Continuity and Differentiability Questions - Mathematics

Continuity and Differentiability Questions

1. Prove that f (x) = sin(1/x2) x=0 is not continuous at x = 0.
0 x=0

Proof: Define xn = √1 , which satisfies xn → 0 as n → ∞. Then f (xn) = sin(πn + π/2) =

πn+π/2

(−1)n+1. Then f (xn) is divergent, and by the sequential characterization of limits, lim f (x) does

x→0

not exists. Therefore f is not continuous at x = 0.

2. Prove that f (x) = x cos(1/x) x=0 is continuous, but not differentiable, at x = 0.
0 x=0

Proof: Since −1 ≤ cos(1/x) ≤ 1 for all x = 0, it follows that −|x| ≤ x cos(1/x) ≤ |x| for all x = 0. By

the squeeze theorem, it follows that lim f (x) = 0 = f (0). That is, f is continuous at x = 0.

x→0

We need to show that

lim f (x) − f (0)

x→0 x−0

does not exist. We again appeal to the sequential characterization of limit. Define xn = 1 , which
πn
tends to zero as n → ∞. Consider the following difference quotient, for x = 0

lim f (xn) − f (0) = xn cos(1/xn) = cos(πn) = (−1)n,
xn −0 xn
n→∞

which does not exist. Therefore f (x) is not differentiable at x = 0.

3. Prove that 3x2 − 1 is continuous on R.

Proof: Let a∈R and ε>0 be arbitrary. Choose δ >0 such that δ < min(1, ε ). Let x ∈ R
3(1+2|a|)

such that |x − a| < δ. Then

|(3x2 − 1) − (3a2 − 1)| = 3|x2 − a2| = 3|x − a| |x + a| < 3δ |(x − a) + 2a|
≤ 3δ(|x − a| + 2|a|) < 3δ(1 + 2|a|) < ε.

4. Assume f is continuous at x = 1 and f (1) = 2. Prove that there exists δ > 0 such that f (x) > 3/2

for all x ∈ (1 − δ, 1 + δ).

Proof: Assume f is continuous at x = 1 and f (1) = 2. Applying the definition of continuity with

ε = 1/2, there exists δ > 0 such that for all |x − 1| < δ, | f (x) − 2| < 1 . Then for x ∈ (1 − δ, 1 + δ),
2

we have

f (x) = 2 − (2 − f (x)) ≥ 2 − | f (x) − 2| > 2 − 1 = 3.
22

5. (T/F) If f is continuous on (a, b), then f is bounded on (a, b).

False: Define f (x) = 1 on (0, 1). It follows that f (x) is continuous and unbounded on (0, 1).
x

6. (T/F) If f is continuous on (a, b), then there exists xM ∈ (a, b) such that f (xM) = sup{ f (x) : x ∈
(a, b)}.

False: Define f (x) = x on (0, 1). It follows that f (x) is continuous on (0, 1). Note that sup{ f (x) :
x ∈ (a, b)} = 1, but clearly f (x) < 1 for all x ∈ (0, 1). So there is no xM ∈ (0, 1) such that
f (xM) = sup{ f (x) : x ∈ (a, b)}.

7. Find the derivative of x3 − 3x2 + 2 for x ∈ R using the limit definition.
Proof: Let x ∈ R. Then

((x + h)3 − 3(x + h)2 + 2) − (x3 − 3x2 + 2) 3x2h + 3xh2 + h3 − 6xh − 3h2
lim = lim
h→0 h h→0 h

= lim 3x2 + 3xh + h2 − 6x − 3h = 3x2 − 6x.

h→0

8. Find the derivative of √1 for x > 0 using the limit definition.
x

Proof: Let x > 0. Then

√1 − √1 √√
x+h x √x − x+h x− (x + h)√
lim = lim √ = lim √ √ √
h→0 h h→0 h x + h x h→0 h x + h x( x + x + h)

= lim √ √ −√1 √ = − √1 .
h→0 x + h x( x + x + h) 2 x3

9. Find the derivative of x for x ∈ R using the limit definition.
x4+1
Proof: Let x ∈ R. Then

x+h − x (x + h)(x4 + 1) − x((x + h)4 + 1)
(x+h)4+1 x4+1
lim = lim
h h(x4 + 1)((x + h)4 + 1)
h→0 h→0

= lim (x + h)x4 + h − x(x4 + 4x3h + 6x2h2 + 4xh3 + h4)

h→0 h(x4 + 1)((x + h)4 + 1)

= lim h − 3x4h − 6x3h2 − 4x2h3 − xh4

h→0 h(x4 + 1)((x + h)4 + 1)

1 − 3x4 − 6x3h − 4x2h2 − xh3 1 − 3x4
= lim = (x4 + 1)2
(x4 + 1)((x + h)4 + 1)
h→0

10. Let f , g : (a, b) → R. If both f and g are differentiable at x0 ∈ (a, b) and g(x0) = 0, then f is
g

differentiable at x0 and f (x0) = g(x0) f (x0)− f (x0)g (x0) .
g g(x0)2

Proof: Assume that both f and g are differentiable at x0 ∈ (a, b) and g(x0) = 0. Then

f (x0+h) − f (x0) f (x0 + h)g(x0) − f (x0)g(x0 + h)
g(x0+h) g(x0)
lim = lim
h→0 h h→0 hg(x0)g(x0 + h)

= lim f (x0 + h)g(x0) − f (x0)g(x0) + f (x0)g(x0) − f (x0)g(x0 + h)
h→0 hg(x0)g(x0 + h) hg(x0)g(x0 + h)

= lim f (x0 + h) − f (x0) g(x0) − f (x0) g(x0 + h) − g(x0)
h→0 h g(x0)g(x0 + h) g(x0)g(x0 + h) h

= f (x0) g(x0) − f (x0) g (x0)
g(x0)2 g(x0)2

= f (x0)g(x0) − f (x0)g (x0) .
g(x0)2

Note that these limits exist since f is differentiable at x0, g is differentiable at x0, and g is continuous
at x0 with g(x0) = 0.

11. If f is increasing and differentiable on (a, b), then f (x) ≥ 0 for all x ∈ (a, b).

Proof: Let x0 ∈ (a, b). Then for any x ∈ (x0, b), we have f (x) ≥ f (x0) and hence f (x)− f (x0) ≥ 0.
x−x0
Since f is differentiable at x0, it follows that

f (x0) = lim f (x) − f (x0) = lim f (x) − f (x0) ≥ 0.
x − x0 x − x0
x→x0 x→x0+