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.
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