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PROBLEM 1
1.a. A sequence ($ x_n $) is said to be a Cauchy sequence if
-Choice 2 by 3.5.1 Definition
1.b. The statement of the Bolzano-Weierstrass theorem is:
-Choice 3 by 3.4.8 Theorem
1.c. Let $ f: A \mapsto \Re $. Suppose that $ (a,\infty) \subset A $ for some $ a \in \Re $. We say the limit of f as $ x \rightarrow \infty $ and write $ \lim_{x\to\infty}f = L $
-Choice 5 by 4.3.10 Definition
1.d. Let $ A \subset \Re $, let $ f: A \mapsto \Re $, and let $ c \in A $. We say that f is continuous at c if
-Choice 4 by 5.5.1 Definition
PROBLEM 2
Let $ x_1 := 8 $ and $ x_{n+1} := \frac{1}{2}x_n + 2 $ for $ n \in N $
a) Use induction to show that($ x_n $) is bounded below by 4.
-Base case $ x_1 = 8 $, so $ x_1 > 4 $ -Assume $ x_n > 4 $; is $ x_{n+1} > 4 $? $ x_n > 4 \Rightarrow \frac{1}{2}x_n > 2 $ $ x_n $ is greater than 4, so half of $ x_n $ is greater than 2 $ \frac{1}{2}x_n > 2 \Rightarrow \frac{1}{2}x_n + 2> 2 + 2 \Rightarrow \frac{1}{2}x_n + 2 > 4 $ Half of $ x_n $ is greater than 2, so adding 2 to both sides, the left hand side is greater than 4. $ \frac{1}{2}x_n + 2 = x_{n+1} $, therefore $ x_{n+1} $ > 4, so the series ($ x_n $) is bounded below by 4
b) Show that sequence ($ x_n $) is monotone
-The sequence is bounded below by 4, from above
$ \forall x > 4 \in \Re, (x - \frac{1}{2}x) > 2 $
-For all real x greater than 4, the distance, or length, between x and half of x is greater than 2 units
$ \forall x > 4 \in \Re, x > \frac{1}{2}x + 2 $
-Then for all real x greater than 4, adding 2 units to half of x is always less than the original x
