Exercise 2.5 P.128
20. Explain why the function is discontinuous at the given number a . Sketch the graph of the function.
f (x) =
{ x2−x
x2−1 , if x̸= 1
1 , if x = 1 a = 1
< pf >
∵ limx→1+f (x) = limx→1−f (x) = limx→1x2−x
x2−1 = limx→1 x
x+1 =12 ̸= f(1) = 1
∴ f(x) is discontinuous at x = 1
30. Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain. State the domain.
B(x) = tan x
√4− x2
< pf >
(1) tan x is continuous in any interval ((2k−1)π2 ,(2k+1)π2 ) ,∀ k ∈ Z (2)√
x is continuous at every x≥ 0
(3) polynomial 4− x2 is continuous at every x∈ R By (1),(2),(3) B(x) is continuous at every x∈ (0,π2) 38. Use continuity to evaluate the limit
xlim→2arctan( x2− 4 3x2− 6x)
< pf >
xlim→2arctan(3xx22−4−6x) = arctan( lim
x→2
(x−2)(x+2)
3x(x−2) ) = arctan(46) = arctan(23) P.129
42. Find the numbers at which f is discontinuous. At which of these numbers is f continuous from the right, from the left, or neither? Sketch the graph of f .
f (x) =
x + 1 , if x≤ 1
1
x , if 1 < x < 3
√x− 3 , if x ≥ 3
< pf >
∵(1) x + 1 is continuous everywhere (2)1x is continuous at every x̸= 0 (3)√
x− 3 is continuous at every x ≥ 3
∴we only need to check x = 1 ∧ 3 lim
x→1+f (x) = lim
x→1+ 1
x= 1̸= lim
x→1−f (x) = lim
x→1−(x + 1) = 2 lim
x→3−f (x) = lim
x→3− 1
x= 13 ̸= lim
x→3+f (x) = lim
x→3+
√x− 3 = 0
⇒ f(x) is discontinuous at x = 1 ∧ 3
1
46. Find the values of a and b that make f continuous everywhere.
f (x) =
x2−4
x−2 , if x < 2 ax2− bx + 3 , if 2 ≤ x < 3 2x− a + b , if x≥ 3
< pf >
It’s easy to check f (x) continuous at every x̸= 2 ∧ 3 So we only need to check x = 2∧ 3
if f (x) is continuous at everywhere then lim
x→2f (x) = f (2)∧ lim
x→3f (x) = f (3)
xlim→2f (x) =
lim
x→2+f (x) = lim
x→2+ax2− bx + 3 = 4a − 2b + 3 lim
x→2−f (x) = lim
x→2− x2−4
x−2 = lim
x→2−
(x−2)(x+2) x−2 = 4
xlim→3f (x) =
{ lim
x→3+f (x) = lim
x→3+2x− a + b = 6 − a + b lim
x→3−f (x) = lim
x→3−ax2− bx + 3 = 9a − 3b + 3
⇒
{ 4a− 2b + 3 = 4
9a− 3b + 3 = 6 − a + b ⇒
{ a = 12 b = 12
54. Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
sin x = x2− x, (1, 2)
< pf >
sin x and x2− x are both continuous at (1, 2) Define f (x) = sin x− (x2− x)
then f (1) = sin 1− (1 − 1) = sin 1 > 0 f (2) = sin 2− (4 − 2) = sin 2 − 2 < 0
∵ f(1)f(2) < 0
∴By I.V.T there exist a c ∈ (1, 2) s.t f(c) = 0 64. For what values of x is g continuous
g(x) =
{ 0 , if x is rational x , if x is irrational
< pf >
Claim : g(x) is continuous only at x = 0 Given ε > 0 take δ = ε > 0
if 0 <|x − 0| < δ = ε then
{ |g(x) − 0| = 0 < δ = ε , if x is rational
|g(x) − 0| = |x| < δ = ε , if x is irrational
⇒ lim
x→0g(x) = 0 = g(0) For any c̸= 0
(1) if c is irrational
Take ε =2c then∀ δ > 0 if 0 <|x − c| < δ then
{ |g(x) − g(c)| = c > c2 = ε , if x is rational
|g(x) − g(c)| = |x − c| < δ , if x is irrational (2) if c is rational
2
Given ε > 0 take δ = ε if 0 <|x − c| < δ then
{ |g(x) − g(c)| = 0 < ε , if x is rational
|g(x) − c| = |x − c| < δ = ε , if x is irrational
⇒ lim
x→c, x∈Qg(x) = 0̸= lim
x→c, x/∈Qg(x) = c By (1) , (2) lim
x→cg(x) does not exist at every c̸= 0
⇒ g(x) is discontinuous at every x ̸= 0
65. Is there a number that is exactly 1 more than its cube?
< pf >
Yes,
Define f (x) = x− (1 + x3) f (−1) = −1 ∧ f(−2) = 5
∵ f(x) is continuous at everywhere
∴By I.V.T ∀ t ∈ (−1, 5) ∃c ∈ (−2, −1) s.t f(c) = t i.e∃ c ∈ (−2, −1) s.t f(c) = 0
⇒ c − (1 + c3) = 0⇒ c = 1 + c3
66. If a and b are positive numbers, prove that the equation a
x3+ 2x2− 1+ b
x3+ x− 2 = 0 has at least one solution in the interval (−1, 1).
< pf >
x3+ 2x2− 1 = (x + 1)(x2+ x− 1) ∧ x3+ x− 2 = (x − 1)(x2+ x + 2) Define f (x) = x3+2xa2−1 +x3+xb−2
lim
x→1−f (x) = a2 + lim
x→1− b
x3+x−2 ⇒ lim
x→1− b
x3+x−2 =−∞
lim
x→(√52−1)+
f (x) = lim
x→(√52−1)+ a
x3+2x2−1+ 3√b5−9
2
⇒ lim
x→(√52−1)+ a
x3+2x2−1 =∞ Given ε > 0 ∃ x1, x2∈ (−1, 1)
s.t 1− x1< ε∧ x2−√52−1 < ε satisfy f (x1) < 0∧ f(x2) > 0
⇒ f(x) has at least one solution in the interval (−1, 1).
3
