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Material Type: Assignment; Class: FIRST-YEAR INTEREST GROUP SMNR; Subject: Nursing; University: University of Texas - Austin; Term: Fall 2006;
Typology: Assignments
1 / 3
M365C (Fall 2006) Due: Friday September 8, 2006 1
Instructions: Solve all the following problems. Your solutions will be graded for accuracy, thoroughness, and clarity.
2
n→∞
2
n→∞
2
M365C (Fall 2006) Due: Friday September 8, 2006 2
Solutions
n 2
. Indeed, observe that
n 2
n 2
n 2
n 2
√ n 2
2 n −
n 2
n −
n 2
n +
n 2
n −
n 2
n +
n 2
n +
n 2
n +
n 2
n 2 − n 2 − n
(n +
n 2
n
n 2
n 2
n
2 n 2
n 2
2 n + 2
n 2
2 n
4 n
Hence, for any > 0, choose any N >
. Then for any n > N , we have n > N >
, which implies the following:
4 n
n 2
This proves that limn→∞
n 2
indeed exists and is equal to
|xn − L| < = 1. Hence, L − 1 < xn < L + 1, for every n > N. Hence,
min{L − 1 , x 1 ,... , xN } < xn < max{L + 1, x 1 ,... , xN }, for every n ∈ N.
Hence {xn}n∈N is bounded.
bound, we have s ≤ s ′ and s ′ ≤ s. Hence, s = s ′ , which proves uniqueness of least upper bound (assuming it
exists).
b) Let X ⊂ R. We say that l ∈ R is a lower bound of X if l ≤ x for each x ∈ X. We say g ∈ R is a greatest
lower bound or an infimum of the set X, if g is a lower bound of X and l ≤ g for every lower bound l of X.
The infimum is unique, if it exists, because: Let g and g
′ be two infimums of the set X ⊂ R. Then, we must
have g ≤ g
′ and g
′ ≤ g; hence g = g
′ .
c) Since ∅ 6 = A ⊂ B, inf B is a lower bound of A and sup B is an upper bound of A. Hence, by definition of
infima and suprema, we immediately have the desired inequalities.