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Chapter 4 n Real Analysis 281
- Disprove the claim: If lim x → a | f ( x )| = L , then either lim x → a f ( x ) = L or lim x → a f ( x ) = − L.
- If lim x → a f ( x ) = ∞ and lim x → a g ( x ) = ∞, then lim x → a f + g = ∞.
- If lim x → a f ( x ) = ∞ and lim x → a g ( x ) = L ∈ R∗, then lim x → a
f ( x ) g ( x )
- If lim x → a f ( x ) = L , then lim x → a ( f ( x ) − L ) = 0.
- Disprove the two claims: (a) If lim x → a f ( x ) = L , then f ( a ) = L. (b) If f ( a ) = L , then lim x → a f ( x ) = L.
- The squeeze theorem (theorem 4.3.4). In exercises 57–68, prove each mathematical statement about continuity.
- The constant function is continuous (from theorem 4.3.5).
- The sum of two continuous functions is continuous (from theorem 4.3.5).
- The difference of two continuous functions is continuous (from theorem 4.3.5).
- The product of two continuous functions is continuous (from theorem 4.3.5).
- If f is continuous at x = a , then − f is continuous at x = a.
- If f is continuous at x = a , then | f | is continuous at x = a.
- Prove that f ( x ) = xn^ is continuous for all n ∈ N via induction (and theorem 4.3.5).
- If lim h → 0 f ( a + h ) = f ( a ), then f is continuous at x = a.
- Disprove the claim: If f and g are not continuous at x = a , then f + g is not continuous at x = a.
- Disprove the claim: If | f | is continuous at x = a , then f is continuous at x = a.
- Disprove the claim: If the composite function f ( g ( x )) is continuous, then f ( x ) and g ( x ) are both continuous.
- The following variation on the characteristic function of Q is continuous at x = 0:
f ( x ) =
x if x ∈ Q 0 if x 6 ∈ Q In exercises 69–70, state both an intuitive description and a definition of each limit.
- (^) x lim→ a f ( x ) = −∞ 70. (^) x lim→∞ f ( x ) = L
4.4 The Derivative
Calculus is the study of change. While Sir Isaac Newton and Gottfried Leibniz are both credited for independently developing calculus in the late 1600s, mathematicians had already been working with derivatives for nearly a half century. The study of change as expressed by the derivative was motivated by a sixteenth and seventeenth century European reflection on and ultimate rejection of ancient Greek astronomy and physics. The European astronomers Nicolaus Copernicus, Tycho Brahe, and Johannes Kepler
282 A Transition to Advanced Mathematics
each had insights that challenged the theories of the ancient Greeks, setting the stage for the ground-breaking work of the Italian scientist Galileo Galilei in the early 1600s. Many of the questions about a moving object (that is, an object changing position and velocity) that these scientists were studying are readily answered by considering lines tangent to curves. A number of mathematicians from many different countries made important contributions to the question of finding the equation of a tangent line. Pierre de Fermat studied maxima and minima of curves via tangent lines, essentially using the approach studied in contemporary calculus courses. This work prompted fellow French mathematician Joseph-Louis Lagrange to assert that Fermat should be credited with the development of calculus! The English mathematician Isaac Barrow, who was Newton’s teacher and mentor, corresponded regularly with Leibniz on these mathematical ideas. As we have mentioned, neither Newton nor Leibniz thought of the derivative as a measure of change in terms of our contemporary definition involving limits. Our study of the derivative follows more closely the work of Fermat and Barrow from the early 1600s, in which we think of a tangent line as a limit of secant lines. Naturally, the contemporary presentation is informed by an understanding of Cauchy’s notion of the limit from the early 1800s. The derivative enables the determination of the equation of a line tangent to a given curve at a given point. Given a function y = f ( x ), the slope of a secant line joining two points ( c , f ( c )) and ( c + h , f ( c + h )) is
m =
rise run
1 y 1 x
f ( c + h ) − f ( c ) ( c + h ) − c
f ( c + h ) − f ( c ) h
In this context, the symbol “ m ” is the first letter in the French word montrer which translates as “to climb.” To find the slope of the line tangent to a function f at the point ( c , f ( c )), we take a limit of the slopes of secant lines, letting h approach 0. Figure 4.13 illustrates why this limit process makes intuitive sense; you can see that the slopes of the secant lines get closer and closer to the slope of the tangent line as the point ( c + h , f ( c + h )) gets closer and closer to ( c , f ( c )). The definition of the derivative reflects these ideas. The following definition expresses the real number c as a variable quantity x to identify a general formula for the derivative, enabling us to determine the slope of the line tangent to f ( x ) whenever this slope is defined.
Definition 4.4.1 Let f ( x ) be a function with domain D. Then the derivative of f ( x ) is
f ′( x ) = lim h → 0
f ( x + h ) − f ( x ) h
whenever this limit exists. We say that f ( x ) is differentiable at x = c when f ′( c ) exists for c ∈ D, and that f ( x ) is differentiable when f ′( x ) exists for all x ∈ D. The ratio f ( x + h ) − f ( x ) h is called the difference quotient of the derivative.
284 A Transition to Advanced Mathematics
then cancels the denominator, simplifying the difference quotient so the limit can be evaluated. If the original function f ( x ) is a rational function, then finding a common denominator will simplify the difference quotient in this way. If f ( x ) contains a square root, multiplying by the conjugate square root function will simplify the difference quotient.
Example 4.4.2 We use the definition of the derivative to find the derivative of f ( x ) = 5
x + 1. Multiplying both the numerator and the denominator of the difference quotient by the conjugate square root function and then simplifying yields the following calculation.
f ′( x ) = lim h → 0
x + h + 1 − 5
x + 1 h
= lim h → 0
x + h + 1 − 5
x + 1 h
x + h + 1 + 5
x + 1 5
x + h + 1 + 5
x + 1
= lim h → 0
25 [( x + h + 1) − ( x + 1)] 5 h (
x + h + 1 +
x + 1)
= lim h → 0
5 h h (
x + h + 1 +
x + 1)
=
x + 1 n
Question 4.4.1 Using the definition of the derivative, differentiate each function. (a) f ( x ) = 2 x + 1
(b) g ( x ) = 7 x^3
(c) s ( x ) =
x + 5 (d) t ( x ) =
x n
While we can use the formal definition of the derivative to compute derivatives of a given function, theoretical applications of the definition are more important. Using the definition, we can prove general theorems that hold for all derivatives, making it easy to differentiate many familiar functions without explicitly applying the definition one function at at time. Many functions are so complicated in structure that directly using the difference quotient becomes unwieldy or impossible. The next theorem states analytic properties of derivatives to facilitate such computations. Using these results is a common exercise in calculus courses, but you may not have considered the underlying proofs that justify them. These proofs are the focus of the remainder of this section.
Theorem 4.4.1 If c ∈ R and both f and g are differentiable functions, then the following hold.
d dx
[ (^) c ] (^) = 0
- The scalar multiple rule: d dx
[ c · f ( x ) ] = c · f ′( x )
Chapter 4 n Real Analysis 285
d dx [ f + g ] = f ′^ + g ′
d dx [ f − g ] = f ′^ − g ′
d dx
[
xn^
]
= n · xn −^1 , for n ∈ R
d dx
[ (^) f · g ] (^) = g · f ′^ + f · g ′
[
f g
]
g · f ′^ − f · g ′ g^2
, provided that g ( x ) 6 = 0
[ f ( g ( x )) ] = f ′( g ( x )) · g ′( x )
A standard goal of a calculus course is to develop a mastery in using these differentiation rules. Before diving into the proofs of various parts of this theorem, the next example provides the opportunity to revisit the skills you learned in calculus.
Question 4.4.2 Using theorem 4.4.1, differentiate each function.
(a) f ( x ) = 10 x^3 − 7 x^2 + 5 (b) g ( x ) =
5 x + 2
(c) h ( x ) = cos (3 x + 1) 5 x^2 + 2
(d) p ( x ) = ( x^5 + x ) tan(2 x ) (e) q ( x ) = ln(4 x^2 + 1) · sin^2 (5 x + 3) (f) r ( x ) = (2 x^5 + 3) 3
4 ex^ + 6 x
n The next three examples give the proofs of some of these differentiation rules. As in the study of limits and continuity, we first consider the scalar multiple and sum rules, and then discuss a couple of different approaches to proving the power rule. Example 4.4.3 We prove the scalar multiple rule from theorem 4.4.1: For any constant c ∈ R and differentiable function f ,
d dx
[ c · f ( x ) ] = c · f ′( x ).
Proof Apply the definition of the derivative and the limit of a scalar multiple rule.
d dx [ c · f ( x ) ] = lim h → 0
c · f ( x + h ) − c · f ( x ) h = lim h → 0
c · [ f ( x + h ) − f ( x )] h
= c · lim h → 0
f ( x + h ) − f ( x ) h
= c · f ′( x ) n
Example 4.4.4 We prove the sum rule: If f and g are differentiable functions, then
d dx [ f + g ] = f ′( x ) + g ′( x ).
Chapter 4 n Real Analysis 287
Question 4.4.3 The following steps outline a proof of the quotient rule: If f ( x ), g ( x ) are differentiable functions with g ( x ) 6 = 0, then d dx
[
f ( x ) g ( x )
]
g ( x ) · f ′( x ) − f ( x ) · g ′( x ) g ( x )^2
(a) What is the difference quotient for the function
f ( x ) g ( x )
(b) Using the common denominator g ( x ) · g ( x + h ) · h , simplify the difference quotient from part (a). (c) In the numerator from part (b), subtract and add the term g ( x ) · f ( x ). Now split the fraction into a difference of two differences, gathering together the two terms with g ( x ) as a common factor and the two terms with f ( x ) as a common factor. (d) What is the limit of the difference of difference quotients from part (c) as h approaches 0? (e) Based on parts (a)–(d), craft a complete proof of the quotient rule as modeled in examples 4.4.3, 4.4.4, and 4.4.5. n
Question 4.4.4 The following steps outline a proof of the chain rule: If f ( x ), g ( x ) are differentiable functions, then d dx [ f [ g ( x )] ] = f ′[ g ( x )] · g ′( x ).
(a) What is the difference quotient h ( t ) − h ( x ) t − x (from the alternate definition of the derivative) for the function h ( x ) = f [ g ( x ) ]? (b) Assuming there are no values x for which g ( x ) = g ( t ), multiply both the numerator and the denominator of the difference quotient from part (a) by g ( t ) − g ( x ). Factor out the resulting difference quotient for g ( x ). (c) Take the limit of the product of difference quotients from part (b) as t approaches x to obtain the chain rule formula. (d) Based on parts (a)–(d), craft a proof of the chain rule under the assumption that g ( x ) 6 = g ( t ) as modeled in examples 4.4.3, 4.4.4, and 4.4.5. The assumption that there are no values for which g ( x ) equals g ( t ) may be unreasonable; a complete proof of the chain rule that does not use this assumption is outlined in exercises 67–70 at the end of this section. n We end this section by considering the relationship between two of the most sig- nificant properties of functions studied in this chapter: continuity and differentiability. Some properties of functions are completely independent of one another, as we saw in our discussion of one-to-one and onto functions; some functions are both, some are neither, while still others have just one of these properties. This observation leads us
288 A Transition to Advanced Mathematics
to ask if continuity and differentiability are independent of one another, or is there a connection between these two properties? As you may recall, every differentiable function is continuous, but not every continuous function is differentiable. We consider the theorem and its proof, along with a counterexample that together justify these assertions. Theorem 4.4.2 If a function f with domain D is differentiable at a ∈ ( b , c ) ⊆ D, then f is continuous at a. Proof By the alternate definition of the derivative, given any ε > 0, there exists a value δ > 0 so that ∣ ∣∣ ∣
f ( x ) − f ( a ) x − a − f ′( a )
∣ < ε
whenever 0 < | x − a | < δ. Multiplying both sides by | x − a |, we see that | f ( x ) − f ( a ) − f ′( a )( x − a )| < ε| x − a |. Applying the second inequality (| x | − | y | ≤ | x − y |) from theorem 4.3.3 in section 4.3, we have | f ( x ) − f ( a )| − | f ′( a )( x − a )| ≤ | f ( x ) − f ( a ) − f ′( a )( x − a )|. This fact implies | f ( x ) − f ( a )| < | f ′( a )( x − a )| + ε| x − a |, and so | f ( x ) − f ( a )| < (| f ′( a )| + ε) · | x − a |. The term on the right can be made arbitrarily small: we restrict values of x in that term so that | x − a | is smaller than both δ (so that the first inequality holds) and ε/(| f ′( a )| + ε). Then | f ( x ) − f ( a )| < ε, which proves the result. n Theorem 4.4.2 asserts that every differentiable function is continuous. Are there continuous functions that are not differentiable? Perhaps you can recall from calculus examples of continuous functions that are not differentiable. The next example provides one such counterexample. Example 4.4.6 We discuss the continuity and differentiability of f ( x ) = | x | at x = 0. We can show that y = | x | is continuous at x = 0, using the definition. Let ε > 0 and choose δ = ε. For any x such that | x − 0 | < δ, the following string of relations holds: | f ( x ) − f (0)| = | | x | − | 0 | | = | | x | | = | x | < ε. By the definition of continuity, | x | is continuous at x = 0. On the other hand, we can show that | x | is not differentiable at x = 0, using the alternate definition of the derivative. The difference quotient for f ( x ) at x = 0 is f ( x ) − f (0) x − 0
| x | − | 0 | x
| x | x
Taking the limit of this difference quotient as x approaches 0,
lim x → 0 −
| x | x
= − 1 and lim x → 0 +
| x | x
290 A Transition to Advanced Mathematics
4.4.2 Exercises for Section 4.
In exercises 1–6, express the slope of a secant line to each function for the designated x -coordinates as a difference quotient, and sketch the corresponding graph.
- f ( x ) = x^2 + 2 at x = 3 and x = 4
- f ( x ) = x^2 + 2 at x = 3 and x = 3. 01
- f ( x ) = x^2 + 2 at x = 3 and x = 3. 0001 4. f ( x ) = x^3 at x = 0 and x = 1 5. f ( x ) = x^3 at x = 0 and x = 0. 01 6. f ( x ) = x^3 at x = 0 and x = 0. 0001
In exercises 7–18, use the definition of the derivative to compute the derivative (if it exists) of each function.
- f ( x ) = 2 x + 3
- g ( x ) = 3 x − 5
- h ( x ) = x^2 + 1
- j ( x ) = x^2 + x
- p ( x ) = 1 / x
- q ( x ) =
x + 1
- r ( x ) =
x − 3
- s ( x ) =
x
- t ( x ) =
2 x + 2
- u ( x ) =
x + 7
- v ( x ) =
4 x if x ≤ 2 2 x^2 if x > 2
- w ( x ) =
4 x + 3 if x ≤ 2 2 x^2 if x > 2
In exercises 19–28, compute the derivative of each function using the analytic differentiation rules from theorem 4.4.1, along with your recollection of the derivatives of functions from calculus.
- f ( x ) = ( x^9 + x^6 )^37
- f ( x ) = ( x + x −^1 )^4
- f ( x ) = (3 x^2 +
6 x + 5 − 4) · (2 x + 1 / x )
- f ( x ) = ( x^2 + 1)^3 ·
(5 x^3 + 2 x )^2 + 1
- f ( x ) = sin^5 ( x^3 + 2 x )
- f ( x ) = ln( x ) · cos(2 x + 7)
- f ( x ) = log 3 (cot(2 x ))
- f ( x ) = ln( x^2 + 2) · log 5 (csc( x ) + 2 x )
- f ( x ) = ( k · x^5 + 2 x ) 3
x , where k ∈ R
- f ( x ) =
3 x^2 + 2 x +
kx
) 3 n , where k , n ∈ R
In exercises 29–34, determine the exact value of h ′(3π/4) and state the equation of the line tangent to h ( x ) at x = 3 π/4 using the information in the following table.
f ( x ) f ′( x ) g ( x ) g ′( x ) x = 3 π/ 4 4 2 5 3
- h ( x ) = 7 · f ( x ) − sec( x ) + π^3
- h ( x ) = g ( x ) · cos( x )
- h ( x ) = g ( x ) + x f ( x ) + 2
- h ( x ) = tan( x ) + π · cot^2 ( g ( x ))
- h ( x ) = sin[π · f ( x )]+cos[π · g ( x )]
- h ( x ) =
f ( x ) x
x^2 g ( x )
Chapter 4 n Real Analysis 291
In exercises 35–38, answer each question about f ( x ) =
x.
- Using the definition of the derivative, find f ′( x ).
- Using the power rule, find f ′( x ).
- Determine the equation of the tangent line to f ( x ) =
x at (9, 3).
- Determine the equation of the tangent line to f ( x ) =
x that is perpendicular to the line determined by 2 y + 8 x = 16.
Exercises 39–43 develop a proof that the derivative of sin θ is cos θ.
- Prove that sin θ · cos θ < θ < tan θ. Hint: Compare the areas of the three nested regions in figure 4.14 and use the fact that a pie-shaped sector of the unit circle with central angle θ (in radians) has an area of θ/2.
- Identify upper and lower bounds on sin θ/θ using the inequalities from exercise 39. Hint: Divide by sin θ and take reciprocals.
- Prove that lim θ→ 0 sin θ/θ = 1. Hint: Apply the squeeze theorem (see theorem 4.3.4 from section 4.3) to the inequalities from exercise 40.
- Prove that lim θ→ 0
(1 − cos θ)/θ = 0. Hint: Multiply both the numerator and the denominator by 1 + cos θ and then use both the Pythagorean identity sin^2 θ + cos^2 θ = 1 and the limit from exercises 41.
- Prove that the derivative of sin θ is cos θ. Hint: Working with the definition of the derivative, simplify the resulting difference quotient using the limits from exercises 41 and 42 along with the trigonometric identity sin( u + v ) = sin u cos v + sin v cos u.
In exercises 44–48, derive the formulas for the derivative of the other trigonometric functions; all but exercise 44 use the quotient rule.
- Prove that the derivative of cos θ is − sin θ. Hint: Use the cofunction identity cos x = sin(π/ 2 − x ) and the derivative from exercises 43.
Figure 4.14 Figure for exercise 39
(cos q , sin q )
(0,0) (1,0)
(1, tan q )
q
Chapter 4 n Real Analysis 293
- If f and g are differentiable functions on ( a , b ) with f − g a constant, then f and g have the same derivative at any x ∈ ( a , b ).
- Define a function f that is nowhere differentiable, while f^2 is everywhere differentiable. Hint: Consider a variation on the characteristic function of Q. Exercises 67–70 develop a proof of the chain rule in a fuller generality than was discussed in question 4.4.4. Throughout these exercises assume that g ( x ) is differentiable at a point x = a and that f ( x ) is differentiable at g ( a ).
- Prove that the following function F is continuous at h = 0; intuitively, we think of F as the derivative of f with respect to t = g ( a ).
f ( h ) =
f [ g ( a ) + h ] − f [ g ( a )] h
if h 6 = 0 f ′[ g ( a )] if h = 0
- Prove that f [ g ( a ) + h ] = f [ g ( a )] + h · F ( h ) for sufficiently small values of h by taking the limit of these two expressions as h approaches 0.
- In a parallel way, we can define a function G so that G (0) = g ′( a ) and g ( a + k ) = g ( a ) + k · G ( k ) for sufficiently small values of k. Use this fact, the result from exercise 68, and the choice of h = g ( a + k ) − g ( a ) = k · G ( k ) to prove that: f [ g ( a ) + h ] = f [ g ( a + k )] and h · F ( h ) = k · G ( k ) · F ( k · G ( k )).
- Using the two equations obtained in exercise 69, substitute the first equation into the second to prove that
f [ g ( a + k )] = f [ g ( a )] + k · G ( k ) · F ( k · G ( k )). The last term on the right is continuous at 0 based on the definitions of F and G. Subtract f [ g ( a )] on both sides of this equation, divide both sides by k and take the limit as k approaches 0 to obtain the chain rule.
4.5 Understanding Infinity
The notion of infinity has been an important element in many cultures’ attempts to understand life: people refer to eternal time; an eternal spiritual afterlife; a boundless universe; an all-powerful deity. Mathematics has a unique and important perspective on infinity; the insights arising from mathematics’ rigorous, logical approach to infinity have had an important influence on Western society’s view of the world. But many advanced mathematical results on infinity (especially those that grew out of Georg Cantor’s work in the late 1860s) are not widely known. In this section, we explore a mathematical understanding of the infinite. We have already taken the first steps in this direction in our study of limits. One major breakthrough in the development of calculus is the harnessing of infinity in the very specific and powerful way expressed by the notion of limit to obtain the derivative (and the integral as discussed in section 4.6). As mathematicians developed and refined their understanding of limits, derivatives, and integrals in the eighteenth and nineteenth
