## Description

Devansh Jain

IIT Bombay

Questions to be Discussed

Sheet 1

2 (iv) – Sandwich Theorem for finding limits

3 (ii) – Checking convergence of a sequence

5 (iii) – Monotonic Bounded sequences are convergent

7 – Proof using definition

9 – Relation between product and convergence

11 – Conditions for exchanging product and limits (Discussed in lecture)

2. (iv) lim (n)1/n.

nโโ

Define hn := n1/n โ 1.

Then, hn โฅ 0 โn โ N. (Why?)

Observe the following for n > 2 :

n = (1 + hn)n > 1 + nhn .

Thus, hn .

Using Sandwich Theorem, we get that lim hn = 0 which gives us that lim n1/n = 1.

nโโ nโโ

3. (ii) To show: is not convergent.

nโฅ1

We will use the following two results: (a) Sum of convergent sequences is convergent.

(b) The sequence {(โ1)n}nโฅ1 is not convergent. We now proceed as follows:

an := (โ1)n .

(โ1)n

It is easy to show that bn := is convergent. (Its absolute value will behave the n

same way as 1/n.)

Now, for the sake of contradiction, let us assume that (an) converges. Then, by (a),

(โ1)n

we have it that cn := an + bn = must be convergent.

2

However, (cn) converging is equivalent to {(โ1)n}nโฅ1 converging. (Why?) However, by (b), we know that the above is false. Thus, we have arrived at a contradiction.

an

5. (iii) a1 = 2, an+1 = 3 + โn โฅ 1.

2

Claim 1. an < 6 n โ N.

Proof. We shall prove this via induction. The base case n = 1 is immediate as 2 < 6. Assume that it holds for n = k. an 6

By principle of mathematical induction, we have proven the claim.

Claim 2. an < an+1 โn โ N.

a

Proof. an+1 โ an = 3 โ n = 6 โ an > 0 =โ an+1 > an.

2 2

ak+1 = 3 + .

2

Thus, (an) is a monotonically increasing sequence that is bounded above. Therefore, it must converge. Use lim an+1 = lim an = L and then solve for L.

nโโ nโโ

7. If lim an = L 6= 0, show that there exists n0 โ N such that nโโ

|L|

|an| โฅ for all n โฅ n0.

2

|L|

Let us choose . Why is this a valid choice of ?)

By hypothesis, there exists n0 โ N such that whenever n โฅ n0.

n0

9. (i) {anbn}nโฅ1 is convergent, if {an}nโฅ1 is convergent.

(ii) {anbn}nโฅ1 is convergent, if {an}nโฅ1 is convergent and {bn}nโฅ1 is bounded.

Both are false.

The sequences, an := 1 โn โ N and bn := (โ1)n โn โ N act as a counterexample for both the statements.

11. (i) We shall show that the statement is false with the help of a counterexample.

Let a = โ1, b = 1, c = 0. Define f and g as follows:

1/x ;x 6= 0 f (x) = x and g(x) = . 0 ;x = 0

It can be seen that lim f (x) = 0 but lim[f (x)g(x)] = lim 1 = 1. xโ0 xโc xโ0

(ii) We shall prove that the given statement is true.

We are given that g is bounded. Thus, โM โ R+ such that |g(x)| โค M โx โ (a, b).

Let > 0 be given. We want to show that there exists ฮด > 0 such that whenever 0 < |x โ c| < ฮด.

Let . As lim f (x) = 0, there exists ฮด > 0 such that xโc

.

Thus, whenever 0 < |x โ c| < ฮด, we have it that

(iii) We shall prove that the given statement is true.

Let > 0 be given. Let l := lim g(x).

xโc

Let .

By hypothesis, there exists ฮด1 > 0 such that 0

Also, there exists ฮด2 > 0 such that 0 .

Let ฮด = min{ฮด1, ฮด2}. Then, whenever 0 < |x โ c| < ฮด, we have that:

Thus, we have it that 0

References

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