Solutions to Analysis - Homework11
30.5
For every > 0, then for x s.t. 0 < |x − 2| < δ, we need to find δ for given such that x2 − 1 < .
Assuming δ < 1 and 0 < |x − 2| < δ, then
2
− 1 = 2 − x = |2 − x| ≤ δ < δ (because |2 − x| ≤ δ, |x| ≥ 2 − δ and δ < 1)
x
x |x|
2−δ
2
So if we choose δ < min{1, }, then x − 1 < , which means limx→2 x2 = 1.
30.8
We can find small enough 0 > 0 such that L − 0 > 0. For given 0 , by limx→a f (x) = L > 0, there
exists δ0 > 0 such that if 0 < |x − a| < δ0 , then |f (x) − L| < 0 . Hence f (x) > L − 0 > 0.
31.1(v)
For every > 0, there exist δ1 > 0 s.t. if 0 < |x − a| < δ1 , then |f (x) − L| <
bounded, so there exists C > 0 s.t. |f (x)| < C for 0 < |x − a| < δ1 .
Also, there exist δ2 > 0 s.t. if 0 < |x − a| < δ2 , then |g(x) − M | < 2(1+C)
.
Let δ = min{δ1 , δ2 }. It follows that if 0 < |x − a| < δ, then
2(1+|M |) .
Since f (x) − L is
|f (x)g(x) − LM | = |f (x)g(x) − f (x)M + f (x)M − LM |
≤ |f (x)||g(x) − M | + |M ||f (x) − L|
<C
+ |M |
2(1 + C)
2(1 + |M |)
< .
Hence limx→a f (x)g(x) = LM .
31.4
For every > 0, there exists 0 < δ1 < δ, s.t. if 0 < |x − a| < δ1 , then |f (x) − L| < and |h(x) − L| < .
It leads to f (x) > L − and h(x) < L + when 0 < |x − a| < δ1 .
Thus we have L − < f (x) ≤ g(x) ≤ h(x) < L + which means |g(x) − L| < when 0 < |x − a| < δ1 .
Hence limx→a g(x) = L.
31.5
00
⇒00 : Suppose limx→a f (x) exists and is L. Assume there exist > 0 s.t. for every δ > 0, there exist
x, y satisfying 0 < |x − a| < δ and 0 < |y − a| < δ, and |f (x) − f (y)| ≥ . Then ≤ |f (x) − f (y)| ≤
|f (x) − L| + |f (y) − L|. So at least one of |f (x) − L|, |f (y) − L| is greater than 2 . So for any δ > 0, there
exist x s.t. 0 < |x − a| < δ and |f (x) − L| ≥ 2 . Contradiction.
1
00
⇐00 , For given sequence {xn } that converges to a and xn 6= a. From right-hand side condition, we know
that f (xn ) is a Cauchy sequence, so it converges.
For two different sequences {xn } → a and {yn } → a. Assume {f (xn )} → L, {f (yn )} → M and L 6= M .
For 0 = 3|L−M |, there exist N ∈ N s.t. for every n ≥ N , we have |f (xn )−L| < 0 and |f (yn )−M | < 0 .
Then |f (xn ) − f (yn )| = |f (xn ) − L − (f (yn ) − M ) + L − M | ≥ |L − M | − |f (xn ) − L| − |f (yn ) − M | ≥
30 − 0 − 0 = 0 . Contradict to the right-hand side condition.
31.6
Define
lim inf f (x) = glb{limits of convergent sequence f (xn ) s.t. xn → a and xn 6= a};
x→a
lim sup f (x) = lub{limits of convergent sequence f (xn ) s.t. xn → a and xn 6= a}.
x→a
Then lim inf x→a f (x) ≤ lim supx→a f (x).
Result: If lim inf x→a f (x) = lim supx→a f (x) = L, then limx→a f (x) = L.
Proof. Let lim inf x→a f (x) = A, then any convergent subsequent of f (xn ) where xn → a converges to a
number that is greater than and equal to A. Also, if lim supx→a f (x) = B, then any convergent subsequence
of f (xn ) where xn → a converges to a number that is less than and equal to B.
Therefore, pick a sequence xn → a, then any convergent subsequence of f (xn ) converges to a number that
is greater than and equal to L and also less than and equal to L, so it is L. Hence f (xn ) converges to L.
32.5
First, we look at the limit from left. For every c ∈ (a, b], let V = {f (x) : x ∈ (a, c)}. V is non-empty
and bounded above by f (c) since f is increasing. Then V has a least upper bound, v, by least upper bound
axiom.
From property of LUB, given > 0. there exists d ∈ (a, c) s.t. f (d) > v − . By monotonicity of
f , v − < f (x) < v for any x ∈ [d, c). Thus let δ = c − d, then if 0 < c − x < δ, it implies that
v − < f (x) < v ⇒ |f (x) − v| < . Hence limx→c− f (x) exists.
Next, we prove the limit from right exist. For every c ∈ [a, b), let U = {f (x) : x ∈ (c, b)}. U is non-empty
and bounded below by f (c) since f is increasing. Then U has a greatest lower bound, u, by greatest lower
bound axiom.
From property of GLB, given > 0. there exists e ∈ (c, b) s.t. f (e) < u + . By monotonicity of
f , u < f (x) < u + for any x ∈ (c, e]. Thus let δ = e − c, then if 0 < x − c < δ, it implies that
u < f (x) < u + ⇒ |f (x) − u| < . Hence limx→c+ f (x) exists.
33.3
Given > 0, choose δ = . Then if |x| < δ = , we have |f (x) − f (0)| = |f (x)| = |x| < if x is rational
and |f (x) − f (0)| = 0 < if x is irrational. In either case, |x| < δ ⇒ |f (x) − f (0)| < . Then f is continuous
at 0.
Let p 6= 0. We can take a sequence {xn } of rationals converging to p. Then f (xn ) = xn → p. Also take
a sequence {yn } of irrationals converging to p. Then f (yn ) = 0 → 0. Since p 6= 0, it follows that limit of f
at p does not exist. Then f is not continuous at p.
33.4
1
1
1. Suppose x = m
n is nonzero rational. Then f (x) = n . There exists 0 < n , for every δ > 0, the interval
1
(x − δ, x + δ) contains irrational points y s.t. f (y) = 0 and |f (x) − f (y)| = n > 0 . Thus f is discontinuous
at x.
2
2. If x = 0, then f (x) = 1. Given 0 = 12 , for for every δ > 0, the interval (x − δ, x + δ) contains irrational
points y s.t. f (y) = 0 and |f (x) − f (y)| = 1 > 0 . Thus f is discontinuous at x = 0.
3. Suppose x is irrational. Given > 0, choose n ∈ N s.t. n1 < . There are finitely many rationals r = pq
in [0, 1] with gcd(p, q) = 1 and 1 ≤ q ≤ n; we can list them as {r1 , r2 , . . . , rm }. Choose
δ = min{|x − rk | : k = 1, 2, . . . , m}.
Then δ > 0 since x ∈
/ Q. Furthermore, if |x − y| < δ, then either y is irrational and f (y) = 0, or y = pq
in lowest terms with q > n and f (y) = 1q < n1 < . In either case, |f (x) − f (y)| = |f (y)| < . Then f is
continuous at irrational points.
33.5
For every c ∈ Q∗ , there exists a rational sequence {rn } converging to c. It follows by continuity that
f (rn ) → f (c). Also f (rn ) → 0 since f (x) = 0 for x is rational. Then f (c) = 0. Hence f (x) = 0 on interval
[a, b].
33.6
From exercise 32.5, we know that for increasing function f on interval [a, b], the right limit and left limit
exist at every point on (a, b).
If f is discontinuous at x0 , then x0 corresponds to an interval I(x0 ) = (limx→x− f (x), limx→x+ f (x))
0
0
since limx→x− f (x) < limx→x+ f (x) by discontinuity and increment of f . For different discontinuous points
0
0
x1 and x2 , I(x1 ) and I(x2 ) must be disjoint because f is increasing.
The range of f on [a, b] is [f (a), f (b)] by monotonicity. Let D = {x ∈ [a, b] : f is discontinuous at point x}.
Let D0 = {x ∈ {a, b} : f is discontinuous at point x}, this is finite. And let
D(n) = {x ∈ (a, b) : lim f (x) − lim f (x) >
x→x+
0
x→x−
0
1
}.
n
(a)
Then cardinal number of D(n) is less than b f (b)−f
c which is finite. Hence D = ∪n∈N D(n)∪D0 is countable.
1/n
3