Calculus 2nd Edition Rogawski

SECTION 2.9 The Formal Definition of a Limit 91
27. Show that if f (x) is continuous and 0 ≤ f (x) ≤ 1 for
0 ≤ x ≤ 1, then f (c) = c for some c in [0, 1] (Figure 6).
1
y
y = x
y = f(x)
x
c 1
FIGURE 6 A function satisfying 0 ≤ f (x) ≤ 1 for 0 ≤ x ≤ 1.
28. Use the IVT to show that if f (x) is continuous and one-to-one
on an interval [a,b], then f (x) is either an increasing or a decreasing
function.
29. Ham Sandwich Theorem Figure 7(A) shows a slice of
ham. Prove that for any angle θ (0 ≤ θ ≤ π), it is possible to cut the
slice in half with a cut of incline θ. Hint: The lines of inclination θ
are given by the equations y = (tan θ)x + b, where b varies from −∞
to ∞. Each such line divides the slice into two pieces (one of which
may be empty). Let A(b) be the amount of ham to the left of the line
minus the amount to the right, and let A be the total area of the ham.
Show that A(b) = −A if b is sufficiently large and A(b) = A if b is
sufficiently negative. Then use the IVT. This works if θ ̸= 0or π 2 .If
θ = 0, define A(b) as the amount of ham above the line y = b minus
the amount below. How can you modify the argument to work when
θ = π 2 (in which case tan θ = ∞)?
30. Figure 7(B) shows a slice of ham on a piece of bread.
Prove that it is possible to slice this open-faced sandwich so that each
part has equal amounts of ham and bread. Hint: By Exercise 29, for
all 0 ≤ θ ≤ π there is a line L(θ) of incline θ (which we assume is
unique) that divides the ham into two equal pieces. Let B(θ) denote
the amount of bread to the left of (or above) L(θ) minus the amount
to the right (or below). Notice that L(π) and L(0) are the same line,
but B(π) = −B(0) since left and right get interchanged as the angle
moves from 0 to π. Assume that B(θ) is continuous and apply the IVT.
(By a further extension of this argument, one can prove the full “Ham
Sandwich Theorem,” which states that if you allow the knife to cut at a
slant, then it is possible to cut a sandwich consisting of a slice of ham
and two slices of bread so that all three layers are divided in half.)
y
θ
(A) Cutting a slice of ham
at an angle θ
x
y
FIGURE 7
π
L( 2 )
L(θ)
(B) A slice of ham on top
of a slice of bread
L(0) = L(π)
x
A “rigorous proof” in mathematics is a
proof based on a complete chain of logic
without any gaps or ambiguity. The formal
limit definition is a key ingredient of
rigorous proofs in calculus. A few such
proofs are included in Appendix D. More
complete developments can be found in
textbooks on the branch of mathematics
called “analysis.”
2.9 The Formal Definition of a Limit
In this section, we reexamine the definition of a limit in order to state it in a more rigorous
and precise fashion. Why is this necessary? In Section 2.2, we defined limits by saying
that lim f (x) = L if |f (x) − L| becomes arbitrarily small when x is sufficiently close
x→c
(but not equal) to c. The problem with this definition lies in the phrases “arbitrarily small”
and “sufficiently close.” We must find a way to specify just how close is sufficiently close.
The Size of the Gap
Recall that the distance from f (x) to L is |f (x) − L|. It is convenient to refer to the
quantity |f (x) − L| as the gap between the value f (x) and the limit L.
Let’s reexamine the trigonometric limit
sin x
lim
x→0 x
= 1 1
In this example, f (x) = sin x and L = 1, so Eq. (1) tells us that the gap |f (x) − 1| gets
x
arbitrarily small when x is sufficiently close, but not equal, to 0 [Figure 1(A)].
Suppose we want the gap |f (x) − 1| to be less than 0.2. How close to 0 must x be?
Figure 1(B) shows that f (x) lies within 0.2 of L = 1 for all values of x in the interval
[−1, 1]. In other words, the following statement is true:
sin x
∣ − 1
x ∣ < 0.2 if 0 < |x| < 1