MATH 409–503
October 10, 2013
Test 1: Solutions
Problem 1 (20 pts.) Prove that for any x ∈ (0, 1) and any natural number n,
(1 − x)n ≤ 1 − nx +
n(n − 1) 2
x.
2
The proof is by induction on n. First we consider the case n = 1. In this case the inequality
reduces to (1 − x)1 ≤ 1 − x, which is true for any x ∈ R. Now assume that the inequality holds for
n = k, that is,
k(k − 1) 2
(1 − x)k ≤ 1 − kx +
x
2
for all x ∈ (0, 1). Multiplying both sides of this inequality by the positive number 1 − x, we get
(k + 1)k 2 k(k − 1) 3
k(k − 1) 2
k
x (1 − x) = 1 − (k + 1)x +
x −
x .
(1 − x) (1 − x) ≤ 1 − kx +
2
2
2
Since x > 0, it follows that
(1 − x)k+1 ≤ 1 − (k + 1)x +
(k + 1)k 2
x ,
2
which means that the inequality holds for n = k + 1 as well. By induction, the inequality holds for
every natural number n and every x ∈ (0, 1).
Problem 2 (25 pts.) For each of the following sequences, find the limit or show that the
sequence is not convergent.
√
√
(i) xn = n + 3 − n − 1, n ∈ N.
√
(ii) y1 = 17, yn = 1 + yn−1 − 1 for n ≥ 2.
1
(ii) zn = n3 sin 2 , n ∈ N.
n
Solution:
lim xn = 0, lim yn = 2. The sequence {zn } diverges to +∞.
n→∞
n→∞
For any n ∈ N we have
√
√
√
√
4
4
( n + 3 − n − 1)( n + 3 + n − 1)
√
√
√
0 < xn = n + 3 − n − 1 =
=√
<√
,
n+3+ n−1
n+3+ n−1
n+3
√
which implies that xn → 0 as n → ∞. Further, yn − 1 = yn−1 − 1 for all n ≥ 2. It follows by
1−n
n−1
n−1
for all n ∈ N. Since 161/m → 1 as
= 162
= 16(1/2)
induction that yn − 1 = (y1 − 1)(1/2)
m → ∞, we obtain that yn − 1 → 1 as n → ∞. Then yn → 2 as n → ∞.
√
√
1
sin x
= 1. As a consequence, n2 sin(1/n2 ) = sin(1/n2 )/(1/n2 ) → 1 as
x→0 x
n → ∞. In particular, n2 sin(1/n2 ) > 1/2 for all sufficiently large n. Then zn = n3 sin(1/n2 ) > n/2
for all sufficiently large n, which implies that zn → +∞ as n → ∞.
Finally, we know that lim
Problem 3 (25 pts.) Prove the Bolzano-Weierstrass Theorem: any bounded sequence
of real numbers has a convergent subsequence.
Suppose {xn } is a bounded sequence of real numbers. We are going to build a nested sequence
of intervals In = [an , bn ], n = 1, 2, . . . , such that each In contains infinitely many elements of {xn }
and |In+1 | = |In |/2 for all n ∈ N. The sequence is built inductively. First we set I1 to be any closed
bounded interval that contains all elements of {xn } (such an interval exists since the sequence {xn } is
bounded). Now assume that for some n ∈ N the interval In is already chosen and it contains infinitely
many elements of the sequence {xn }. Then at least one of the subintervals I 0 = [an , (an + bn )/2)]
and I 00 = [(an + bn )/2, bn ] also contains infinitely many elements of {xn }. We set In+1 to be such a
subinterval. By construction, In+1 ⊂ In and |In+1 | = |In |/2. Since |In+1 | = |In |/2 for all n ∈ N, it
follows by induction that |In | = |I1 |/2n−1 for all n ∈ N. As a consequence, |In | → 0 as n → ∞. By
the nested intervals property, the intersection I1 ∩ I2 ∩ I3 ∩ . . . consists of a single number c.
Next we are going to build a strictly increasing sequence of natural numbers n1 , n2 , . . . such that
xnk ∈ Ik for all k ∈ N. The sequence is built inductively. First we choose n1 so that xn1 ∈ I1 . Now
assume that for some k ∈ N the number nk is already chosen. Since the interval Ik+1 contains infinitely
many elements of the sequence {xn }, there exists m > nk such that xm ∈ Ik+1 . We set nk+1 = m.
Now we claim that the subsequence {xnk }k∈N of the sequence {xn } converges to c. Indeed, for any
k ∈ N the points xnk and c both belong to the interval Ik . Hence |xnk − c| ≤ |Ik |. Since |Ik | → 0 as
k → ∞, it follows that xnk → c as k → ∞.
Problem 4 (20 pts.) Consider a function f : R → R defined by

2π


if − 1 < x < 0,
sin


x
f (x) = |x + 1| − |x − 1| if 0 ≤ x < 1,



 e−x sin(πx)
if |x| ≥ 1.
(i) Determine all points at which the function f is continuous.
(ii) Determine if the limits lim f (x) and lim f (x) exist (limits may be finite or infinite).
x→+∞
x→−∞
Solution:
The function f is continuous on R \ {0, 1} and discontinuous at 0 and 1.
lim f (x) = 0. There is no limit (finite or infinite) of f (x) as x → −∞.
x→+∞
The function g1 (x) = 2π/x is continuous on R \ {0} while the function g2 (x) = sin x is continuous
on R. It follows that the composition g(x) = g2 (g1 (x)) = sin(2π/x) is continuous on R \ {0}. Note
that the left-hand limit of g at 0 does not exists. Indeed, let xn = −4/(4n + 1) and yn = −4/(4n − 1)
for all n ∈ N. Then {xn } and {yn } are two sequences of negative numbers converging to 0. We have
2
g(xn ) = sin(2π/xn ) = sin(−π/2 − 2πn) = −1 and g(yn ) = sin(2π/yn ) = sin(π/2 − 2πn) = 1 for all
n ∈ N. It follows that there is no limit of g(x) as x → 0−.
Further, the functions h1 (x) = x + 1, h2 (x) = x − 1, and h3 (x) = |x| are continuous on R. Hence
the composition functions h4 (x) = h3 (h1 (x)) = |x + 1| and h5 (x) = h3 (h2 (x)) = |x − 1| are also
continuous on R as well as the sum h(x) = h4 (x) + h5 (x) = |x + 1| + |x − 1|.
Finally, the functions j1 (x) = −x, j2 (x) = ex , j3 (x) = πx, and j4 (x) = sin x are continuous on R.
Therefore the compositions j5 (x) = j2 (j1 (x)) = e−x and j6 (x) = j4 (j3 (x)) = sin(πx) are continuous
on R as well as the product j(x) = j5 (x)j6 (x) = e−x sin(πx).
By definition, the function f coincides with g on (−1, 0), with h on [0, 1), and with j on the union
(−∞, 1] ∪ [1, ∞). Hence it follows from the above that f is continuous on R \ {−1, 0, 1}. Moreover, f
has no left-hand limit at 0 since g does not have such a limit. Therefore f is discontinuous at 0. It
remains to determine whether f is continuous at 1 and −1. We have
lim f (x) = lim j(x) = j(1) = 0,
x→1+
lim f (x) = lim g(x) = g(−1) = 0,
x→−1+
lim f (x) = lim h(x) = h(1) = 2,
x→1+
x→−1+
x→1−
x→1−
lim f (x) = lim j(x) = j(−1) = 0,
x→−1−
x→−1−
f (1) = 0,
f (−1) = 0.
Thus f is continuous at −1 and has a jump discontinuity at 1.
Since f (x) = j(x) for |x| ≥ 1, the behaviour of the function f at infinity is the same as that
of j(x) = e−x sin(πx). We know that | sin(πx)| ≤ 1 and e−x → 0 as x → +∞. It follows that
j(x) → 0 as x → +∞. To show that lim j(x) does not exist, consider two sequences Xn = −n
x→−∞
and Yn = 1/2 − 2n, n ∈ N. Both sequences diverge to −∞. However j(Xn ) = 0 for all n while
j(Yn ) = e−Yn sin(πYn ) = e−1/2+2n → +∞ as n → ∞.
Bonus Problem 5 (15 pts.) Suppose E is an uncountable subset of R. Prove that for
any M > 0 one can find distinct elements x1 , x2 , . . . , xn ∈ E such that |x1 + x2 + · · · + xn | > M.
For any ε > 0 let Aε = E ∩ (ε, +∞) and Bε = E ∩ (−∞, −ε). If x1 , x2 , . . . , xn are distinct elements
of Aε , then x1 + x2 + · · · + xn > nε. In the case Aε is an infinite set, we can find arbitrarily many
such elements so that nε can be made larger than any prescribed M > 0. Similarly, if the set Bε is
infinite, we can find distinct elements x1 , x2 , . . . , xn ∈ Bε such that |x1 + x2 + · · · + xn | > nε > M .
It remains to show that for some ε > 0 at least one of the sets Aε and Bε is infinite. Indeed, the
set E can be represented as the union of countably many sets:
E = (E ∩ {0}) ∪ A1 ∪ B1 ∪ A1/2 ∪ B1/2 ∪ · · · ∪ A1/n ∪ B1/n ∪ . . .
Since the countable union of finite sets is at most countable, it follows that for some n ∈ N one of the
sets A1/n and B1/n is infinite.
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