Consider the following proposed proof that if f: (0, 1) → R is uniformly continuous, then lim f(x) exists [special case of Proposition 3.4.6]. 2-0 1. Let {n} be a sequence in (0, 1) that converges to 0. In particular, {n} is a Cauchy sequence, so Lemma 3.4.5 implies that the image sequence {f(x)} is a Cauchy sequence, which by Theorem 2.4.5 converges to some limit in R. 2. If {n} is another sequence in (0, 1) that converges to 0, then the same reasoning shows that the sequence {f(yn)} potentially different from the limit of {f(x)}=1. converges to some limit, 3. Define a sequence {n}1 by setting 22k-1 equal to x and %2% equal to y for each natural number k. This sequence converges to 0, so the same reasoning as above implies that {f(zn)}1 converges. 4. By Proposition 2.1.17, all subsequences of a convergent sequence converge to the same limit. Since {f(x)}1 and {f(yn)}1 are subsequences of the convergent sequence {f(zn)}1, the sequence {f(yn)}a_1 converges to the same limit as does {f(x)}x=1. 5. Thus there exists a number L such that for every sequence {n} in (0, 1) converging to 0, the image sequence {f(an)}1 converges to L. So lim f(x) exists (and equals L) by Lemma 3.1.7. x-0 Which one of the following statements best describes this proposed proof? Step 4 is faulty. The proof is valid. Step 1 is faulty. Step 5 is faulty. Step 3 is faulty. Step 2 is faulty.

(5) Consider the sequence defined by -( + 2) for n21 S1 = 1 Sn+1 = (Sn 4 (a) Prove by induction for all n E N (b) Prove the sequence is monotone. (c) Use the previous two parts to explain whether or not the limit of s, exists. If the limit exists, find the limit.

Q2 pls

Consider the function f : R → R, and define the sequence {an} such that an = f(n). True or False? If lim n→∞ an = L, then lim x→∞ f(x) = L If true prove it, if false,provide a counterexample (i.e. a specific function for which the statement does not hold).