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.