SOLUTION OF HW3
MINGFENG ZHAO
September 24, 2012
1. [10 Points] Let {xn } be a sequence. If there exists an a ∈ (0, 1) such that
|xn+1 − xn | ≤ an ,
(1)
∀n ∈ N.
Then xn → x for some x ∈ R.
Proof. For any n, k ∈ N, then we know that
|xn+k − xn | = |(xn+k − xn+k−1 ) + (xn+k−1 − xn+k−2 ) + · · · + (xn+1 − xn )|
≤
|xn+k − xn+k−1 | + |xn+k−1 − xn+k−2 | + · · · + |xn+1 − xn )|
≤ an+k−1 + an+k−2 + · · · + an
By (1)
= an · ak−1 + ak−2 + · · · + a + 1
= an ·
1 − ak
1−a
≤ an ·
1
1−a
=
Since 0 < a
an
.
1−a
Since 0 < a < 1, then an → 0, as n → ∞. Hence for any > 0, there exists some N ∈ N such that
for all n ≥ N , we have that
an < · (1 − a).
Now for any n, m ≥ N , without loss of generality, assume that m ≥ n, then
|xm − xn | = |xn+n−m − xn |
1
2
MINGFENG ZHAO
≤
an
1−a
<
(1 − a)
1−a
= .
Which implies that {xn } is Cauchy in R. Therefore, we can conclude that xn → x for some x ∈ R.
2. [10 Points] a. For a real valued function f , if
L = lim f (x),
x→a
Then |f (x)| → |L| as x → a.
b. Show that there is a function such that |f (x)| → |L|, as x → a, but the limit of f (x) does not
exist.
Proof. a. Claim I: ||a| − |b|| ≤ |a − b| for all a, b ∈ R.
For any a, b ∈ R, then we know that
|a| = |a − b + b| ≤ |a − b| + |b| and |b| = |b − a + a| ≤ |b − a| + |a| = |a − b| + |a|.
Which implies that
|a| − |b| ≤ |a − b|,
and |b| − |a| ≤ |a − b|.
Hence we can conclude that
||a| − |b|| ≤ |a − b|.
Since L = lim f (x), then for any > 0, there exists some δ > 0 such that for all 0 < |x − a| < δ,
x→a
we have
|f (x) − L| < .
SOLUTION OF HW3
3
By the Claim I, then for all 0 < |x − a| < δ, we have
||f (x)| − |L|| ≤ |f (x) − L| < .
Which implies that |f (x)| → |L| as x → a.
b. Let a = 0, L = 1 and
f (x) =




 1,



 −1,
if x ≥ 0
if x < 0
Then we know that |f (x)| = 1 for all x ∈ R, which implies that |f (x)| = 1 → |L| = 1, as x → 0.
But the limit of f (x) does not exist, as x → 0. In fact, if we say that f (x) → A for some A ∈ R, as
x → 0, then for any n ∈ N, there exists some δn > 0 such that for all 0 < |x| < δ, we have
|f (x) − A| <
1
.
n
Taking xn > 0 such that 0 < xn < δn , then 0 < |xn | < δn , which implies that
|f (xn ) − A| <
1
n
and |f (−xn ) − A| <
1
,
n
∀n ∈ N.
That is, we get
|1 − A| <
1
,
n
and | − 1 − A| <
1
,
n
∀n ∈ N.
By taking n → ∞, we know that |1 − A| = 0 and | − 1 − A| = 0, that is, A = 1 and A = −1,
contradiction.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu