Download Section 2.5 Continuity EX.20 EX.30

Section 2.5 Continuity
EX.20
Explain why the function is discontinuous at the given number a .
Sketch the graph of the function.
f (x) =


x2 −x
x2 −1
 1
, if x 6= 1
a=1
, if x = 1
< pf >
∵ limx→1+ f (x) = limx→1− f (x) = limx→1
x2 −x
x2 −1
= limx→1
x
x+1
=
1
2
6= f (1) = 1
∴ f (x) is discontinuous at x = 1
EX.30
Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain.
State the domain.
tan x
B(x) = √
4 − x2
< pf >
∵ The domain of tan x is R \ {2kπ ± π2 |k ∈ Z} and the domain of
√ 1
4−x2
is the interval (−2, 2),
∴ the domain of B(x) is the intersection of the above two domains, namely, (−2, 2) \ {− π2 , π2 }.
(2k−1)π (2k+1)π
, 2
for all k ∈ Z.
(1) tan x is continuous in any interval of the form
2
(2)
(3)
1
x
√
is continuous on R \ {0}.
x is continuous at every x ≥ 0.
(4) polynomial 4 − x2 is continuous at every x ∈ R.
By (1), (2), (3), and (4), and using Theorems 4, 5, 7, and 9, B(x) is continuous at every number
in its domain (−2, 2) \ {− π2 , π2 }.
1
EX.38
Use continuity to evaluate the limit
lim arctan(
x→2
x2 − 4
)
3x2 − 6x
< pf >
2
−4
lim arctan( 3xx2 −6x
) = arctan(lim (x−2)(x+2)
) = arctan( 46 ) = arctan( 23 )
3x(x−2)
x→2
x→2
EX.42
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
, if 1 < x < 3
x


√

 x − 3 , if x ≥ 3
< pf >
∵(1) x + 1 is continuous everywhere
(2) x1 is continuous at every x 6= 0
√
(3) x − 3 is continuous at every x ≥ 3
∴we only need to check x = 1 ∧ 3
lim f (x) = lim+ x1 = 1 6= lim− f (x) = lim− (x + 1) = 2
x→1
x→1
x→1
√
lim− f (x) = lim− x1 = 13 6= lim+ f (x) = lim+ x − 3 = 0
x→1+
x→3
x→3
x→3
x→3
⇒ f (x) is discontinuous at x = 1 ∧ 3
EX.46
Find the values of a and b that make f continuous everywhere.
2
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 x 6= 2 ∧ 3
So we only need to check x = 2 ∧ 3
if f (x) is continuous at everywhere
then lim f (x) = f (2)∧ lim f (x) = f (3)
x→2 
x→3

 lim f (x) = lim ax2 − bx + 3 = 4a − 2b + 3
x→2+
x→2+
lim f (x) =
2 −4
x→2

 lim f (x) = lim xx−2
= lim− (x−2)(x+2)
=4
x−2
−
−
x→2
x→2
x→2


 lim f (x) = lim 2x − a + b = 6 − a + b
x→3+
x→3+
lim f (x) =
x→3

 lim f (x) = lim ax2 − bx + 3 = 9a − 3b + 3
x→3−
x→3−


 4a − 2b + 3 = 4
 a= 1
2
⇒
⇒
 9a − 3b + 3 = 6 − a + b
 b= 1
2
EX.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
3
EX.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,
then
|g(x) − g(0)| =

|0 − 0| = 0 < ε
, if x is rational
.
|x − 0| < δ = ε , if x is irrational
∴ lim g(x) = g(0).
x→0
For any c 6= 0
(1) If c is irrational,
take ε =
|c|
2
> 0. For all δ > 0, there exists a rational number x such that 0 < |x − c| < δ
(because the rational numbers are dense in the real numbers) and
|g(x) − g(c)| = |0 − c| = |c| > ε.
.
(2) If c is rational and positive,
take ε = |c| > 0. For all δ > 0, there exists a irrational number x ∈ (c, c + δ) (because the
irrational numbers are dense in the real numbers) and
|g(x) − g(c)| = |x − 0| = |x| > |c| = ε.
.
(3) If c is rational and negative,
take ε = |c| > 0. For all δ > 0, there exists a irrational number x ∈ (c − δ, c) (because the
irrational numbers are dense in the real numbers) and
|g(x) − g(c)| = |x − 0| = |x| > |c| = ε.
.
By (1), (2), and (3), ∴ g(x) is discontinuous at every x 6= 0.
4
EX.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
EX.66
If a and b are positive numbers, prove that the equation
a
b
+ 3
=0
2
+ 2x − 1 x + x − 2
x3
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) =
lim f (x) =
x→1−
lim
√
x→(
5−1 +
)
2
a
2
a
x3 +2x2 −1
+
b
x3 +x−2
b
b
+ lim− x3 +x−2
⇒ lim− x3 +x−2
= −∞
f (x) =
x→1
lim
√
x→(
5−1 +
)
2
x→1
a
+
x3 +2x2 −1
b
3
√
5−9
2
⇒
lim
√
x→(
5−1 +
)
2
a
x3 +2x2 −1
Given ε > 0 ∃ x1 , x2 ∈ (−1, 1)
s.t 1 − x1 < ε ∧ x2 −
√
5−1
2
<ε
satisfy f (x1 ) < 0 ∧ f (x2 ) > 0
⇒ f (x) has at least one solution in the interval (−1, 1).
5
=∞