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Hints/Solutions for Tutorial Problem Set - 4

1. Since f  is continuous at 0 and since 1n

→ 0, we have f (0) = limn→∞

f ( 1n

) = 0. Also,

since f (0) exists and since 1n

→ 0, we have f (0) = limn→∞

f ( 1n)−f (0)

1/n= 0. Again, since

f (0) exists, there exists δ > 0 such that f (x) exists for each x ∈ (−δ, δ). We can

choose n0 ∈N

such that1

n0 < δ. By Rolle’s theorem, for each n ≥ n0, there existsxn ∈ ( 1

n+1, 1n

) such that f (xn) = 0. By the sandwich theorem, xn → 0. Since f (0)

exists, we have f (0) = limn→∞

f (xn)−f (0)xn

= 0.

2. The given condition implies that |f (x) − f (y)| ≤ |x − y|2 for all x, y ∈ R. Let y ∈ R.

Then for all x(= y) ∈ R, we have |f (x)−f (y)x−y

| ≤ |x − y| ⇒ limx→y

f (x)−f (y)x−y

= 0, i.e.

f (y) = 0. Thus f (y) = 0 for all y ∈ R. Consequently f  : R → R is a constantfunction.

3. Let p(x) = 1 − x + x2

2− x3

3+ · · · + (−1)n xn

nfor all x ∈ R. Then p(x) = −1 + x − x2 +

· · · + (−1)nxn−

1 for all x ∈ R.We first assume that n is odd. By Ex.4 of Tutorial Problem Set-3, the equation

 p(x) = 0 has at least one real root. Also, p(−1) = −n = 0 and p(x) = −(1+xn

1+x) = 0

for all x ∈ R \ {−1}. As a consequence of Rolle’s theorem, the equation p(x) = 0 canhave at most one real root. Therefore the equation p(x) = 0 has exactly one real root.We now assume that n is even. Then p(−1) = −n < 0 and p(x) = −(1−x

n

1+x) for all

x ∈ R \ {−1}. So p(x) > 0 for all x > 1 and p(x) < 0 for all x < 1. Hence p isstrictly increasing in [1, ∞) and p is strictly decreasing in (−∞, 1]. So p(x) > p(1) forall x > 1 and also p(x) > p(1) for all x < 1, i.e. p(x) > p(1) for all x(= 1) ∈ R. Since

 p(1) = ( 12

− 13

) + (14

− 15

) + · · · + ( 1n−2

− 1n−1

) + 1n

> 0, we get p(x) > 0 for all x ∈ R.

Therefore the equation p(x) = 0 has no real root.

4. Since f (0) > 0, there exists δ1 ∈ (0, 12

) such that f (x) > f (0) = 0 for all x ∈ (0, δ1).

Also, since f (1) > 0, there exists δ2 ∈ (0, 12

) such that f (x) < f (1) = 0 for all

x ∈ (1 − δ2, 1). By the intermediate value theorem, there exists c ∈ ( δ12

, 1 − δ22

) suchthat f (c) = 0. Now, by Rolle’s theorem, there exists c1 ∈ (0, c) and c2 ∈ (c, 1) suchthat f (c1) = f (c2) = 0.

Alternative: We first assume that there exists c ∈ (0, 1) such that f (c) < 0. Then bythe intermediate value property of derivatives, there exist c1 ∈ (0, c) and c2 ∈ (c, 1)such that f (c1) = f (c2) = 0. We now assume that f (x) ≥ 0 for all x ∈ (0, 1). Then

f  is an increasing function on [0, 1]. So 0 = f (0) ≤ f (x) ≤ f (1) = 0 for all x ∈ [0, 1],i.e. f (x) = 0 for all x ∈ [0, 1]. Hence f (x) = 0 for all x ∈ [0, 1] and we can takec1 = 1

3and c2 = 1

2.

5. Let α > 1 and let f (x) = (1 + x)α − (1 + αx) for all x ≥ −1. Then f  : [−1, ∞) → R

is differentiable and f (x) = α[(1 + x)α−1 − 1] for all x ≥ −1. Clearly f (x) ≤ 0 forall x ∈ [−1, 0] and f (x) ≥ 0 for all x ∈ [0, ∞). Hence f  is decreasing on [−1, 0] andincreasing on [0, ∞). So f (x) ≥ f (0) = 0 for −1 ≤ x ≤ 0 and also f (x) ≥ f (0) = 0for x ≥ 0. Therefore f (x) ≥ 0 for all x ≥ −1, which proves the required inequality.

6. Since f (c) exists, there exists δ > 0 such that f (x) exists for each x ∈ (c − δ, c + δ).

Hence by L’Hopital’s rule, limh→0

f (c+h)−2f (c)+f (c−h)h2 = lim

h→0f 

(c+h)−f 

(c−h)2h , provided the

second limit exists. Now limh→0

f (c+h)−f (c−h)2h

= 12

[limh→0

f (c+h)−f (c)h

+ limh→0

f (c−h)−f (c)−h

] =

12

[f (c) + f (c)] = f (c). Hence limh→0

f (c+h)−2f (c)+f (c−h)h2

= f (c).

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If  f (x) =

1 if  x > 0,

0 if  x = 0,

−1 if  x < 0,

then f  : R → R is not continuous at 0 and hence f (0)

does not exist, but limh→0

f (0+h)−2f (0)+f (0−h)h2

= 0, since f (h) + f (−h) = 0 for all h(= 0) ∈

R.