Download Math 220 February 5 I. Determine the values of a for which lim x→a f

Math 220
February 5
I. Determine the values of a for which limx→a f (x) exists.


1 + sin(x) if x < 0
f (x) = cos(x)
if 0 ≤ x ≤ π


sin(x)
if x > π
II. For what value of the constants a,b is the function f continuous on (−∞, ∞).


cos(x) + 2 sin(x) if x < 0
f (x) = x2 + 6x + b
if 0 ≤ x ≤ 2

 x
a
if x > 2
III. Use the Intermediate Value Theorem to show that there is a root of the given equation
in the specified interval.
1.
√
3
x = 1 − x, (0, 1)
2. sin(x) = x + 3, (−4, 0)
3. x3 − 2x2 + 5 = ex , (−2, 1)
IV. Determine the infinite limit.
1.
lim−
x+3
x−4
lim+
x+3
x−4
x→4
2.
x→4
3.
3−x
x→4 (x − 4)2
lim
4.
lim ln(x2 − 16)
x→4+
5.
lim cot(x)
x→π +
6.
lim−
x→2
x2 + 5x + 6
x2 − 3x + 2
1
7.
lim+
x→2
x2 − 4
x2 + 2x − 8
V. Find the limit or show that it does not exist.
1.
x3 + 2x + 1
x→∞
x+3
lim
2.
x3 + 2x + 1
x→∞
x4 + 3
lim
3.
2x + 1 − 3x3
x→∞
x2 + 5
lim
4.
lim
x→−∞
5.
√
4x2 + 5x − 2x
ex + 1
x→−∞ e2x + 5
lim
6.
ex + 1
x→∞ e2x + 5
lim
7.
lim ex sin(x + 3)
x→−∞
8.
√
16x8 − x3
x→−∞ 3x4 + 5x + 1
lim
9.
sin(x2 ) + sin2 (x) + ex
x→−∞
3x4 + 5ex + 1
lim
VI. Simplify the expression.
1. sin(2 cos−1 (x))
2. cos(2 tan−1 (x))
3. tan(cos−1 (x))
4. sec(sin−1 (x))
2
1
Solutions
I. Determine the values of a for which limx→a f (x) exists.


1 + sin(x) if x < 0
f (x) = cos(x)
if 0 ≤ x ≤ π


sin(x)
if x > π
Answer:
The functions 1 + sin(x), cos(x), sin(x) are continuous so the limit exists for at least all a
except a = 0, π.
At a = 0
lim f (x) = 1 + sin(0) = 1 + 0 = 1
x→0−
lim f (x) = cos(0) = 1 = 1
x→0+
lim f (x) = lim+ f (x) = lim f (x)
x→0−
x→0
x→0
The limit exists at a = 0
At a = π
lim f (x) = cos(π) = −1
x→π −
lim f (x) = sin(π) = 0
x→π +
lim f (x) 6= lim+ f (x)
x→π −
x→π
The limit does not exist at a = 0
The limit exist for all values of a except a = π
II. For what value of the constants a,b is the function f continuous on (−∞, ∞).


cos(x) + 2 sin(x) if x < 0
f (x) = x2 + 6x + b
if 0 ≤ x ≤ 2

 x
a
if x > 2
Answer:
f (x) is continuous at x = a if
lim f (x) = f (a)
x→a
The functions cos(x) + 2 sin(x), x2 + 6x + b, ax are continuous.
lim f (x) = cos(0) + 2 sin(0) = 1 + 0 = 1
x→0−
lim+ f (x) = 02 + 6 · 0 + b = b
x→0
3
lim f (x) = lim+ f (x) =⇒ b = 1
x→0−
x→0
lim f (x) = 22 + 6(2) + 1 = 4 + 12 + 1 = 17
x→2−
lim+ f (x) = a2
x→2
lim− f (x) = lim+ f (x) =⇒ a2 = 17 =⇒ a =
x→0
√
17
x→0
Thus the function will be continuous if b = 1 and a =
√
17
III. Use the Intermediate Value Theorem to show that there is a root of the given equation in the specified interval.
1.
√
3
x = 1√
− x, (0, 1) Answer:
f (x) = 3 x − 1 + x
We want to show f (x) = 0 for some x ∈ (0, 1)
f (x) is continuous for all real numbers.
f (0) = −1
f (1) = 1
f (0) = −1 < 0 < 1 = f (1)
Therefore by the Intermediate Value Theorem there exists c ∈ (0, 1) such that f (c) = 0.
Thus the equation has a root.
2. sin(x) = x + 3, (−4, 0) Answer:
f (x) = sin(x) − x − 3
We want to show f (x) = 0 for some x ∈ (−4, 0)
f (x) is continuous for all real numbers.
f (−4) = sin(−4) + 1 > 0
f (0) = −3
f (−4) < 0 < f (0)
Therefore by the Intermediate Value Theorem there exists c ∈ (−4, 0) such that
f (c) = 0. Thus the equation has a root.
3. x3 − 2x2 + 5 = ex , (−2, 1) Answer:
f (x) = x3 − 2x2 + 5 − ex
We want to show f (x) = 0 for some x ∈ (−2, 1)
4
f (x) is continuous for all real numbers.
f (−2) = (−2)3 − 2(−2)2 + 5 − e−2 = −8 − 6 + 5 + e−2 = −9 + e−2 < 0
f (1) = 13 − 2(1)2 + 5 + e1 = 4 + e > 0
f (−2) < 0 < f (1)
Therefore by the Intermediate Value Theorem there exists c ∈ (−2, 1) such that
f (c) = 0. Thus the equation has a root.
IV. Determine the infinite limit.
1.
lim−
x→4
x+3
x−4
Answer:
x+3
7
lim−
= −
x→4 x − 4
0
2.
lim+
x→4
+
= −∞
−
x+3
x−4
Answer:
7
x+3
lim+
= +
x→4 x − 4
0
3.
+
=∞
+
3−x
x→4 (x − 4)2
lim
Answer:
3−x
−1
lim
= +
2
x→4 (x − 4)
0
−
= −∞
+
4.
lim ln(x2 − 16)
x→4+
Answer:
ln
lim+ x − 16 = ln(0+ ) = −∞
2
x→4
5
5.
lim cot(x)
x→π +
Answer:
−1
cos(π + )
= −
lim+ cot(x) =
+
x→π
sin(π )
0
6.
lim−
x→2
−
= −∞
−
x2 + 5x + 6
x2 − 3x + 2
Answer:
x2 + 5x + 6
x2 + 5x + 6
20
lim− 2
= lim−
= −
x→2 x − 3x + 2
x→2 (x − 2)(x − 1)
0
7.
lim+
x→2
+
= −∞
−
x2 − 4
x2 + 2x − 8
Answer:
x2 − 4
(x − 2)(x + 2)
(x + 2)
4
2
lim+ 2
= lim+
= lim+
= =
x→2 x + 2x − 8
x→2 (x + 4)(x − 2)
x→2 (x + 4)
6
3
V. Find the limit or show that it does not exist.
1.
x3 + 2x + 1
x→∞
x+3
lim
Answer:
x3 + 2x + 1 1/x
lim
x→∞
x+3
1/x
2
x + 2 + 1/x
= lim
x→∞
1 + 3/x
= lim
x→∞
>
x2 + 2 + 1/x
0
>
1+
3/x
= −∞
2.
x3 + 2x + 1
x→∞
x4 + 3
lim
6
0
Answer:
x3 + 2x + 1 1/x4
lim
x→∞
x4 + 3
1/x4
1/x + 2/x3 + 1/x4
= lim
x→∞
1 + 3/x4
0
= lim
0
*
0
*
>
3 + 4
1/x
+
2/x
1/x
0
*
x→∞
4
1 +
3/x
=0
3.
2x + 1 − 3x3
x→∞
x2 + 5
lim
Answer:
2x + 1 − 3x3 1/x2
x→∞
x2 + 5
1/x2
2/x + 1/x2 − 3x
= lim
x→∞
1 + 5/x2
lim
0
= lim
x→∞
0
*
>
2 − 3x
2/x
+
1/x
0
*
2
1 +
5/x
= −∞
4.
lim
x→−∞
√
4x2 + 5x − 2x
7
Answer:
lim
x→−∞
√
4x2 + 5x − 2x
!
√
2
4x + 5x + 2x
√
4x2 + 5x − 2x
(4x2 + 5x − 4x2
= lim √
x→−∞
4x2 + 5x − 2x
5x
= lim √
x→−∞
4x2 + 5x − 2x
5x
= lim √
x→−∞
4x2 + 5x − 2x
5
p
= lim
x→−∞ −
4 + 5/x − 2
5
q
= lim
0
x→−∞
>
− 4+
5/x
−2
5
= √
− 4−2
5
=
−4
5
=−
4
5.
ex + 1
x→−∞ e2x + 5
lim
Answer:
ex + 1
x→−∞ e2x + 5
0+1
=
0+5
1
=
5
lim
6.
ex + 1
x→∞ e2x + 5
lim
8
1/x
√
−1/ x2
!
Answer:
e−2x
e−2x
e−x + e−2x
= lim
x→−∞ 1 + 5e−2x
ex + 1
lim 2x
x→−∞ e
+5
= lim
x→−∞
*0
* 0 −2x
e−x
+
e
0
*
1 + 5
e−2x
=0
7.
lim ex sin(x + 3)
x→−∞
Answer:
−ex ≤ ex sin(x + 3) ≤ ex sin(x + 3)
lim −ex ≤ lim ex sin(x + 3) ≤ lim ex
x→−∞
x→−∞
x→−∞
x
0 ≤ lim e sin(x + 3) ≤ 0
x→−∞
x
lim e sin(x + 3) = 0
x→−∞
8.
√
16x8 − x3
x→−∞ 3x4 + 5x + 1
lim
Answer:
√ !
16x8 − x3 −1/ x8
lim
x→−∞ 3x4 + 5x + 1
1/x4
p
− 16 − 1/x5
= lim
x→−∞ 3 + 5/x3 + 1/x4
q
0
*
5
− 16 − 1/x
= lim
0
0
x→−∞
*
*
3
4
3 +
5/x
+
1.x
√
− 16
=
3
4
=−
3
√
9
9.
sin(x2 ) + sin2 (x) + ex
x→−∞
3x4 + 5ex + 1
lim
Answer:
−2 + ex
sin(x2 ) + sin2 (x) + ex
2 + ex
≤
≤
3x4 + 5ex + 1
3x4 + 5ex + 1
3x4 + 5ex + 1
2
2
x
sin(x ) + sin (x) + ex
2 + ex
−2 + e
≤
lim
≤
lim
lim
x→−∞
x→−∞ 3x4 + 5ex + 1
x→−∞ 3x4 + 5ex + 1
3x4 + 5ex + 1
−2 + ex
=0
x→−∞ 3x4 + 5ex + 1
2 + ex
lim
=0
x→−∞ 3x4 + 5ex + 1
lim
sin(x2 ) + sin2 (x) + ex
=0
x→−∞
3x4 + 5ex + 1
lim
VI. Simplify the expression.
1. sin(2 cos−1 (x))
Answer:
sin(2 cos−1 (x)) = 2 sin(cos−1 (x)) cos(cos−1 (x))
= 2 sin(cos−1 (x))x
√
= 2 1 − x2 x
√
= 2x 1 − x2
2. cos(2 tan−1 (x))
Answer:
cos(2 tan−1 (x)) = cos2 (tan−1 (x)) − sin2 (tan−1 (x))
2 2
1
x
= √
− √
1 + x2
1 + x2
1
x2
=
−
1 + x2 1 + x2
1 − 2x2
=
1 + x2
10
3. tan(cos−1 (x))
Answer:
sin(cos−1 (x))
cos(cos−1 (x))
√
1 − x2
=
x
tan(cos−1 (x)) =
4. sec(sin−1 (x))
Answer:
1
cos(sin−1 (x))
1
=√
1 − x2
sec(sin−1 (x)) =
11