MA 109 : Calculus-I D1 T4, Tutorial 2 Solved

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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 Rolleโ€™s Theorem
5 – Mean Value Theorem (MVT)
7. If lim an = L 6= 0, show that there exists n0 โˆˆ N such that nโ†’โˆž
|L|
|an| โ‰ฅ for all n โ‰ฅ n0.
2
|L|
Let us choose . Why is this a valid choice of ?)
By hypothesis, there exists n0 โˆˆ N such that whenever n โ‰ฅ n0.

n0

10. To show:
{an}nโ‰ฅ1 is convergent โ‡โ‡’ {a2n}nโ‰ฅ1 and {a2n+1}nโ‰ฅ1 converge to the same limit. Proof. ( =โ‡’ ) Let bn := a2n and cn := a2n+1. We are given that lim an = L. We nโ†’โˆž
must show that lim bn = lim cn.
nโ†’โˆž nโ†’โˆž
Let > 0 be given. By hypothesis, there exists n0 โˆˆ N such that for n โ‰ฅ n0.
Note that 2n > n and 2n + 1 > n for all n โˆˆ N. Thus, we have that and for all n โ‰ฅ n0.
Thus, lim bn = lim cn = L. nโ†’โˆž nโ†’โˆž
( โ‡= ) Let (bn) and (cn) be as defined before. We are given that lim bn = lim cn = L. We must show that (an) converges.
nโ†’โˆž nโ†’โˆž
Let > 0 be given. By hypothesis, there exists n1, n2 โˆˆ N such that
for all n โ‰ฅ n1 (1) and for all n โ‰ฅ n2. (2)
Choose n0 = max{2n1, 2n2 + 1}.
Let n โ‰ฅ n0 be even. Then, n โ‰ฅ 2n1 or n/2 โ‰ฅ n1 and an = bn/2. By (1), we have it that

Similarly, let n โ‰ฅ n0 be odd. Then, n โ‰ฅ 2n2 + 1 or (n โˆ’ 1)/2 โ‰ฅ n2 and an = c(nโˆ’1)/2.
By (2), we have it that
Thus, we have shown that whenever n โ‰ฅ 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 โˆˆ 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) โˆ’ f (0) h2 sin(1/h)
f 0(0) = lim = lim = lim h sin(1/h) hโ†’0 h hโ†’0 h hโ†’0
|f 0(0)| โ‰ค lim |h| = 0 =โ‡’ f 0(0) = 0 hโ†’0
We can compute f 0(x) = 2x sin(1/x) โˆ’ 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 โˆˆ 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 itโ€™s differentiable everywhere with derivative 0.
Verify that f 0(c) = f 0(0)f (c) does hold for all x โˆˆ 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) โˆ’ f (0) = f 0(0) =โ‡’ lim f (h) โˆ’ 1 = f 0(0). (2)
hโ†’0 h hโ†’0 h
Now, let us show that f is differentiable everywhere.
Let c โˆˆ R. We must show that the following limit exists: f (c + h) โˆ’ f (c) lim
hโ†’0 h
Using (1), we can write the above expression as:
f (c)f (h) โˆ’ f (c)f (c)(f (h) โˆ’ 1)f (h) โˆ’ 1 lim .
hโ†’0 hhh
By (2), we know that the above limit exists. Thus, we have it that f is differentiable at c for every c โˆˆ 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 โˆˆ (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) โˆˆ J and J is an interval, we
have it that 0 โˆˆ J = f (I). Thus, 0 = f (x0) for some x0 โˆˆ 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 Rolleโ€™s Theorem, there exists x2 โˆˆ (x0,x1) such that f 0(x2) = 0. But this
contradicts the hypothesis that f 0(x) 6= 0 for all x โˆˆ (a,b).
Tutorial Sheet 2
5. To prove that |sina โˆ’ sinb| โ‰ค |a โˆ’ b| for all a,b โˆˆ 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 โˆˆ (a,b) such
0(c) = f (b) โˆ’ f (a). (By MVT)
that f
b โˆ’ a
Also, we know that |f 0(c)| = |cosc| โ‰ค 1.
Thus, we have it that .
b โˆ’ a
This is equivalent to what we wanted to prove.
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