Answers & Hints for Selected Exercises 653(c) If B ~A, then An B = B , and by (μ1) and (μ5),
0::; μ(A - B) =μ(A) - μ(An B) =μ(A) - μ(B), so μ(B) ::; μ(A).- By (μ5), μ ([O, 1] - C) = μ ([O, 1]) - μ(C) = length[O, 1] - 0 = 1.
Chapter 4
EXERCISE SET 4.1- (a) 0.002, 0.0001, e/5 (b) 0.003, 0.00016, e/3
(c) 0.001, 0.00008, min{l,e/ 7}, (d) 0.0008, 0.00004, min{l, e/19} - Choose e as follows: (a) e/2 ( c) min{l, e/11} (e) min{l, e/9}
(g) min{l, e/12} (i) e/3 - (e) If {xn} is any sequence in V(f) - {-3} 3 Xn--+ -3, then by the algebra
of limits of sequences, 2 x~ + 5xn + 1 --+ 2(-3)^2 + 5( -3) + 1 = 4.
(i) If {xn} is any sequence in V(f) - {-2} 3 Xn--+ -2, then by the algebra
of limits of sequences,^3 ~:~~^2 = 3xn - 6--+ 3(-2) - 6 = -12.
- f(x ) = ax+ b. If a = 0, choose 8 = e. Then 0 < Jx - xol < 8 =? J(ax +
b) - (axo + b)J = 0 < e. If a =f. 0, choose 8 = e/JaJ. Then 0 < Jx - xol < 8 =?
J(ax + b) - (axo + b)J = Ja(x -xo)I = JaJJx - xol <Jal· e/JaJ = e. - First prove that X->lim 0 f(x ) = 0. Let e > 0 and choose 8 = e. Then 0 < Jx -OJ <
8 =? lf(x) - OJ = Jf(x)J = JxJ < e. Next, show that if Xo =f. 0, lim f(x) does
X-+Xo
not exist.
By Thm. 2.3.6, ::3 sequences {rn} of rationals and {zn} of irrationals con-
verging to xo. Then f(rn) = rn --+ Xo, whereas f(zn) = -Zn --+ -xo. Since
x 0 =f. 0, lim f(rn) =f. lim f(zn), so by Cor. 4.1.11 lim f(x) does not exist.
x-+xo x--+xo x--+xo
- 1 is not a cluster point of V(f) = (-oo, O] U {1} U [2, +oo).
EXERCISE SET 4.2- (b) lim f(x ) = L {:}Ve> 0 3 0 < Jx - OJ< 8 =? lf(x ) - LI < e
X--+Xo
{:}Ve> 0 3 0 < Jx - OJ < 8 =? llf(x ) - LI - OJ < e
{:} lim Jf(x ) -LI= 0.
X--+Xo
(c) Use the inequality in Thm. 1.2. 15 (c). A counterexample for the converse
is f(x ) = {
1
i~ x ~ O } and L = l.
-l 1f x < 0
- Ve> 0, take 8 = e. Then 0 < Jx - xol < 8 =? lf(x ) -xol = Jx - xol < e.