## Description

Devansh Jain

IIT Bombay

Questions to be Discussed

Sheet 1

7 – Proof using definition (Optional, Discussed last time)

10 – Proof of Even-odd convergence (Optional, Talked about last time)

13 (ii) – Check Continuity of a function

15 – Check differentiability of a function

18 – Evaluate derivative of Multiplicative Cauchy functional equation Sheet 2

3 – Intermediate Value Property (IVP) and Rolleโs Theorem

5 – Mean Value Theorem (MVT)

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

10. To show:

{an}nโฅ1 is convergent โโ {a2n}nโฅ1 and {a2n+1}nโฅ1 converge to the same limit. Proof. ( =โ ) Let bn := a2n and cn := a2n+1. We are given that lim an = L. We nโโ

must show that lim bn = lim cn.

nโโ nโโ

Let > 0 be given. By hypothesis, there exists n0 โ N such that for n โฅ n0.

Note that 2n > n and 2n + 1 > n for all n โ N. Thus, we have that and for all n โฅ n0.

Thus, lim bn = lim cn = L. nโโ nโโ

( โ= ) Let (bn) and (cn) be as defined before. We are given that lim bn = lim cn = L. We must show that (an) converges.

nโโ nโโ

Let > 0 be given. By hypothesis, there exists n1, n2 โ N such that

for all n โฅ n1 (1) and for all n โฅ n2. (2)

Choose n0 = max{2n1, 2n2 + 1}.

Let n โฅ n0 be even. Then, n โฅ 2n1 or n/2 โฅ n1 and an = bn/2. By (1), we have it that

Similarly, let n โฅ n0 be odd. Then, n โฅ 2n2 + 1 or (n โ 1)/2 โฅ n2 and an = c(nโ1)/2.

By (2), we have it that

Thus, we have shown that whenever n โฅ n0. This is precisely what it

means for (an) to converge to L.

13. (ii) The function is continuous everywhere.

Proof. For x 6= 0, it simply follows from the fact that product and composition of continuous functions is continuous.

To show continuity at x = 0 :

Let (xn) be any sequence of real numbers such that xn โ 0. We must show that f (xn) โ 0.

Let > 0 be given.

Observe that .

Now, we shall use the fact xn โ 0. By this hypothesis, there must exist n1 โ N such that .

Choosing n0 = n1, we have it that .

15. For x 6= 0, it simply follows from the fact that product and composition of differentiable functions is differentiable.

To show differentiable at x = 0 and evaluating f 0(0)

f (0 + h) โ f (0) h2 sin(1/h)

f 0(0) = lim = lim = lim h sin(1/h) hโ0 h hโ0 h hโ0

|f 0(0)| โค lim |h| = 0 =โ f 0(0) = 0 hโ0

We can compute f 0(x) = 2x sin(1/x) โ cos(1/x) for x = 06 and f 0(0) = 0. f 0 is not continuous as limit at x = 0 is not defined (Why?).

18. Given: f (x + y) = f (x)f (y) for all x,y โ R. (1)

Let x = y = 0. This gives us that f (0) = (f (0))2 . Thus, f (0) = 0 or f (0) = 1.

Case 1. f (0) = 0.

Substitute y = 0 in (1). Thus, f (x) = f (0)f (x) = 0.

Therefore, f is identically 0 which means itโs differentiable everywhere with derivative 0.

Verify that f 0(c) = f 0(0)f (c) does hold for all x โ R. (We did not need to use the fact that f is differentiable at 0, it followed from definition.)

Case 2. f (0) = 1.

As f is differentiable at 0, we know that:

lim f (0 + h) โ f (0) = f 0(0) =โ lim f (h) โ 1 = f 0(0). (2)

hโ0 h hโ0 h

Now, let us show that f is differentiable everywhere.

Let c โ R. We must show that the following limit exists: f (c + h) โ f (c) lim

hโ0 h

Using (1), we can write the above expression as:

f (c)f (h) โ f (c)f (c)(f (h) โ 1)f (h) โ 1 lim .

hโ0 hhh

By (2), we know that the above limit exists. Thus, we have it that f is differentiable at c for every c โ R. Moreover, f 0(c) = f 0(0)f (c).

(Optional) We have gotten that the derivative of f is a scalar multiple of f . Use this to conclude.

Tutorial Sheet 2

3. Part 1. We will first show the existence of such an x0 โ (a,b).

Proof. I := [a,b] is an interval and f is continuous. Thus, f has the intermediate value property on I. Thus, the range J := f (I) must be an interval. As f (a) and f (b) are of different signs, 0 lies between them. As f (a),f (b) โ J and J is an interval, we

have it that 0 โ J = f (I). Thus, 0 = f (x0) for some x0 โ I = (a,b).

(i) f is continuous on [x0,x1], (ii) f is differentiable on (x0,x1), and (iii) f (x0) = f (x1).

Thus, by Rolleโs Theorem, there exists x2 โ (x0,x1) such that f 0(x2) = 0. But this

contradicts the hypothesis that f 0(x) 6= 0 for all x โ (a,b).

Tutorial Sheet 2

5. To prove that |sina โ sinb| โค |a โ b| for all a,b โ R. Case 1. a = b. Trivial.

Case 2. a 6= b. Without loss of generality, we can assume that a < b.

As f (x) := sin(x) is continuous and differentiable on R, there exists c โ (a,b) such

0(c) = f (b) โ f (a). (By MVT)

that f

b โ a

Also, we know that |f 0(c)| = |cosc| โค 1.

Thus, we have it that .

b โ a

This is equivalent to what we wanted to prove.

References

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