Real Analysis Fall 2004 Take Home Test 1 SOLUTIONS 1. Use the ...

Real Analysis
Fall 2004
Take Home Test 1
SOLUTIONS
1. Use the definition of a limit to show that
(a) lim
n→∞
sin n
n = 0
(b)
Proof. Let ε > 0 be given. Define N > 1 , where N is a positive integer. Then for
∣ ε n > N,
sin n ∣∣∣
∣ n
− 0 ≤ 1 n < 1 sin n
< ε. Hence lim
N n→∞ n = 0 .
1
lim
n→3 + x − 3 = ∞
Proof. Let A > 0 be given. Define δ = 1 1
. Then for x−3 < δ, f(x) =
A x − 3 > 1 δ = A.
1
Hence lim
x→3 + x − 3 = ∞.
2. (a) Let {x n } be a bounded sequence of real numbers and {x kn } be a monotone subsequence.
Prove that {x kn } converges to a limit.
Proof. By Bolzano-Weierstrass Theorem, the bounded subsequence {y n } = {x kn }
has a convergent subsequence {y ln } which converges to x 0 . Now either {x kn } is
nondecreasing or nonincreasing. Without loss of generality, assume that {y n } is
monotonically nondecreasing. Given ε > 0, there exists N such that n > N implies
x 0 − ε < y ln < x 0 + ε. In particular, x 0 − ε < y lN+1 < x 0 + ε. For n > l N+1 we have
x 0 − ε < y lN+1 < y n < x 0 + ε. Thus lim y n = lim x kn = x 0 .
n→∞ n→∞
1
(b) If lim x n = L = ∅, prove that lim = 1
n→∞ n→∞ x n L
Proof. Let ε 1 = |L|
2 . Then there exists N 1 > 0 such that for n > N 1 , |x n − L| < |L|
2 .
Consequently, for n > N 1 ,
|x n | = |x n − L + L| ≥ |L| − |x n − L| > |L| − |L|
2 = |L|
2 .

For any given ε > 0 there exists N 2 > 0 such that for n > N 2 , |x n − L| < 2ε
L . 2
Let N = max{N 1 , N 2 }. Then for n > N,
1
Thus lim
n→∞ x n
= L.
1
∣ − 1 x n L∣ = |x n − L|
|x n ||L|
≤ |x n − L|
L 2 / 2
< 2ε/L2
L 2 / 2
= ε.
1
3. Let x n+1 = , x 1 > 0. Prove that the sequence {x n } converges and then compute
3 + x n
the limit of the sequence.
1
< 1 3 + x n 3
= 3
10 for all n ≥ 1. Hence 3 10 < x n < 1 3
First Proof (My Preference). Since x 1 > 0, x n+1 =
x n+1 = 1
3 + x n
< 1
3 + 1 3
1
(b) If lim x n = L exists, then L = lim x n+1 = lim =
n→∞ n→∞ n→∞ 3 + x n
Thus
L = −3 + √ 13
.
2
(a) If x 1 > L, then x 2 =
1
3 + L
for all n ≥ 1. Moreover,
for all n > 1.
1
lim (3 + x n) = 1
3 + L
n→∞
= L. (1)
L 2 + 3L − 1 = 0. (2)
1
< 1
3 + x 1 3 + L = L from (1). More generally, if x 2n−1 > L,
1
then x 2n = < 1
3 + x 2n−1 3 + L = L. On the other hand, if x 1 < L, then x 2n−1 < L
and x 2n > L. Without loss of generality, assume x 1 > L. Define the sequence {s n }
as s n = |x n − L|. Then claim that s n+1 < s n for all n ≥ 1. If not, then s n+1
≥ 1.
s n
Assume n = 2k. Then
s 2k+1
s 2k
= x 2k+1 − L
L − x 2k
=
1
3+x 2k
− L
≥ 1
L − x 2k
1 − 3L − Lx 2k ≥ 3L − 3x 2k + Lx 2k − x 2 2k
(x 2 2k − 2Lx 2k + L 2 ) − L 2 − 6L + 3x 2k + 1 ≥ 0
(x 2k − L) 2 − 1 − 3L + 3x 2k + 1 ≥ 0 (using (2))
(x 2k − L)[x 2k − L + 3] ≥ 0.

Since x 2k − L < 0, x 2k − L + 3 ≤ 0 or x 2k ≤ L − 3 < 0, a contradiction.
A similar conclusion is obtained for n = 2k − 1.
Since s n is bounded and monotonically decreasing, lim
s n = α. Thus lim s 2n = α =
n→∞ n→∞
lim s 2n+1. For the case x 1 > L, x 2n+1 > L and x 2n < L and consequently
n→∞
(i) s 2n = L − x 2n → α ⇒ x 2n → L − α as n → ∞,
(ii) s 2n+1 = x 2n+1 − L → α ⇒ x 2n+1 → L + α as n → ∞.
Now
lim x 1
2n+1 = lim
n→∞ n→∞
1
3 + L − α
⇒ L + α =
3 + x 2n
L 2 + 3L − 1 + 3α + αL − α 2 − αL = 0
α(3 − α) = 0 ⇒ α = 0 or α = 3.
If α = 3, L − α < 0 and L + α > 1 3 , a contradiction.
lim x n = L.
n→∞
Hence α = 0 and thus
Second Proof. Let f(x) =
1
(3 + x) 2 ≤ 1 3
for all x > 0. Now1
1
3 + x . Then f ′ (x) =
−1
(3 + x) 2 . Since x > 0, |f ′ (x)| =
|x 2 − L|
|x 1 − L| = |f(x 1) − L|
|x 1 − L|
= |f ′ (ξ 1 )| ≤ 1 3
by the Mean Value Theorem, where ξ 1 is between x 1 and L. |x 2 −L| ≤ 1 3 |x 1−L|. Similarly,
|x 3 − L|
|x 2 − L| = |f(x 2) − L|
= |f ′ (ξ 2 )| ≤ 1 |x 2 − L|
3 ,
ξ 2 between x 2 and L. |x 3 − L| ≤ 1 3 |x 2 − L| ≤
( 1
3
) 2
|x 1 − L|. By induction, |x n − L| ≤
( 1
3
) n−1
|x 1 − L| → 0 as n → ∞. Thus lim
n→∞
x n = L.
4. Prove that the function f(x) = 1 is continuous for all real numbers x ≠ 0.
x2 1 L is defined as the fixed point of f, i.e., f(L) = L = 1
3 + L ⇒ L2 + 3L − 1 = 0 ⇒ L = −3 + √ 13
2

Method 1. Let ε > 0 be given. Let x 0 ∈ R\{0}. Put δ = |x 0|
2 . Then if |x − x 0| < δ = |x 0|
2 ,
then |x 0|
2 < x < 3|x 0|
2 . Let δ 2 = ε
10 |x 0|x 2 0 . For δ = min(δ 1, δ 2 ), if |x − x 0 | < δ, then
1
∣x − 1 2 x 2 ∣ = |x2 0 − x2 |
0 x 2 0x 2
= |x − x 0||x + x 0 |
x 2 0x 2
< |x − x 0| 5 |x 2 0|
x 2 x 2 0
0
4
≤ |x − x 0| · 10
|x 0 |x 2 0
= ε
Method 2. Consider the function g(x) = x 2 . Given ε > 0, let δ 1 = |x 0|
2 . Then |x 0|
2 < x <
3|x 0 |
for all |x−x 0 | < δ 1 . Let δ 2 =
2ε
2
5|x 0 | . Then for δ = min(δ 1, δ 2 ), and x ∈ (x 0 −δ, x 0 +δ),
we have
|x 2 − x 2 0 | = |x − x 0||x + x 0 |
< |x − x 0 | 5|x 0|
2
≤ ε.
Then use a theorem about the reciprocal of a continuous function is continuous.
5. Suppose that f is continuous on [a, b], one-to-one (if x 1 ≠ x 2 , then f(x 1 ) ≠ f(x 2 )), and
f(a) < f(b). Prove that f is monotonically increasing (strictly) on [a, b].
Proof. Suppose that f is not monotonically increasing on [a, b].
b
a
a
b

Then there exists x 1 , x 2 ∈ [a, b] with x 1 < x 2 but f(x 1 ) > f(x 2 ). We have two cases to
consider:
Case (i): If f(a) < f(x 1 ), then f(a) lies between f(x 1 ) and f(x 2 ). Hence by the Intermediate
Value Theorem, there exists c between x 1 and x 2 with f(c) = f(a), a
contradiction to the assumption that f is one-to-one.
Case (ii): If f(a) > f(x 1 ), then since f(a) < f(b), the value f(a) lies between f(x 1 ) and
f(b). Again, by the Intermediate Value Theorem, there exists d between x 1 and b
with f(d) = f(a), a contradiction. Hence f is increasing.
6. Suppose that S 1 , S 2 , . . . , S n are sets in R 1 and that S = S 1 ∩ S 2 ∩ · · · ∩ S n , S ≠ ∅. Let
B i = sup S i , b i = inf S i , 1 ≤ i ≤ n. Find a formula relating sup S and inf S in terms of
the {b i } and {B i }.
Proof. S = ∩ n i=1 S i. Claim that sup S = min{B i , 1 ≤ i ≤ n}. Let B r = min{B i : 1 ≤ i ≤
n}. If x ∈ S, then x ∈ S r , and hence x ≤ B r . Thus sup S ≤ B r . On the other hand,
sup S ≥ x ∈ S r and by the definition of B r , sup S ≥ B r . Hence sup S = B r .
One may show in a similar fashion that inf S = max{b i |1 ≤ i ≤ n}.
7. Suppose that lim f(x) = ∞ and lim g(x) = a. Suppose that for some positive number
x→a x→−∞
M, we have g(x) ≠ a for x < −M. Prove that lim f(g(x)) = ∞.
x→−∞
Proof. Since lim f(x) = ∞, for any A > 0, there exists δ > 0 such that |x − a| < δ ⇒
x→a
|f(x)| > A. For this δ there exists ˜M > 0 such that x < −˜M ⇒ |g(x) − a| < δ. Let
K = max{˜M, M}. Then for x < −K, |g(x) − a| < δ and thus |f(g(x))| > A. Hence
lim
x→−∞
f(g(x)) = ∞.
8. If f(x) is continuous on [a, b], if a < c < d < b, and M = f(c) + f(d), prove there exists a
number ξ between a and b such that M = 2f(ξ).
Proof. Let M = f(c)+f(d). If f(c) ≤ f(d), then M 2
and M f(c) + f(d) f(c) + f(c)
= ≥
2 2
2
=
f(c) + f(d)
2
≤
f(d) + f(d)
2
= f(d),
= f(c). By the Intermediate Value Theorem (since
f(c) ≤ M ≤ f(d)), there exists ξ between c and d (hence between a and b) such that
2
f(ξ) = M . Thus M = 2f(ξ).
2

9. Suppose that f(x) and g(x) are functions defined for x > 0, lim g(x) exists and is finite,
x→0 +
and |f(b) − f(a)| ≤ |g(b) − g(a)| for all positive real number a and b. Prove that lim f(x)
x→0 +
exists and is finite.
Proof. Suppose that lim g(x) = L and |f(b) − f(a)| ≤ |g(b) − g(a)|. If lim f(x) does
x→0 + x→0 +
not exist, there exists ε > 0 and two sequences x n > 0, z n > 0 such that lim x n = 0,
n→∞
lim z n = 0 and for all n |f(x n ) − f(z n )| ≥ ε. For this ε > 0 there exists δ > 0 such that
n→∞
0 < x < δ ⇒ |g(x) − L| < ε 2 . Now for 0 < ̂x 1, ̂x 2 < δ,
|g( ̂x 1 ) − g( ̂x 2 )| = |g( ̂x 1 ) − L − g( ̂x 2 ) + L|
≤ |g( ̂x 1 ) − L| + |g( ̂x 2 ) − L|
< ε 2 + ε 2
= ε.
For this δ there exists N such that n > N, 0 < x n , z n < δ. Hence for n > N |f(x n ) −
f(z n )| ≤ |g(x n ) − g(z n )| < ε, a contradiction. Thus lim f(x) exists.
x→0+
2 + 2 1 2 + 2 1 3 + · · · + 2 1 n
10. Evaluate lim
.
n→∞ n
(You need not give a proof but you should show some work or justification. Quote a
theorem or what have you. Calculator results or graphical analysis are not acceptable.)
Proof. Let s n = 2 + 2 1 2 + · · · + 2 1 n
. Then 1 < n2 1 n
n
n < s n < n · 2 = 2. Notice lim
n
2 1 n = 1.
n→∞
Thus s n is bounded. Moreover, we claim that s n+1 < s n . Suppose the contrary, that is
s n+1 ≥ s n . Then
Hence
or
(2 + 2 1 2 + · · · + 2 1 n + 2 1
n+1 )
n + 1
≥ (2 + 2 1 2 + · · · + 2 1 n )
.
n
n(2 + 2 1 2 + · · · + 2
1
n ) + n2
1
n+1 ≥ n(2 + 2
1
2 + · · · + 2
1
n ) + 2 + 2
1
2 + · · · + 2
1
n
which is a contradiction.
n2 1
n+1 ≥ 2 + 2
1
2 + · · · + 2
1
n ≥ n2
1
n , n > 1
Hence {s n } is a monotone bounded sequence and thus is converges by the Bolzano-
Weierstrass Theorem. Moveover, lim
n→∞
s n = 1.