# So the sequence is not bounded as we can find terms

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, so the sequence is not bounded, as we can find terms of the sequence larger than any constant M (by taking k N large enough so that 2 k > M ). But on the other hand, x 2 k +1 = 0, for any k N , so the subsequence { x 2 k +1 } k trivially converges to 0. Observation: Of course many other examples like this could be constructed and you should come up with your own.

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3. (20 pts) Let f be a real function. Write the definition for each of the following: lim x a - f ( x ) = L, where a, L R . Solution: > 0 , δ > 0 such that x, ( a - δ < x < a → | f ( x ) - L | < ) . lim x + f ( x ) = -∞ . Solution: m R , M R such that x, ( x > M f ( x ) < m ) . 4. (20 pts) (a) (5 pts) Prove that if x is an upper bound for a set E R and x E , then x is the supremum of E . Solution: It is given that x is an upper bound for E . Suppose M is another upper bound for E . Since x E , it follows that x M . Thus x is the smallest upper bound of E , thus, it is the supremum of E . (b) (5 pts) State without proof an analogous statement for the infimum of E . Solution: If x is a lower bound for a set E R and x E , then x is the infimum of the set E . (c) (10 pts) Find the supremum and the infimum of the set E = { 1 + ( - 1) n (1 + 1 n ) | n N } . Solution: The set E can be alternatively described as E = {- 1 2 k - 1 , 2 + 1 2 k | k N } . The following inequalities are obvious - 1 ≤ - 1 2 k - 1 < 2 + 1 2 k 5 2 , k N . Thus the set E is bounded from below by - 1 and bounded from above by 5 / 2. But - 1 and 5 / 2 are elements of the set E . Thus from parts (a) and (b), it follows that sup E = 5 / 2, inf E = - 1.
5. (20 pts) Suppose 2 x 1 < 3 and x n +1 = 2 + x n - 2 for n N . Study the monotonicity and the convergence of the sequence { x n } n . Completely justify all your claims. Solution: First we prove by induction that 2 x n < 3 for any n N . For n = 1 this is given by hypothesis. Assume that 2 x n < 3 for a given n . Then 0 x n - 2 < 1, thus 0 x n - 2 < 1, so 2 x n +1 = 2 + x n - 2 < 3. Thus 2 x n < 3 for any n N , so we proved that the sequence is bounded.

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