Calculate the following quantities:(a) lim infn→∞ (−1)n3 + 1n

im supn→∞(−1)n3 + 1n

. Prove, using the ϵ − N definition oflimit, that the sequence (an) given byan = n2n − 1converges to 12 .2. Prove using the definition of limit thatthe sequence (an) given byan = 1(3n − 1)converges to 0.3. Evaluate the following limits(a) limn→∞n4 + 3n1 + n3 + n5 .(b) limn→∞3n − 13n − 3n−1(c) limn→∞√n + 1 − √n(d) limn→∞ nr4 − 1n − 2!4. Let (an) be a sequence. Show that iflimn→∞ an exists, then (an) is bounded.5. Show that the sequence (an) given byan = n2n + 1does not converge.6. Use the sandwich theorem to findlimn→∞1(3n − 1)7. (a) Let n ∈ N. Define functions f, gbyf (x) = (1 + x)nandg(x) = 1 + nxShow by induction that for all nat-ural numbers n ≥ 1 and real num-bers x ≥ −1f (x) ≥ g(x).(b) Sketch graphs f and g case for thecase n = 3.(c) Let a > 1. Show that the sequence(an) is unbounded above.(d) If |a| < 1, show that limn→∞ an = 08. Let x ∈ R. Define a sequence of partialsums by (sn) bys1 = 1s2 = 1 + xs3 = 1 + x + x2... = ...sn = 1 + x + x2 · · · + xn−1Show that (sn) converges if and only if|x| < 1. Find the limit of the sequence(sn) if x = 910 .9. Evaluate:(a) limn→∞1√n(b) limn→∞5n + (−1)n4n(c) limn→∞n34n(d) limn→∞√n√n − 2(e) limn→∞sin n + n3n3(f) limn→∞(−1)n√n1https://pennance.us1

For  n∈N, Let xn be defined as (1+1n)n+xn=e then, limn→∞ xn is equal to12limx→∞ tan3x−tanx3x52⋅limx→∞ tan3x−tanx3x5limx→∞ ddx√x∫√3 r3(r+1)(r−1)dr2⋅limx→∞ ddx√x∫√3 r3(r+1)(r−1)dr

Let (xn)n≥1 and (yn)n≥1 be bounded sequence. Prove thatlim infn→∞ (xn + yn) ≥ lim infn→∞ xn + lim infn→∞ yn. (0.1)NOTE) Let xn := (−1)n and yn := (−1)n+1 for n ≥ 1. Thenlim infn→∞ (xn + yn) = 0 > −2 = lim infn→∞ xn + lim infn→∞ yn.Hence, the equality in (0.1) does not hold in general.(3-4) Recall that the Fibonacci sequence (zn)n≥1 is defined byz1 = z2 = 1 and zn+2 = zn+1 + zn for all n ≥ 1.Define a sequence (xn)n≥1 byxn = znzn+1

Evaluate :  limn→∞ 1.2+2.3+3.4+...+n(n+1)n3