comparison
and the Squeeze Theorems) We also assume familiarity with one-sided limits lim f(x) x→a+
and lim f(x) as well as asymptotic limits lim f(x) and lim f(x). Here is one theorem x→a− x→−∞ x→∞
that relates limits of functions and limits of sequences which students may not have seen. 1.32 Theorem: (Sequential Characterization of Limits of Functions) Let f : A ⊆ R → R,
leta∈RbealimitpointofA,andletb∈R. Then limf(x)=bifandonlyifforevery x→a
sequence(xk)inA\{a}withxk →awehavef(xk)→b.
Proof: Suppose that lim f(x) = b. Let (xk) be a sequence in A \ {a} with xk → a. Let
x→a
ε>0. Since limf(x)=b,wecanchooseδ>0sothat0<|x−a|<δ=⇒|f(x)−b|<ε.
x→a
Sincexk →awecanchoosem∈Zsothatk≥m=⇒|xk −a|<δ. Thenfork≥m,we
have |xk −a| < δ and we have xk ̸= a (since the sequence (xk) is in the set A\{a}) so that 0 < |xk − a| < δ and hence |f(xk) − b| < ε. This shows that f(xk) → b.
Conversely, suppose that lim f(x) ̸= b. Choose ε0 > 0 so that for all δ > 0 there x→a
existsx∈Awith0<|x−a|<δand|f(x)−b|≥ε0. Foreachk∈Z+,choosexk ∈A with 0 < |xk −a| ≤ 1 and |f(xk)−b| ≥ ε0. In this way we obtain a sequence (xk)k≥1 in
k
A\{a}.Since|xk−a|≤1 forallk∈Z+,itfollowsthatxk→a(indeed,givenε>0 k
wecanchoosem∈Zwithm>1andthenk≥m=⇒|xk−a|≤1≤1 <ε).Since εkm
|f(xk) − b| ≥ ε0 for all k, it follows that f(xk) ̸→ b (indeed if we had f(xk) → b we could choosem∈Zsothatk≥m=⇒|f(xk)−b|<ε0 andthenwecouldchoosek=mtoget |f(xk) − b| < ε0).
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1.33Definition: Letf:A⊆R→R. Fora∈A,wesaythatfiscontinuousata when
∀ε>0 ∃δ>0 ∀x∈A |x−a|<δ=⇒|f(x)−f(a)|<ε.
We say that f is continuous (on A) when f is continuous at every point a ∈ A.
1.34 Note: Let f : A ⊆ R → R and let a ∈ A. Verify, as an exercise, that (1) if a is an isolated point of A then f is continuous at a, and
(2)ifaisalimitpointofAthenf iscontinuousataifandonlyif limf(x)=f(a). x→a
1.35 Note: We assume the reader is familiar with continuity. In particular, we assume the reader knows that every elementary function is continuous in its domain (an elementary function is any function which can be obtained from the basic elementary functions x, √n x, ex , ln x, sin x and sin−1 x using addition, subtraction, multiplication, division, and composition of functions).
1.36 Theorem: (The Sequential Characterization of Continuity) Let f : A ⊆ R → R and let a ∈ A. Then f is continuous at a if and only if for every sequence (xk) in A with xk → a we have f(xk) → f(a).
Proof: Suppose that f is continuous at a. Let (xk) be a sequence in A with xk → a. Let ε>0. Chooseδ>0sothatforallx∈Awehave|x−a|<δ=⇒|f(x)−f(a)|<ε. Choosem∈Zsothatforallindiceskwehavek≥m=⇒|xk−a|<δ. Thenwhenk≥m we have |xk − a| < δ and hence |f(xk) − f(a)| < ε. Thus we have f(xk) → f(a).
Conversely, suppose that f is not continuous at a. Choose ε0 > 0 so that for all δ > 0
there exists x ∈ A with |x−a| < δ and |f(x)−f(a)| ≥ ε0. For each k ∈ Z+, choose xk ∈ A
with |xk − a| ≤ 1 and |f(xk) − f(a)| ≥ ε0. Consider the sequence (xk) in A (we remark k
that the Axiom of Choice is being used here). Since |xk − a| ≤ 1 for all k ∈ Z+, it follows k
that xk → a. Since |f(xk) − f(a)| ≥ ε0 for all k ∈ Z+, it follows that f(xk) ̸→ f(a).
1.37 Theorem: (Intermediate Value Theorem) Let I be an interval in R and let f : I → R be continuous. Let a,b ∈ I with a ≤ b and let y ∈ R. Suppose that either f(a) ≤ y ≤ f(b) or f(b) ≤ y ≤ f(a). Then there exists x ∈ [a,b] with f(x) = y.
Proof: We prove the theorem in the case that f(a) ≤ y ≤ f(b). If y = f(a) then we can takex=aandify=f(b)thenwecantakex=b. Supposethatf(a)
x
thenforx=1 anda=1wehave|x−a|=2 <δbut|f(x)−f(a)|=2>ε.If0<δ<1 33
thenforx=δ anda=δwehave|x−a|=2δ<δbut|f(x)−f(a)|=2 >2>ε.This 33δ
proves that f is not uniformly continuous (but f is continuous because it is elementary). 10
1.44 Theorem: (Closed Bounded Intervals and Uniform Continuity) Let a, b ∈ R with a < b and let f : [a, b] → R. If f is continuous then f is uniformly continuous (on [a, b]).
Proof: Suppose, for a contradiction, that f : [a, b] → R is continuous but not uniformly
continuous on [a,b]. Choose ε > 0 so that for all δ > 0 there exist x,y ∈ [a,b] such
that |x−y| < δ but |f(x)−f(y)| ≥ ε. For each k ∈ Z+ choose xk and yk in [a,b] with
|xk −yk| ≤ 1 and |f(xk)−f(yk)| ≥ ε. By the Bolzano Weierstrass Theorem, we can choose k
a convergent subsequence (ykj ) of (yk). Let c = lim ykj . For all j we have |xkj −ykj | ≤ 1 j→∞ kj
henceykj − 1 ≤xkj ≤ykj + 1 . Sinceykj →cand 1 →0wehaveykj ± 1 →candhence kj kj kj kj
xkj → c by the Squeeze Theorem. Since f is continuous at c and xkj → c and ykj → c, we have f(xkj ) → f(c) and f(ykj ) → f(c) by the Sequential Characterization of Continuity. Sincef(xkj)→candf(ykj)→cwehavef(xkj)−f(ykj)→0. Butthisimpliesthatwe can choose j so that |f (xkj ) − f (ykj )| < ε, giving the desired contradiction.
Differentiation in R
1.45 Definition: For a subset A ⊆ R, we say that A is open when it is a union of open intervals. Let A ⊆ R be open, let f : A → R. For a ∈ A, we say that f is differentiable
at a when the limit
lim f(x) − f(a) x→a x−a
exists in R. In this case we call the limit the derivative of f at a, and we denote to by f′(a), so we have
f′(a) = lim f(x) − f(a). x→a x−a
We say that f is differentiable (on A) when f is differentiable at every point a ∈ A. In this case, the derivative of f is the function f′ : A → R defined by
f′(x) = lim f(u) − f(x). u→x u−x
When f ′ is differentiable at a, denote the derivative of f ′ at a by f ′′ (a), and we call f′′(a) the second derivative of f at a. When f′′(a) exists for every a ∈ A, we say that f is twice differentiable (on A), and the function f′′ : A → R is called the second derivative of f. Similarly, f′′′(a) is the derivative of f′′ at a and so on.
1.46 Remark: Note that
lim f(x)−f(a) = lim f(a+h)−f(a).
x→a x−a h→0 h
To be precise, the limit on the left exists in R if and only if the limit on the right exists in
R, and in this case the two limits are equal.
1.47 Note: The student should be familiar with derivatives from first year calculus, and should be able to calculate the derivatives of elementary functions using differentiation rules including the Product Rule, the Quotient Rule and the Chain Rule. We shall provide proofs of some of the theorems whose proofs are often omitted in calculus courses.
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1.48Exercise: LetA⊆Rbeopen,letf:A→R,andleta∈A. Showthatfis differentiable at a with derivative f′(a) if and only if
∀ε > 0 ∃δ > 0 ∀x ∈ A |x − a| ≤ δ =⇒ f (x) − f (a) − f ′ (a)(x − a) ≤ ε |x − a|.
1.49 Theorem: (Differentiability Implies Continuity) Let A ⊆ R be open, let f : A → R
and let a ∈ A. If f is differentiable at a then f is continuous at a.
Proof: The proof is left as an exercise (the proof is often given in first year calculus).
1.50Theorem: (ChainRule)LetA,B⊆Rbeopen,letf:A→B,letg:B→Rand let h = g ◦ f : A → R. Let a ∈ A and let b = f(a) ∈ B. Suppose that f is differentiable at a and g is differentiable at b. Then h is differentiable at a with
h′(a) = g′f(a) f′(a).
Proof: We shall use the ε-δ formulation of the derivative given in Exercise 1.48. Note first
that for x ∈ A and y = f(x) ∈ B we have h(x) − h(a) − g′(f(a))f′(a)(x − a)
and also
= g(f(x)) − g(f(a)) − g′(f(a))f′(a)(x − a)
= g(y) − g(b) − g′(b)f′(a)(x − a)
= g(y) − g(b) − g′(b)(y − b) + g′(b)(y − b) − g′(b)f′(a)(x − a)
≤ g(y) − g(b) − g′(b)(y − b) + |g′(b)| |y − b − f′(a)(x − a)|
= g(y) − g(b) − g′(b)(y − b) + |g′(b)| f(x) − f(a) − f′(a)(x − a)
|y − b| = f(x) − f(a) = f(x) − f(a) − f′(a)(x − a) + f′(a)(x − a) ≤ f (x) − f (a) − f ′ (a)(x − a) + |f ′ (a)| |x − a|.
Let ε > 0. Since g is differentiable at b, we can choose δ0 > 0 so that |y−b|≤δ0 =⇒g(y)−g(b)−g′(b)(y−b)≤ ε′ |y−b|.
2(1+|f (a)|) Since f is continuous at a (by Theorem 1.49), we can choose δ1 so that
|x−a|≤δ1 =⇒|f(x)−f(a)|≤δ0 =⇒|y−b|≤δ0. Since f is differentiable at a we can choose δ2 > 0 and δ3 > 0 so that
|x−a|≤δ2 =⇒f(x)−f(a)−f′(a)(x−a)≤|x−a|and |x−a|≤δ3 =⇒f(x)−f(a)−f′(a)(x−a)≤ ε′ |x−a|.
2(1+g (b)|) Letδ=min{δ1,δ2,δ3}. Letx∈Aandlety=f(x)∈B. Thenwhen|x−a|≤δwehave
h(x) − ≤ ≤
≤ ≤
h(a) − g′(f(a))f′(a)(x − a)
g(y) − g(b) − g′(b)(y − b) + |g′(b)| f(x) − f(a) − f′(a)(x − a)
ε 2(1+|f ′ (a)|)
ε 2(1+|f ′ (a)|)
ε 2(1+|f ′ (a)|)
|y − b| + (1 + |g′(b)|) · ε |x − a| 2(1+|g′ (b)|)
f(x)−f(a)−f′(a)(x−a)+|f′(a)||x−a|+ ε |x−a| 2
|x−a|+|f′(a)||x−a|+ ε |x−a| 2
ε |x − a| + ε |x − a| = ε|x − a|. 22
=
Thus h is differentiable at a with h′(a) = g′(f(a))f′(a), as required.
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1.51 Exercise: Let f : A ⊆ R → R. Show that f is strictly monotonic if and only if f has the property that for all a,b,c ∈ A, if b lies strictly between a and c then f(b) lies strictly between f(a) and f(c).
1.52 Theorem: (The Inverse Function Theorem) Let A be an interval in R, let f : A → R be injective and continuous, let B = f(A) (and note that f : A → B is bijective) and let g = f−1 : B → A be the inverse function. Then
(1) the functions f and g are strictly monotonic and g is continuous, and
(2) if A is an open interval then so is B, and if f is differentiable at a ∈ A with f′(a) ̸= 0,
then g is differentiable at b = f(a) with g′(b) = 1 . f′(a)
Proof: To prove Part 1, suppose that f is injective and continuous. Let a, b, c ∈ A with a
To complete the proof of Part 1, it remains to show that g is continuous. Suppose that
f and g are strictly increasing (the case that f and g are strictly decreasing is similar). Let
b ∈ B and let a = g(b) so that f(a) = b. Since f and g are strictly increasing, it follows
that b is the left (or right) endpoint of B if and only if a is the left (or right) endpoint of
A. To show that g is continuous at b, it suffices to show that if b is not the right endpoint
of B then lim g(y) = g(b) and that if b is not the left endpoint then lim g(y) = g(b). We y→b+ y→b−
shall prove the first of these two statements (the proof of the second is similar). Suppose
that b is not the right endpoint of B and hence a is not the right endpoint of A. Let
ε > 0 be small enough that a+ε ∈ A. Choose δ = f(a+ε)−b = f(a+ε)−f(a) and
notethatδ>0sincefisstrictlyincreasing. Thenforally∈B,ifb
that f has a local minimum value at a when ∃δ>0 ∀x∈A |x−a| ≤ δ =⇒ f(x) ≥ f(a).
1.54Theorem: (Fermat’sTheorem)LetA⊆Rbeopen,letf:A→R,andleta∈A. Suppose that f is differentiable at a and that f has a local maximum or minimum value at a. Then f′(a) = 0.
Proof: The proof is left as an exercise (you probably saw the proof in first year calculus).
1.55 Theorem: (Mean Value Theorems) Let a, b ∈ R with a < b.
(1) (Rolle’s Theorem) If f : [a, b] → R is differentiable in (a, b) and continuous at a and b
with f(a) = 0 = f(b) then there exists a point c ∈ (a,b) such that f′(c) = 0.
(2) (The Mean Value Theorem) If f : [a, b] → R is differentiable in (a, b) and continuous
at a and b then there exists a point c ∈ (a, b) with f′(c) = f(b) − f(a).
b−a
(3) (Cauchy’s Mean Value Theorem) If f,g : [a,b] → R are differentiable in (a,b) and
continuous at a and b, then there exists a point c ∈ (a, b) such that f′(c)g(b) − g(a) = g′(c)f(b) − f(a).
Proof: To Prove Rolle’s Theorem, let f : [a, b] → R be differentiable in (a, b) and contin- uousataandbwithf(a)=0=f(b). Iff isconstant,thenf′(x)=0forallx∈[a,b], so we can choose any c ∈ (a,b) and we have f′(c) = 0. Suppose that f is not constant. Eitherf(x)>0forsomex∈(a,b)orf(x)<0forsomex∈(a,b). Supposethatf(x)>0 for some x ∈ (a,b) (the case that f(x) < 0 for some x ∈ (a,b) is similar). By the Extreme Value Theorem, f attains its maximum value at some point, say c ∈ [a, b]. Since f (x) > 0 forsomex∈(a,b),wemusthavef(c)>0. Sincef(a)=f(b)=0andf(c)>0,wehave c ∈ (a,b). By Fermat’s Theorem, we have f′(c) = 0. This completes the proof of Rolle’s Theorem.
To prove the Mean Value Theorem, suppose that f : [a,b] → R is differentiable in (a,b) and continuous at a and b. Let g(x) = f(x) − f(a) − f(b)−f(a) (x − a). Then g is
b−a
differentiable in (a,b) with g′(x) = f′(x) − f(b)−f(a) and g is continuous at a and b with
b−a ′
g(a) = 0 = g(b). By Rolle’s Theorem, we can choose c ∈ (a,b) so that g (c) = 0, and then
g′(c) = f(b)−f(a), as required. b−a
Finally, we use the Mean Value Theorem to Prove Cauchy’s Mean Value Theorem.
Suppose f,g : [a,b] → R are both differentiable in (a,b) and continuous at a and b.
Let h(x) = f (x)g(b) − g(a) − g(x)f (b) − f (a). Then h is differentiable in (a, b) and
continuous at a and b with h(a) = f(a)g(b) − g(a)f(b) = h(b). By the Mean Value
Theorem, we can choose c ∈ (a,b) so that h′(c) = h(b)−h(a) = 0, and then we have b−a
f′(c)g(b) − g(a) − g′(c)f(b) − f(a) = 0, as required.
1.56Corollary: Leta,b∈Rwitha
Proof: The proof is left as an exercise (the proof is often given in first year calculus) 14
1.57 Theorem: (l’Hˆopital’s Rule) Let A be a nonempty open interval in R. Let a ∈ A, or
let a be an endpoint of A. Let f,g : A\{a} → R. Suppose that f and g are differentiable
in A\{a} with g′(x) ̸= 0 for all x ∈ A\{a}. Suppose either that lim f(x) = 0 = lim g(x) x→a x→a
or that lim g(x) = ±∞. Suppose that lim f′(x) = u ∈ R. Then lim f(x) = u. x→a x→a g′(x) x→a g(x)
Similar results hold for limits x → a+, x → a−, x → ∞ and x → −∞ and also when the limit is u = ±∞.
Proof: We give the proof for x → a+ (assuming that a ∈ A or a is the left endpoint of A) withu∈R. Supposefirstthatlim f(x)=0= lim g(x). Chooseb∈Awitha
so we have
f′(ckl) g(xk) − f′(ckl) = f(xk) − f(xl). g′(ckl) g(xl) g′(ckl) g(xl) g(xl)
f(xl) = f′(ckl) + f(xk) − f′(ckl) g(xk). g(xl) g′(ckl) g(xl) g′(ckl) g(xl)
f′(x)
Letε>0.Sincelim ′ =uwecanchooseδ>0sothat|x−a|≤δ=⇒g′(x)−u≤3.
f ′ ( x ) Sincexk →awecanchoosem∈Z+ sok≥m=⇒|xk−a|≤δ. Notethatwhenk,l≥m,
ε
x→a g (x)
since ckl lies between xk and xl we also have |ckl −a| ≤ δ so g′(ckl) −u ≤ min 1, 3 . Fix
f ′ ( c k l ) ε f(xk) ε f′(ckl) g(xk) ε
k≥m.Choosellargeenoughsothatg(xl)≤3 andg′(ckl) g(xl)≤3.Thenwehave f(xl) f′(ckl) f(xk) f′(ckl) g(xk)
g(xl) −u≤g′(ckl) −u+g(xl)+g′(ckl) g(xl)≤ε.
15