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.

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.

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

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)

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:

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/ n

2)

= (1 – 4/ ∞)/ ( 1 – 3/∞ -5/ ∞)

=1/1

=1 hence prove

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/ N

2+ 1/ N

2 = 2/N

2 =2(ε/2) = ε

so the sequence is cauchy

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 such that

m, n > N ⇒ | a m – a n | < ε

• Let ε = 1, and m = N, then : n ≥ N ⇒ | a N – a n | < 1

So | an | =

≤

≤

| a n - a N + a N |

| a n - a N | + | a N

|

1+ | a N |

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.

Therefore,

A sequence { a n } is a Cauchy sequence ⇒ A sequence { a n } is a bounded sequence.

Q.E.D.

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.

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

left hand limit = right hand limit = 2 so limit exist but question is wrong

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

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)

And x , x+π ∈ [0, 2π] for which sinx =sin (π+x)

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:

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 .

Also, and , since they do not depend on h. So, we have

, which is what we wanted to show.