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- 1. Question1: Use induction to prove that 2n ≤n! . Determine for which n values this is true and use the appropriate starting point. Soln. 2n ≤n! For n=1. 2 ≤ 1 not true For n=2 4 ≤ 2.1 4 ≤ 2 not true For n=3 8 ≤ 3.2.1 8 ≤ 6 not true For n=4 16 ≤ 4.3.2.1 16 ≤ 24 true For n=5 25 ≤ 5.4.3.2.1 true So this is true for n ≥ 4. Here our first case to check is n = 4 is true. So we're good here. Now for the inductive step, we must assume that for n=k 2k ≤ k! …………….(1) is true.
- 2. And show that this assumption implies that: 2k+1 ≤ (k+1)!.........................................(2) Now (k+1)! Can be written as (k+1)!= (k+1)k! On using equation 1 (k+1)2k ≤ k!(k+1) 2.2k ≤ k!(k+1) 2k+1 ≤ k!(k+1) 2k+1 ≤ (k+1)! Hence proved that 2n ≤n! is true for n ≥ 4.
- 3. Question 2: If S is denumerable, then S is equinumerous with a proper subset of itself. Since S is denumerable, there exists a bijection f: N S. Considerthe set T = S/ {f(1)}. Then for fn g: N T given by g(n) = f (n + 1). It is injective since if g(m) = g(n), then f(m+1) = f(n+1) and since f is injective, m+1 = n+1. Thus m = n. Furthermore, it is surjective, ie x=g(m). Thus T is denumerable and equinumerous to S. Question 3: Let A and B be nonempty, bounded subsets of R. Prove that inf (AUB) = inf { inf A, inf B} Soluti on: Assume without loss of generality that inf(A)< or = inf (B) so that (inf(A), inf(B)) = inf(A).We need to prove that inf (A) is lower bound for AUB and if t is any lower bound for A U B then t < or = inf (A). A < Or = inf(A) for a belongs to A. Same with b. Since inf (B) > or = inf (A) that means b> or = inf (A). Since every
- 4. element c of A U B satisfies c belongs to A or B or both we seethat c > or = inf (A) so inf(A) is lower bound for A U B. Let T be any lower bound for A U B. Since t < or = c we also have t < or = c. In particular t is lower bound for A U B ie > or = any lower bound for A U B. ie inf (A U B) = inf (A)
- 5. Question 4: Prove that an accumulation point of a set S is either an interior point of S or a boundary point of S. Let x ε R be accumulation point of S, if x is not an interior point of S; so let ɛ > 0; since x is accumulation point of S, N (x; ɛ) intersection S is not = 0. Also x isnot an interior point of S so we cannot have N ( x; ɛ) subset S ie N ( x; ɛ) ∩ ( R / S is not eq to 0). This proves our assumption. Question 5: Solution:
- 6. Question 6. Use the definition of compact and NOT Heine –Borelto prove that if A and B are compact then AUB is compact. Let U = {uα}αεA be an open cover of AUB. Therefore U is also an open cover of A and B. Since A and B are compact, there exist finite subcovers {U1..,Un} C {Uα} αεA and { Un+1,…Un}C{Uα}αεA of A and B resp..ie A C U Uj and BC U Uj Then it follows that the finite collection { U1, U2,…Un} C {Uα}αεA is an open cover of A U B. Therefore A U B is compact. Question 7: Lim n→∞ (n2 – 4)/ ( n2 – 3n -5) Taking n2 common from numerator and denominator both Or Lim n→∞ (1 – 4/ n2)/ ( 1 – 3/n -5/ n2) = (1 – 4/ ∞)/ ( 1 – 3/∞ -5/ ∞)
- 7. =1/1 =1 hence prove
- 8. Question 8: 1) For the sequence to be Cauchy, for every ε>0, there exists a positive integer N such that n, m>N ⇒ |an - am| < ε. Suppose ε>0. Then choose N so that if k > N, 1/k2< ε/2. Then Notice that for any m, n > N , we have |an - am| = |1/m2 - 1/n2 | ≤ 1/m2 + 1/n2 < 1/ N2+ 1/ N2 = 2/N2 =2(ε/2) = ε so the sequence is cauchy
- 9. Question: 9 A sequence { a n } is a Cauchy sequence ⇒ A sequence { a n } is a bounded sequence Proof1: * We need to prove: | an | ≤ M for n ∈ N. • A sequence { a n } is a Cauchy sequence ⇔ ∀ε > 0, ∃ N m, n > N ⇒ • Let ε = 1, and m = N, then : |am –an|<ε ⇒ n ≥ N So | an | such that = |aN –an|<1 | an - a N + aN | ≤ | an - a N | + | aN Let M = Max { | a 1 |, | a 2 | , .. , | a N-1 |, 1+ | a N | } | hence | an | ≤ M Therefore, a sequence { a n } is a bounded by M. 1+ | a N | Therefore, A sequence { a n } is a Cauchy sequence ⇒ A sequence { a n } is a bounded sequence. Q.E.D.
- 10. Proof2: If { an} is a Cauchy sequence then with ɛ =1 there exist N such that if n,m> = N then lan – aml < 0. For no > N and for any n we have |an| <= B where B = max{ |a1|, |a2|…|ano| + 1} Case 1 : if n < no then |an| <= B Case 2: if n > no then n,no > N so |an – ano| < 1. |an| = | ano + (an – ano| < |ano| + |an – ano| < |ano| + 1 <= B So {an} is bounded.
- 11. Question 10: Now left hand limit calculation: X→4-h where h→0 so = Lim h →0( (4-h)2 + 4)/ ( (4-h)2 -6) = 20/10 =2 Now right hand limit calculation: = Lim h →0( (4+h)2 + 4)/ ( (4+h)2 -6) = 20/10 =2
- 12. left hand limit = right hand limit = 2 so limit exist but question is wrong
- 13. Question11: Suppose f is continuous on [a,b] and f(a) = f(b). Prove that there are c1 and c2 ∈(a,b), c1 ≠c2 f(c1)=f(c2). Solution: f(a) = f(b) implies that f is a) Either a constant function on [a, b] b) Or first increasing on [a, c] then decreasing [c, b] or vice versa a) Now If it is a constant and continuous function on [a, b] then For c1 , c2 ∈(a,b) c1 ≠c2 , f(c1)=f(c2) is true . b) First increasing on [a, c] then decreasing [c, b] Let consider the increasing continuous function on [a,c] c ∈(a, b) such that f ‘ (c)=0 f is a increasing contiuos function on [a; c] and that f(a) ≠ f(c) ……………...(1) so from Intermediate Value Theorem
- 14. For all α between f (a) and f (c) there exist c1 : a < c1< c …………….…..(2) for which f (c1) = α ………………………………………………………….…..(3) Similarly if f is a decreasing continuos function on [c, b] and f(c) ≠ f(b)=f(a) ….(4) so from Intermediate Value Theorem For all β between f (c) and f (b) there exist c2 : c < c2< b …………….….(5) for which f (c2) = β= …………………………………..……….………….(6) such that α=β Simplest example is Sinx which continuos [0, 2π] and Sin(0)= Sin (2π)
- 15. And x , x+π ∈ [0, 2π] for which sinx =sin (π+x)
- 16. Question 12: prove the product rule for derivative Product Rule If f and g are two differentiable functions, then Let us begin with the definition of a derivative. The key is to subtract and add a term: . You need to know to do this to make any progress. Doing this, we get the following: . From the property of limits, we can break the limit into two pieces because the limit of a sum is the sum of the limits. And so, we have:
- 17. Factoring a on the first limit and from the second limit we get: Another property of limits says that the limit of a product is a product of the limits. Using this fact, we can rewrite the limit as: By definition, though, and .
- 18. Also, and , since they do not depend on h. So, we have , which is what we wanted to show.

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