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

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.