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 Rolle’s Theorem
5 – Mean Value Theorem (MVT)
7. If lim an = L 6= 0, show that there exists n0 ∈ N such that n→∞
|L|
|an| ≥ for all n ≥ n0.
2
|L|
Let us choose . Why is this a valid choice of ?)
By hypothesis, there exists n0 ∈ N such that whenever n ≥ n0.

n0

10. To show:
{an}n≥1 is convergent ⇐⇒ {a2n}n≥1 and {a2n+1}n≥1 converge to the same limit. Proof. ( =⇒ ) Let bn := a2n and cn := a2n+1. We are given that lim an = L. We n→∞
must show that lim bn = lim cn.
n→∞ n→∞
Let > 0 be given. By hypothesis, there exists n0 ∈ N such that for n ≥ n0.
Note that 2n > n and 2n + 1 > n for all n ∈ N. Thus, we have that and for all n ≥ n0.
Thus, lim bn = lim cn = L. n→∞ n→∞
( ⇐= ) Let (bn) and (cn) be as defined before. We are given that lim bn = lim cn = L. We must show that (an) converges.
n→∞ n→∞
Let > 0 be given. By hypothesis, there exists n1, n2 ∈ N such that
for all n ≥ n1 (1) and for all n ≥ n2. (2)
Choose n0 = max{2n1, 2n2 + 1}.
Let n ≥ n0 be even. Then, n ≥ 2n1 or n/2 ≥ n1 and an = bn/2. By (1), we have it that

Similarly, let n ≥ n0 be odd. Then, n ≥ 2n2 + 1 or (n − 1)/2 ≥ n2 and an = c(n−1)/2.
By (2), we have it that
Thus, we have shown that whenever n ≥ 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 → 0. We must show that f (xn) → 0.
Let > 0 be given.
Observe that .
Now, we shall use the fact xn → 0. By this hypothesis, there must exist n1 ∈ 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) − f (0) h2 sin(1/h)
f 0(0) = lim = lim = lim h sin(1/h) h→0 h h→0 h h→0
|f 0(0)| ≤ lim |h| = 0 =⇒ f 0(0) = 0 h→0
We can compute f 0(x) = 2x sin(1/x) − 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 ∈ 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 it’s differentiable everywhere with derivative 0.
Verify that f 0(c) = f 0(0)f (c) does hold for all x ∈ 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) − f (0) = f 0(0) =⇒ lim f (h) − 1 = f 0(0). (2)
h→0 h h→0 h
Now, let us show that f is differentiable everywhere.
Let c ∈ R. We must show that the following limit exists: f (c + h) − f (c) lim
h→0 h
Using (1), we can write the above expression as:
f (c)f (h) − f (c)f (c)(f (h) − 1)f (h) − 1 lim .
h→0 hhh
By (2), we know that the above limit exists. Thus, we have it that f is differentiable at c for every c ∈ 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 ∈ (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) ∈ J and J is an interval, we
have it that 0 ∈ J = f (I). Thus, 0 = f (x0) for some x0 ∈ 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 Rolle’s Theorem, there exists x2 ∈ (x0,x1) such that f 0(x2) = 0. But this
contradicts the hypothesis that f 0(x) 6= 0 for all x ∈ (a,b).
Tutorial Sheet 2
5. To prove that |sina − sinb| ≤ |a − b| for all a,b ∈ 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 ∈ (a,b) such
0(c) = f (b) − f (a). (By MVT)
that f
b − a
Also, we know that |f 0(c)| = |cosc| ≤ 1.
Thus, we have it that .
b − a
This is equivalent to what we wanted to prove.
References