MATH 409–501
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.
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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
x
(1 − x)k ≤ 1 − kx +
2
for all x ∈ (0, 1). Multiplying both sides of this inequality by the positive number 1 − x, we get
k(k − 1) 2
(k + 1)k 2 k(k − 1) 3
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 + 2 − n, n ∈ N.
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(ii) yn = n2 sin , n ∈ N.
n
√
(iii) z1 = 7, zn = 3 + zn−1 − 3 for n ≥ 2.
Solution:
lim xn = 0, lim zn = 4. The sequence {yn } diverges to +∞.
n→∞
n→∞
For any n ∈ N we have
0 < xn =
√
n+2−
√
√
√ √
√
1
2
2
( n + 2 − n)( n + 2 + n)
√
=√
n=
√
√ < √ =√ ,
2 n
n
n+2+ n
n+2+ n
sin x
= 1. As a consequence,
x
n sin(1/n) = sin(1/n)/(1/n) → 1 as n → ∞. In particular, n sin(1/n) > 1/2 for all sufficiently large
n. Then yn = n2 sin(1/n) > n/2 for all sufficiently large n, which implies that yn → +∞ as n → ∞.
which implies that xn → 0 as n → ∞. Further, we know that lim
x→0
1
√
n−1
=
Finally, zn − 3 = zn−1 − 3 for all n ≥ 2. It follows by induction that zn − 3 = (z1 − 3)(1/2)
1−n
n−1
1/m
2
(1/2)
for all n ∈ N. Since 4
→ 1 as m → ∞, we obtain that zn − 3 → 1 as n → ∞.
=4
4
Then zn → 4 as n → ∞.
Problem 3 (25 pts.) Prove the following version of the Intermediate Value Theorem: if
f : [a, b] → R is a continuous function on a closed bounded interval [a, b] such that f (a) < 0
and f (b) > 0, then f (x) = 0 for some x ∈ (a, b).
Consider a set E = {x ∈ [a, b] | f (x) < 0}. The set E is nonempty since a ∈ E and bounded since
E ⊂ [a, b]. By the Completeness Axiom, the supremum c = sup E exists. Note that a ≤ c ≤ b because
a is an element of the set E, b is an upper bound of E, and c is the least upper bound of E. Therefore
the function f is continuous at c. We are going to show that f (c) = 0.
For any n ∈ N the number c − 1/n is not an upper bound of the set E. Hence there exists xn ∈ E
such that xn > c − 1/n. At the same time, xn ≤ c for all n ∈ N. The Squeeze Theorem implies that
the sequence {xn } converges to c. Since the function f is continuous at c, it follows that f (xn ) → f (c)
as n → ∞. Notice that f (xn ) < 0 for all n ∈ N since xn ∈ E. The Comparison Theorem for sequences
implies that f (c) ≤ 0. In particular, this means that c 6= b. Then (c, b) is a nonempty interval.
By definition of the set E, the function f is nonnegative everywhere on that interval: f (x) ≥ 0 for
x ∈ (c, b). The Comparison Theorem for functions implies that the right-hand limit of f at the point
c is also nonnegative. That limit is f (c) since f is continuous at c. Thus f (c) ≥ 0 as well as f (c) ≤ 0.
That is, f (c) = 0.
Problem 4 (20 pts.) Consider a function f

|x| + |x − 1|



2π
sin
f (x) =

x

 x
e sin(πx)
: R → R defined by
if − 1 < x ≤ 0,
if 0 < x < 1,
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 \ {−1, 0} and discontinuous at −1 and 0.
lim f (x) = 0. There is no limit (finite or infinite) of f (x) as x → +∞.
x→−∞
The functions g1 (x) = |x| and g2 (x) = x − 1 are continuous on R. Hence the composition function
g3 (x) = g1 (g2 (x)) = |x−1| is also continuous on R as well as the sum g(x) = g1 (x)+g3 (x) = |x|+|x−1|.
Further, the function h1 (x) = 2π/x is continuous on R \ {0} while the function h2 (x) = sin x
is continuous on R. It follows that the composition h(x) = h2 (h1 (x)) = sin(2π/x) is continuous on
R \ {0}. Note that the right-hand limit of h 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 positive numbers converging to
0. We have h(xn ) = sin(2π/xn ) = sin(π/2+2πn) = 1 and h(yn ) = sin(2π/yn ) = sin(−π/2+2πn) = −1
for all n ∈ N. It follows that there is no limit of h(x) as x → 0+.
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Finally, the functions j1 (x) = πx, j2 (x) = sin x, and j3 (x) = ex are continuous on R. Therefore the
composition j4 (x) = j2 (j1 (x)) = sin(πx) is continuous on R as well as the product j(x) = j3 (x)j4 (x) =
ex 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 right-hand limit at 0 since h 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+
x→1+
lim f (x) = lim g(x) = g(−1) = 3,
x→−1+
x→−1+
lim f (x) = lim h(x) = h(1) = 0,
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) = ex sin(πx). We know that | sin(πx)| ≤ 1 and ex → 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 and Yn = 1/2 + 2n,
x→+∞
n ∈ N. Both sequences diverge to +∞. However j(Xn ) = 0 for all n while j(Yn ) = eYn → +∞ 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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