Ism et chapter_3

−1/2 1
f (x0 ) = f (0) = (2 · 0 + 9) =
3
So,
L(x) = f (x0 ) + f (x0 ) (x − x0 )
1
= 3 + (x − 0)
3
Chapter 3 =3+ x
1
3
(b) Using √ the approximation L(x) to esti-
√
mate 8.8, we get 8.8 = f (−0.1) ≈
Applications of 1
L(−0.1) = 3 + (−0.1) = 3 − 0.033 =
3
2.967
Diﬀerentiation 4. (a) f (x) =
2
, x0 = 1
x
f (x0 ) = f (1) = 2
2
f (x) = − 2 and so f (1) = −2
x
3.1 Linear Approximations The linear approximation is
L(x) = 2 + (−2) (x − 1)
and Newtons Method
√ (b) Using the approximation L(x) to estimate
1. (a) f (x) = x, x0 = 1 2 2
√ , we get = f (0.99) ≈ L(0.99) =
f (x0 ) = f (1) = 1 = 1 0.99 0.99
1 2 + (−2)(0.99 − 1) = 2.02
f (x) = x−1/2
2
1 5. (a) f (x) = sin 3x, x0 = 0
f (x0 ) = f (1) = f (x0 ) = sin(3 · 0) = 0
2
So, f (x) = 3 cos 3x
L(x) = f (x0 ) + f (x0 ) (x − x0 ) f (x0 ) = f (0) = 3 cos(3 · 0) = 3
1 So,
= 1 + (x − 1) L(x) = f (x0 ) + f (x0 ) (x − x0 )
2
1 1 = 0 + 3 (x − 0)
= + x
2 2 = 3x
(b) √
Using the approximation L(x) to estimate
√ (b) Using the approximation L(x) to esti-
1.2, we get 1.2 = f (1.2) ≈ L(1.2) =
1 1 mate sin(0.3), we get sin(0.3) = f (0.1) ≈
+ (1.2) = 1.1 L(0.1) = 3(0.1) = 0.3
2 2
2. (a) f (x0 ) = f (0) = 1 and 6. (a) f (x) = sin x, x0 = π
1 −2/3
f (x0 ) = sin π = 0
f (x) = (x + 1)
3 f (x) = cos x
1
So, f (0) = f (x0 ) = f (π) = cos π = −1
3 The linear approximation is,
The Linear approximation is,
1 1 L(x) = f (x0 ) + f (x0 ) (x − x0 )
L(x) = 1 + (x − 0) = 1 + x
3 3 = 0 + (−1) (x − π) = π − x
(b) Using the approximation L(x) to estimate
√ √
3 3
(b) Using the approximation L(x) to esti-
1.2, we get 1.2 = f (0.2) ≈ L(0.2) = mate sin(3.0), we get sin(3.0) = f (3.0) ≈
1
1 + (0.2) = 1.066 L(3.0) = π − 3.0
3
√ √
4
3. (a) f (x) = 2x + 9, √0 = 0
x 7. (a) f (x) = √ 16 + x, x0 = 0
4
f (x0 ) = f (0) = 2 · 0+9 = 3 f (0) = 16 + 0 = 2
1 1
f (x) = (2x + 9)
−1/2
·2 f (x) = (16 + x)−3/4
2 4
1 1
= (2x + 9)
−1/2 f (0) = (16 + 0)−3/4 =
4 32

150

3.1. LINEAR APPROXIMATIONS AND NEWTONS METHOD 151

L(x) = f (0) + f (0)(x − 0) 36
L(72) = 120 + (72 − 80)
1 20
=2+ x = 120 + 1.8(−8)
32
1 = 105.6 cans
= 2 + (0.04) = 2.00125
32 168 − 120
1 (b) L(x) = f (100) + (x − 100)
(b) L(0.08) = 2 + (0.08) = 2.0025 100 − 80
32 48
1 L(94) = 168 − (94 − 100)
(c) L(0.16) = 2 + (0.16) = 2.005 20
32 = 168 − 2.4(−6)
8. (a) f (x) = sin x, x0 = 0 = 182.4 cans
f (0) = 0
f (x) = cos x 142 − 128
f (0) = cos 0 = 1 11. (a) L(x) = f (200) + (x − 200)
220 − 200
L(x) = f (0) + f (0) (x − 0) 14
L(208) = 128 + (208 − 200)
=0+1·x 20
L(0.1) = 0.1 = 128 + 0.7(8) = 133.6
(b) f (x) = sin x, x0 = π 142 − 136
√ 3
π 3 (b) L(x) = f (240) + (x − 240)
f = 220 − 240
3 2 6
π π 1 L(232) = 136 − (232 − 240)
f = cos = 20
3 3 2 = 136 − 0.3(−8) = 138.4
π π π
L(x) = f +f x−
√ 3 3 3
3 1 π 14 − 8
L(1) = + 1− ≈ 0.842 12. (a) L(x) = f (10) + (x − 10)
2 2 3 10 − 5
2π 6
(c) f (x) = sin x, x0 = L(8) = 14 + (−2) = 11.6
√ 3 5
2π 3 14 − 8
f = (b) L(x) = f (10) + (x − 10)
3 2 10 − 5
2π 2π 1 6
f = cos =− L(12) = 14 + (2) = 16.4
3 3 2 5
2π 2π 2π
L(x) = f +f x− 13. f (x) = x3 + 3x2 − 1 = 0, x0 = 1
3 3 3
√ f (x) = 3x2 + 6x
3 1 2π
= − x−
2√ 2 3
f (x0 )
9 3 1 9 2π (a) x1 = x0 −
L = − − ≈ 0.788 f (x0 )
4 2 2 4 3
13 + 3 · 12 − 1
18 − 14 =1−
9. (a) L(x) = f (20) + (x − 20) 3 · 12 + 6 · 1
20 − 30 3 2
4 =1− =
L(24) ≈ 18 − (24 − 20) 9 3
10 f (x1 )
= 18 − 0.4(4) x2 = x1 −
f (x1 )
= 16.4 games 2 3 2 2
2 3 +3 3 −1
14 − 12 = −
3 2 2 2
(b) L(x) = f (40) + (x − 40) 3 3 +6 3
30 − 40
2 79
f (36) ≈ 12 − (36 − 40) = ≈ 0.5486
10 144
= 12 − 0.2(−4) (b) 0.53209
= 12.8 games
120 − 84 14. f (x) = x3 + 4x2 − x − 1, x0 = −1
10. (a) L(x) = f (80) + (x − 80) f (x) = 3x2 + 8x − 1
80 − 60

152 CHAPTER 3. APPLICATIONS OF DIFFERENTIATION

f (x0 ) 30
(a) x1 = x0 −
f (x0 )
3 1
= −1 − =− 20
−6 2
f (x1 ) y
x2 = x1 −
f (x1 ) 10

1 0.375
=− − = −0.4117647
2 −4.25
0
−5.0 −2.5 0.0 2.5 5.0
(b) The root is x ≈ −0.4064206546. x
−10

Start with x0 = −5 to find the root near −5:
15. f (x) = x4 − 3x2 + 1 = 0, x0 = 1 x1 = −4.718750, x2 = −4.686202,
f (x) = 4x3 − 6x x3 = −4.6857796, x4 = −4.6857795

f (x0 )
(a) x1 = x0 − 18. From the graph, we see two roots:
f (x0 )
14 − 3 · 12 + 1 1
=1− = 15

4 · 13 − 6 · 1 2
10
f (x1 )
x2 = x1 −
f (x1 ) 5

-1 0 1 2 3 4
1 4 1 2 0
1 2 −3 2 +1
= −
2 1 3 1 -5
4 2 −6 2
-10
5
=
8 -15

-20

(b) 0.61803

16. f (x) = x4 − 3x2 + 1, x0 = −1 f (xi )
Use xi+1 = xi − with
f (x) = 4x3 − 6x f (xi )
f (x) = x4 − 4x3 + x2 − 1, and
f (x) = 4x3 − 12x2 + 2x
f (x0 ) Start with x0 = 4 to find the root below 4:
(a) x1 = x0 − x1 = 3.791666667, x2 = 3.753630030, x3 =
f (x0 )
−1 1 3.7524339, x4 = 3.752432297
= −1 − =− Start with x = −1 to find the root just above
2 2
f (x1 ) −1:
x2 = x1 − x1 = −0.7222222222,
f (x1 )
x2 = −0.5810217936,
1 0.3125 x3 = −0.5416512863,
=− − = −0.625
2 2.5 x4 = −0.5387668233,
x5 = −0.5387519962
(b) The root is x ≈ −0.6180339887.

f (xi ) f (xi )
17. Use xi+1 = xi − with 19. Use xi+1 = xi − with
f (xi ) f (xi )
f (x) = x3 + 4x2 − 3x + 1, and f (x) = x5 + 3x3 + x − 1, and
f (x) = 3x2 + 8x − 3 f (x) = 5x4 + 9x2 + 1


10 x1 = −0.644108, x2 = −0.636751
x3 = −0.636733, x4 = −0.636733
Start with x0 = 1.5 to find the root near 1.5:
5
x1 = 1.413799, x2 = 1.409634
x3 = 1.409624, x4 = 1.409624
0

−1.0 −0.5 0.0 0.5 1.0
22. Use xi+1 = xi − f (xii)) with
f
(x
x
f (x) = cos x2 − x, and
y −5 f (x) = 2x sin x2 − 1
3

−10

2
Start with x0 = 0.5 to find the root near 0.5: y

x1 = 0.526316, x2 = 0.525262, 1

x3 = 0.525261, x4 = 0.525261
0
f (xi ) -2 -1 0 1 2

20. Use xi+1 = xi − with x
f (xi ) -1

f (x) = cos x − x, and
f (x) = − sin x − 1 -2

5.0

Start with x0 = 1 to find the root between 0
and 1:
2.5
x1 = 0.8286590991, x2 = 0.8016918647,
x3 = 0.8010710854, x4 = 0.8010707652
0.0
3
−5 −4 −3 −2 −1 0 1 2 3 4 5
x
2
y −2.5 y

1

−5.0
0
Start with x0 = 1 to find the root near 1: -2 -1 0 1
x
2

x1 = 0.750364, x2 = 0.739113, -1

x3 = 0.739085, x4 = 0.739085
-2

21. Use xi+1 = xi − f (xii)) with
f
(x
f (x) = sin x − x2 + 1, and f (xi )
f (x) = cos x − 2x 23. Use xi+1 = xi − with
f (xi )
5.0
f (x) = ex + x, and
f (x) = ex + 1
20
2.5

15
0.0

−5 −4 −3 −2 −1 0 1 2 3 4 5
y 10
x

y −2.5
5

−5.0
0

−3 −2 −1 0 1 2 3
x
Start with x0 = −0.5 to find the root near −5

−0.5: Start with x0 = −0.5 to find the root between


0 and -1: zeros of f ), Newton’s method will succeed.
x1 = −0.566311, x2 = −0.567143 Which root is found depends on the starting
x3 = −0.567143, x4 = −0.567143 place.
f (xi ) 33. f (x) = x2 + 1, x0 = 0
24. Use xi+1 = xi − with
√ f (xi ) f (x) = 2x
f (x) = e−x − x, and f (x0 ) 1
1 x1 = x0 − =0−
f (x) = −e−x − √ f (x0 ) 0
2 x The method fails because f (x0 ) = 0. There
are no roots.
1
34. Newton’s method fails because the function
0.5
does not have a root!
4x2 − 8x + 1
0
0 0.5 1 1.5 2
35. f (x) = = 0, x0 = −1
4x2 − 3x − 7
Note: f (x0 ) = f (−1) is undefined, so New-
-0.5
ton’s Method fails because x0 is not in the do-
main of f . Notice that f (x) = 0 only when
-1
4x2 − 8x + 1 = 0. So using Newton’s Method
on g(x) = 4x2 − 8x + 1 with x0 = −1 leads to
x ≈ .1339. The other root is x ≈ 1.8660.
Start with x0 = 0.5 to find the root close to
36. The slope tends to infinity at the zero. For ev-
0.5:
ery starting point, the sequence does not con-
x1 = 0.4234369253, x2 = 0.4262982542,
verge.
x3 = 0.4263027510
√ 37. (a) With x0 = 1.2,
25. f (x) = x2 − 11; x0 = 3; 11 ≈ 3.316625
√ x1 = 0.800000000,
26. Newton’s method for x near x = 23 is xn+1 = x2 = 0.950000000,
1
2 (xn + 23/xn ). Start with x0 = 5 to get:
x3 = 0.995652174,
x1 = 4.8, x2 = 4.7958333, and x3 = 4.7958315. x4 = 0.999962680,
√ x5 = 0.999999997,
27. f (x) = x3 − 11; x0 = 2; 3 11 ≈ 2.22398 x6 = 1.000000000,
√ x7 = 1.000000000
28. Newton’s method for 3 x near x = 23 is
xn+1 = 1 (2xn + 23/x2 ). Start with x0 = 3
3 n (b) With x0 = 2.2,
to get: x0 = 2.200000, x1 = 2.107692,
x1 = 2.851851851, x2 = 2.843889316, and x2 = 2.056342, x3 = 2.028903,
x3 = 2.884386698 x4 = 2.014652, x5 = 2.007378,
√ x6 = 2.003703, x7 = 2.001855,
29. f (x) = x4.4 − 24; x0 = 2; 4.4 24 ≈ 2.059133
x8 = 2.000928, x9 = 2.000464,
√
30. Newton’s method for 4.6 x near x = 24 is x10 = 2.000232, x11 = 2.000116,
1
xn+1 = 4.6 (3.6xn +24/x3.6 ). Start with x0 = 2
n x12 = 2.000058, x13 = 2.000029,
to get: x14 = 2.000015, x15 = 2.000007,
x1 = 1.995417100, x2 = 1.995473305, and x16 = 2.000004, x17 = 2.000002,
x3 = 1.995473304 x18 = 2.000001, x19 = 2.000000,
x20 = 2.000000
31. f (x) = 4x3 − 7x2 + 1 = 0, x0 = 0 The convergence is much faster with x0 =
f (x) = 12x2 − 14x 1.2.
f (x0 ) 1
x1 = x0 − =0−
f (x0 ) 0 38. Starting with x0 = 0.2 we get a sequence that
The method fails because f (x0 ) = 0. Roots converges to 0 very slowly. (The 20th itera-
are 0.3454, 0.4362, 1.659. tion is still not accurate past 7 decimal places).
Starting with x0 = 3 we get a sequence that
32. Newton’s method fails because f = 0. As long
7 quickly converges to π. (The third iteration is
as the sequence avoids xn = 0 and xn = (the already accurate to 10 decimal places!)
6


√
39. (a) With x0 = −1.1 43. f (x) = √ 4 + x
x1 = −1.0507937, f (0) = 4 + 0 = 2
x2 = −1.0256065, 1
f (x) = (4 + x)−1/2
x3 = −1.0128572, 2
1 1
x4 = −1.0064423, f (0) = (4 + 0)−1/2 =
x5 = −1.0032246, 2 4
1
x6 = −1.0016132, L(x) = f (0) + f (0)(x − 0) = 2 + x
4
x7 = −1.0008068, 1
x8 = −1.0004035, L(0.01) = 2 + (0.01) = 2.0025
√ 4
x9 = −1.0002017, f (0.01) = 4 + 0.01 ≈ 2.002498
x10 = −1.0001009, 1
L(0.1) = 2 + (0.1) = 2.025
x11 = −1.0000504, √ 4
x12 = −1.0000252, f (0.1) = 4 + 0.1 ≈ 2.0248
x13 = −1.0000126, 1
L(1) = 2 + (1) = 2.25
x14 = −1.0000063, √ 4
x15 = −1.0000032, f (1) = 4 + 1 ≈ 2.2361
x16 = −1.0000016,
x17 = −1.0000008,
x18 = −1.0000004,
x19 = −1.0000002,
x20 = −1.0000001, 44. The linear approximation for ex at x = 0 is
x21 = −1.0000000, L(x) = 1 + x. The error when x = 0.01 is
x22 = −1.0000000 0.0000502, when x = 0.1 is 0.00517, and when
(b) With x0 = 2.1 x = 1 is 0.718.
x0 = 2.100000000,
x1 = 2.006060606,
x2 = 2.000024340,
x3 = 2.000000000,
x4 = 2.000000000 45. (a) f (0) = g(0) = h(0) = 1, so all pass
The rate of convergence in (a) is slower through the point (0, 1).
than the rate of convergence in (b). f (0) = 2(0 + 1) = 2,
g (0) = 2 cos(2 · 0) = 2, and
40. From exercise 37, f (x) = (x − 1)(x − 2)2 . New-
h (0) = 2e2·0 = 2,
ton’s method converges slowly near the double
so all have slope 2 at x = 0.
root. From exercise 39, f (x) = (x − 2)(x + 1)2 .
The linear approximation at x = 0 for all
Newton’s method again converges slowly near
three functions is L(x) = 1 + 2x.
the double root. In exercise 38, Newton’s
method converges slowly near 0, which is a zero
of both x and sin x but converges quickly near
π, which is a zero only of sin x. (b) Graph of f (x) = (x + 1)2 :
5
41. f (x) = tan x, f (0) = tan 0 = 0
f (x) = sec2 x, f (0) = sec2 0 = 1 4

L(x) = f (0) + f (0)(x − 0) L(0.01) = 0.01
3
= 0 + 1(x − 0) = x y

f (0.01) = tan 0.01 ≈ 0.0100003 2

L(0.1) = 0.1 1
f (0.1) = tan(0.1) ≈ 0.1003
L(1) = 1 0

f (1) = tan 1 ≈ 1.557 −3 −2 −1 0 1 2 3
−1
√ x
42. The linear approximation for 1 + x at x = 0
1
is L(x) = 1 + 2 x. The error when x = 0.01 is
0.0000124, when x = 0.1 is 0.00119, and when
x = 1 is 0.0858. Graph of f (x) = 1 + sin(2x):


5
2

4

1
3
y
2 0
-2 -1 0 1 2
x
1
-1

0
−3 −2 −1 0 1 2 3
-2
x −1

Graph of h(x) = sinh x:

Graph of f (x) = e2x :
3
5
2

4
1

3 0
-2 -1 0 1 2
y
-1 x
2

-2

1
-3

0

−3 −2 −1 0 1 2 3
x −1
sin x is the closest ﬁt, but sinh x is close.

√
4
47. (a) 16.04 = 2.0012488
L(0.04) = 2.00125
|2.0012488 − 2.00125| = .00000117
46. (a) f (0) = g(0) = h(0) = 0, so all pass √
4
through the point (0, 0). (b) 16.08 = 2.0024953
f (0) = cos 0 = 1, L(.08) = 2.0025
1 |2.0024953 − 2.0025| = .00000467
g (0) = = 1, and
1 + 02 √
h (0) = cosh 0 = 1, (c) 4
16.16 = 2.0049814
so all have slope 1 at x = 0. L(.16) = 2.005
The linear approximation at x = 0 for all |2.0049814 − 2.005| = .0000186
three functions is L(x) = x.

(b) Graph of f (x) = sin x: 48. If you graph | tan x − x|, you see that the dif-
ference is less than .01 on the interval −.306 <
2 x < .306 (In fact, a slightly larger interval
would work as well).
1

49. The ﬁrst tangent line intersects the x-axis at a
0
-2 -1 0 1 2 point a little to the right of 1. So x1 is about
x
1.25 (very roughly). The second tangent line
-1
intersects the x-axis at a point between 1 and
x1 , so x2 is about 1.1 (very roughly). Newton’s
-2 Method will converge to the zero at x = 1.
Starting with x0 = −2, Newton’s method con-
Graph of g(x) = tan−1 x: verges to x = −1.


f (x) = 2x − 1
3
3
At x0 =
2
2
2
y 3 3 1
f (x0 ) = − −1=−
1 2 2 4
and
3
-2 -1
0
0 1 2
f (x0 ) = 2 −1=2
x
2
-1 By Newton’s formula,
f (x0 ) 3 −1 13
x1 = x0 − = − 4 =
-2 f (x0 ) 2 2 8

Starting with x0 = 0.4, Newton’s method con- (b) f (x) = x2 − x − 1
verges to x = 1. f (x) = 2x − 1
5
At x0 = 3
3 2
5 5 1
f (x0 ) = − −1=
2 3 3 9
y
and
5 7
1 f (x0 ) = 2 −1=
3 3
0
By Newton’s formula,
-2 -1 0 1 2
f (x0 )
x x1 = x0 −
-1 f (x0 )
1
5 9 5 1 34
-2 = − 7 = − =
3 3
3 21 21

50. It wouldn’t work because f (0) = 0. x0 = 0.2 (c) f (x) = x2 − x − 1
works better as an initial guess. After jumping f (x) = 2x − 1
8
to x1 = 2.55, the sequence rapidly decreases At x0 = 5
2
toward x = 1. Starting with x0 = 10, it takes 8 8 1
f (x0 ) = − −1=−
several steps to get to 2.5, on the way to x = 1. 5 5 25
and
f (xn ) 8 11
51. xn+1 = xn − f (x0 ) = 2 −1=
f (xn ) 5 5
x2 − c
n By Newton’s formula,
= xn − f (x0 )
2xn x1 = x0 −
x2 c f (x0 )
= xn − n + 8 − 25 1
8 1 89
2xn 2xn = − 11 = + =
xn c 5 5 55 55
= + 5
2 2xn
1 c (d) From part (a),
= xn + F4 F7
2 xn sincex0 = , hence x1 = .
√ √ √ F3 F6
If x0 < a, then a/x0 > a, so x0 < a < From part (b),
a/x0 . F5 F9
√ since x0 = hence x1 = .
52. The root of xn − c is n c, so Newton’s method F4 F8
From part (c),
approximates this number. F6 F11
Newton’s method gives since x0 = hence x1 = .
f (xi ) xn − c F5 F10
xi+1 = xi − = xi − i n−1 Fn+1
f (xi ) nxi Thus in general if x0 = , then x1 =
Fn
1 F2n+1
= (nxi − xi + cx1−n ),
i implies m = 2n + 1 and k = 2n
n F2n
as desired.
3 Fn+1
53. (a) f (x) = x2 − x − 1 (e) Given x0 = , then lim will be
2 n→∞ Fn


the zero of the function f (x) = x2 − 2P x
L(x) = 120 − .01(120) = P −
x − 1 which is 1.618034. Therefore, R
Fn+1 2 · 120x
lim = 1.618034 = 120 −
n→∞ Fn R
2x
.01 =
54. The general form of functionf (x) is, R
1 n+2 1 1 x = .005R = .005(20,900,000)
fn (x) = 2 x − 3 for n < x < n−1 .
5 2 2 = 104,500 ft
Hence
2n+2 1 1 58. If m = m0 (1 − v 2 /c2 )1/2 , then
f (x) = fn (x) = for n < x < n−1 .
5 2 2 m = (m0 /2)(1 − v 2 /c2 )−1/2 (−2v/c2 ), and
By Newton’s method,
m = 0 when v = 0. The linear approxima-
3 f 34 3 f1 34
x1 = − = − tion is the constant function m = m0 for small
4 f 3 4
4 f1 4 3
values v.
3 (3/5 ) 3 x0
= − = = 59. The only positive solution is 0.6407.
4 (8/5 ) 8 2
x1 x0 x0
Similarly, x2 = = 2 and x3 = 3 60. The smallest positive solution of the first equa-
2 2 2 tion is 0.132782, and for the second equa-
x0
Continuing this, we get, xn−1 = n−1 It may tion the smallest positive solution is 1, so the
2
also be observed that, for each fn (x) species modeled by the second equation is cer-
(1/2n ) + 1/2n+1 3 tain to go extinct. This is consistent with the
x0 = = n+1 ,
2 2 models, since the expected number of offspring
x0 3 3 for the population modeled by the first equa-
xn = n = 2n+1 ⇒ xn+1 = 2n+2 which
2 2 2 tion is 2.2, while for the second equation it is
is the zero of F . Therefore Newton’s method
only 1.3
converges to zero of F .
61. The linear approximation for the inverse tan-
55. For small x we approximate ex by x + 1 gent function at x = 0 is
(see exercise 44) f (x) ≈ f (0) + f (0)(x − 0)
Le2πd/L − e−2πd/L tan−1 (x) ≈ tan−1 (0) + 1+02 (x − 0)
1

e2πd/L + e−2πd/L tan−1 (x) ≈ x
L 1 + 2πd − 1 − 2πd
L L
Using this approximation,
≈ 3[1 − d/D] − w/2
1 + 2πd + 1 − 2πd
L L φ = tan−1
4πd
D−d
L L
≈ = 2πd 3[1 − d/D] − w/2
2 φ≈
4.9 D−d
f (d) ≈ · 2πd = 9.8d If d = 0, then φ ≈ 3−w/2 . Thus, if w or D
π D
increase, then φ decreases.
8πhcx−5
56. If f (x) = , then using the linear 62. d (θ) = D(w/6 sin θ)
ehc/(kT x) − 1 d(0) = D(1 − w/6) so
approximation we see that
8πhcx−5 d(θ) ≈ d(0) + d (0)(θ − 0)
f (x) ≈ hc
= 8πkT x−4 = D(1 − w/6) + 0(θ) = D(1 − w/6),
(1 + kT x ) − 1
as desired. as desired.

P R2
57. W (x) =
(R + x)2
, x0 = 0 3.2 Indeterminate Forms and
W (x) =
−2P R2 L’Hˆpital’s Rule
o
(R + x)3
L(x) = W (x0 ) + W (x0 )(x − x0 ) x+2
1. lim
x→−2 x2 − 4
P R2 −2P R2 x+2
= + (x − 0) = lim
(R + 0)2 (R + 0)3 x→−2 (x + 2)(x − 2)
2P x 1 1
=P− = lim =−
R x→−2 x − 2 4

ˆ
3.2. INDETERMINATE FORMS AND L’HOPITAL’S RULE 159

x2 − 4 sin x − x 0
2. lim 11. lim 3
is type ;
x→2 x2− 3x + 2 x→0 x 0
(x − 2)(x + 2) we apply L’Hˆpital’s Rule thrice to get
o
= lim cos x − 1 − sin x
x→2 (x − 2)(x − 1) = lim = lim
x+2 x→0 3x2 x→0 6x
= lim =4 − cos x 1
x→2 x − 1 = lim =− .
x→0 6 6
3x2 + 2
3. lim tan x − x 0
x→∞ x2 − 4 12. lim is type ;
2
3 + x2
x→0 x3 0
= lim we apply L’Hˆpital’s Rule to get
o
x→∞ 1 − 4 sec2 x − 1
x2
3 lim .
= =3 x→0 3x2
1
Apply L’Hˆpital’s Rule twice more to get
o
x+1 ∞ 2 sec2 x tan x
4. lim is type ; lim
x→−∞ x2 + 4x + 3 ∞ x→0 6x
4 sec2 x tan2 x + 2 sec4 x 1
we apply L’Hˆpital’s Rule to get
o = lim = .
1 x→0 6 3
lim = 0. √ √ √
x→−∞ 2x + 4
t−1 t−1 t+1
13. lim = lim · √
e2t − 1 0 t→1 t − 1 t→1 t − 1 t+1
5. lim is type ;
t→0 t 0 (t − 1)
o = lim √
t→1 (t − 1) t + 1
d
e2t − 1
lim dt d 1 1
= lim √ =
dt t
t→0
t→1 t+1 2
2e2t 2
lim = =2 ln t 0
t→0 1 1 14. lim is type ;
sin t 0
t→1 t −1 0
6. lim is type ;
t→0 e3t−1 0 we apply L’Hˆpital’s Rule to get
o
o d 1
dt (ln t)
d
(sin t) cos t 1 lim d = lim t = 1
t→1 1
dt (t − 1)
t→1
lim ddt 3t = lim 3t =
t→0 3e 3
dt (e − 1)
t→0
x3 ∞
tan−1 t 0 15. lim x is type ;
7. lim is type ; x→∞ e ∞
t→0 sin t 0 we apply L’Hˆpital’s Rule thrice to get
o
o
d
tan−1 t 1/(1 + t2 ) 3x2 6x
lim dt d = lim =1 lim = lim x
t→0
dt (sin t)
t→0 cos t x→∞ ex x→∞ e
6
sin t 0 = lim x = 0.
8. lim is type ; x→∞ e
t→0 sin−1 t 0
ex ∞
o 16. lim is type ;
d x→∞ x4 ∞
dt (sin t) cos t
lim = lim √ =1 we apply L’Hˆpital’s Rule four times to get
o
t→0 d sin−1 t t→0 1/( 1 − t2 )
dt ex ex
lim 3
= lim
sin 2x 0 x→∞ 4x x→∞ 12x2
9. lim is type ; ex ex
x→π sin x 0 = lim = lim = ∞.
x→∞ 24x x→∞ 24
o
x cos x − sin x ∞
2 cos 2x 2(1) 17. limx→0 2 is type ;
lim = = −2. x sin x ∞
x→π cos x −1 we apply L’Hˆpital’s Rule twice to get
o
cos x − x sin x − cos x
cos−1 x limx→0
10. lim is undeﬁned (numerator goes to sin2 x + 2x sin x cos x
x→−1 x2 − 1 −x sin x
π, denominator goes to 0). = lim
x→0 sin x (sin x + 2x cos x)


−x x − π cos x
= lim = lim 2
= 0
x→0 sin x + 2x cos x
−1
π
x→ 2 cos x − x − π sin x
2
= lim
x→0 cos x + 2 cos x − 2x sin x ln x ∞
1 21. lim 2 is type
=− . x→∞ x ∞
3 we apply L’Hˆpital’s Rule to get
o
1/x 1
18. Rewrite as one fraction, we have lim = lim = 0.
x→∞ 2x x→∞ 2x2
1 x cos x − sin x
lim cot x − = lim ln x ∞
x→0 x x→0 x sin x 22. lim √ is type ;
0 x→∞ x ∞
which is of type we apply L’Hˆpital’s Rule to get
o
0 1
o 2
cos x − x sin x − cos x lim x = lim √ = 0.
1
= lim
x→∞ √
2 x
x→∞ x
x→0 sin x + x cos x
t ∞
d
(−x sin x) 23. lim t is type
= lim dx t→∞ e ∞
x→0 d
(sin x + x cos x) we apply L’Hˆpital’s Rule to get
o
dx d
(t) 1
− sin x − x cos x lim dt = lim t = 0.
= lim =0 t→∞ d (et ) t→∞ e
x→0 cos x + cos x − x sin x dt

sin 1
t 0
19. Rewrite as one fraction, we have 24. lim 1 is type
t→∞
t
0
x+1 2 we apply L’Hˆpital’s Rule to get
o
lim −
x→0 x sin 2x - 1 cos 1
2 1
(x + 1) sin 2x − 2x 0 = lim t 1 t = lim cos = 1.
= lim is type ; t→∞ − t2 t→∞ t
x→0 x sin 2x 0
we apply L’Hˆpital’s Rule four times to get
o ln (ln t)
d 25. lim
(x + 1) sin 2x − 2x t→1 ln t
lim dx d As t approaches ln from below, ln t is a small
dx (x sin 2x)
x→0
negative number. Hence ln (ln t) is undeﬁned,
sin 2x + 2(x + 1) cos 2x − 2 so the limit is undeﬁned.
= lim
x→0 sin 2x + 2x cos 2x
d
sin (sin t) 0
(sin 2x + 2(x + 1) cos 2x − 2) 26. lim is type
= lim dx d t→0 sin t 0
dx (sin 2x + 2x cos 2x)
x→0 we apply L’Hˆpital’s Rule to get
o
2 cos 2x + 2 cos 2x − 4(x + 1) sin 2x cos (sin t) cos t
= lim lim = 1.
x→0 2 cos 2x + 2 cos 2x − 4x sin 2x t→0 cos t
4
= =1 sin (sinh x) 0
4 27. lim is type
x→0 sinh (sin x) 0
o
1 cos (sinh x) cosh x
20. lim tan x + lim =1
π
x→ 2 x− π 2
x→0 cosh (sin x) cos x
In this case the limit has the form (∞ - ∞). sin x − sinh x
sin x 28. lim
Rewrite tan x as and then as one frac- x→0 cos x − cosh x
cos x
tion, we get 2 sin x − ex + e−x
= lim
1 x→0 2 cos x − ex − e−x
lim tan x +
x→ 2π
x− π 2
2ex sin x − e2x + 1 0
= lim is type
sin x 1 x→0 2ex cos x − e2x − 1 0
= lim + we apply L’Hˆpital’s Rule twice to get
o
x→ π
2 cos x x − π 2
2ex cos x + 2ex sin x − 2e2x
x − π sin x + cos x
2 0 lim
= lim is type x→0 −2ex sin x + 2ex cos x − 2e2x
x→ π
2 x − π cos x
2
0
cos x + sin x − 1 0
o = lim is type
x→0 cos x − sin x − 1 0
sin x + x − π cos x − sin x
2 − sin x + cos x
= lim = lim = −1
x→ π
2 cos x − x − π sin x
2 x→0 − sin x − cos x

ˆ

ln x ∞ x+1
 
29. lim is type
x→0 + cot x ∞  ln x−2 
o = lim  
x→∞  √ 1 
1/x x2 −4
lim
x→0+ − csc2 x 0
sin x This last limit has indeterminate form , so
= lim+ − sin x · = (0)(1) = 0. 0
x→0 x we can apply L’Hˆpital’s Rule and simplify to
o
ﬁnd that the above is equal to
√
x −3(x2 − 4)3/2
30. lim+ = 0 (numerator goes to 0 and de- lim and this is equal to 3. So
x→0 ln x x→∞ −x3 + x2 + 2x
nominator goes to −∞). lim ln y = 3.
x→∞
Thus lim y = lim eln y = e3 ≈ 20.086.
x→∞ x→∞
31. lim x2 + 1 − x √
x→∞
√ 1 x
x2 + 1 + x 35. lim+ √ −√
= lim x 2+1−x √ x→0 x√ x+1 √
x→∞ x2 + 1 + x x + 1 − ( x)2
2
x +1−x 2 = lim+ √ √
= lim √ x→0
√ x x+1
x→∞ x2 + 1 + x x+1−x
1 = lim √ √
= lim √ =0 x→0+ x x+1
x→∞ x 2+1+x = ∞.
√
5−x−2 0
ln x
−1 36. lim √ is type
32. lim ln x − x = lim x
= −∞ since the x→1 10 − x − 3 0
1
x→∞ x→∞
x
o
numerator goes to −1 and the denominator 1 −1/2
2 (5 − x) (−1)
goes to 0+ . (Recall Example 2.8 which shows lim 1
x→1 (10 − x)−1/2 (−1)
2 √
ln x
lim = 0.) 10 − x 3
x→∞ x = lim √ = .
x→1 5−x 2
x x
1 37. Let y = (1/x) . Then ln y = x ln(1/x). Then
33. Let y = 1+
x lim+ ln y = lim+ x ln(1/x) = 0, by Exercise
x→0 x→0
1
⇒ ln y = x ln 1 + . Then 27. Thus lim+ y = lim+ eln y = 1.
x x→0 x→0
1
lim ln y = lim x ln 1 + 38. Let y = lim+ (cos x)1/x . Then
x→∞ x→∞ x x→0
ln 1 + x 1 1
= lim ln y = lim+ ln cos x
x→0 x
x→∞ 1/x
1 1 ln(cos x) 0
1+ x1 − x2 = lim is type
= lim x→0 + x 0
x→∞ −1/x2 so apply L’Hˆpital’s Rule to get
o
1 − tan x
= lim = 1. lim+ = 0.
x→∞ 1 + 1 x→0 1
x
Hence lim y = lim eln y = e. Therefore the limit is y = e0 = 1.
x→∞ x→∞
t t
t−3 (t − 3)
39. lim = lim
34. Notice that the limit in question has the inde- t→∞ t+2 t→∞ (t + 2)t
terminate form 1∞ . Also, note that as x gets 3 t lim 1 − 3
t
x+1 x+1 1− t t→∞ t
large, = . = lim =
t→∞ 2 t 2 t
x−2 x −√2 1+ t lim 1 + t
t→∞
x2 −4 −3 t
x+1 lim 1 +
Deﬁne y = . Then t→∞ t e−3
x−2 = = = e−5
2 t e2
√ x+1 lim 1 + t
t→∞
ln y = x2 − 4 ln and
x−2 t t
3
x+1 t−3 1− t
lim ln y = lim x2 − 4 ln 40. lim = lim 1
x→∞ x→∞ x−2 t→∞ 2t + 1 t→∞ 2+ t


3 t
1− t e−3 we apply L’Hˆpital’s Rule to get
o
= lim t = lim =0
t→∞ 1/2 t→∞ 2t e1/2
2t 1 + n cos nx n
t lim = .
x→0 m cos mx m
41. L’Hˆpital’s rule does not apply. As x → 0, the
o sin x2 2x cos x2
numerator gets close to 1 and the denominator 50. (a) lim 2
= lim
x→0 x x→0 2x
is small and positive. Hence the limit is ∞.
= lim cos x2 = 1,
ex − 1 0 ex x→0
42. lim is type , but lim is not, so sin x
x→0 x 2 0 x→0 2x which is the same as lim .
L’Hˆpital’s Rule does not apply to this limit.
o x→0 x

1 − cos x2
43. L’Hˆpital’s rule does not apply. As x → 0, the
o (b) lim
x→0 x4
numerator is small and positive while the de- 2x sin x2 sin x2
nominator goes to −∞. Hence the limit is 0. = lim 3
= lim
x→0 4x x→0 2x2
2x 1 sin x 2
1
Also lim , which equals lim x2 , is not of = lim = (by part (a)),
x→0 2/x x→0
2 x→0 x2 2
0
the form so L’Hˆpital’s rule doesn’t apply
o
0 while
here either.
sin x 0 cos x 1 − cos x sin x 1 1
44. lim is type , but lim is not, so lim 2
= lim = (1) =
x→0 x2 0 x→0 2x x→0 x x→0 2x 2 2
L’Hˆpital’s rule does not apply. This limit is
o so both of these limits are the same.
undefined because the numerator goes to 1 and (c) Based on the patterns found in exercise
the denominator goes to 0. 45, we should guess
csc x sin x3 1 − cos x3 1
45. lim+ √ lim = 1 and lim = .
x→0 x x→0 x3 x→0 x6 2
∞
In this case limit has the form , L’Hˆspital’s
o
0 (x + 1)(2 + sin x)
Rule should not be used. 51. (a)
x(2 + cos x)
x−3/2 ∞ x
46. lim+ is type . In this case (b) x
x→0 ln x −∞ e
L’Hˆspital’s Rule should be used.
o 3x + 1
(c)
x2 − 3x + 1 x−7
47. lim = ∞. In this case limit has 3 − 8x
x→∞ tan−1 x (d)
the form ∞. So L’Hˆspital’s Rule should not
o 1 + 2x
be used.
52. (a) lim x − ln x = ∞ (see exercise 32).
ln x2 ∞ x→∞
48. lim is type . So L’Hˆspital’s Rule
o √
x→∞ ex/3 ∞ (b) lim x2 + 1 − x = 0 (see exercise 31).
should be used. x→∞
√
sin 3x (c) lim x2 + 4x − x
49. (a) Starting with lim , we cannot x→∞ √
sin 2x
x→0 = lim ( x2 + 4x − x)
3x x→∞
“cancel sin”to get lim . We can cancel 4x
x→0 2x = lim √
the x’s in the last limit to get the final an- x→∞ x2 + 4x + x
swser of 3/2. The first step is likely to give 1
4x x
a correct answer because the linear ap- = lim √
x→∞ 1
proximation of sin 3x is 3x, and the linear ( x2 + 4x + x)
x
approximation of sin 2x is 2x. The linear 4
= lim = 2,
approximations are better the closer x is x→∞ 4
1+ x +1
to zero, so the limits are likely to be the
where to get from the second to
same.
the third line, we have multiplied by
√
sin nx
(b) lim is type 0 ;
0
( x2 + 4x + x)
x→0 sin mx √ .
( x2 + 4x + x)

ˆ

53. lim ex = lim xn = ∞ In general,when the degree of exponential term
x→∞ x→∞
ex in the numerator and denominator are diﬀer-
lim n = ∞. Since n applications of ln ekx + p(x)
x→∞ x ent, then the lim for polyno-
L’Hˆpital’s rule yields
o x→∞ ln (ecx + q(x))
ex mials p and q and positive numbers. k and c
lim = ∞.
x→∞ n! will be the fraction of degrees that is k .
c
Hence e dominates xn .
x

54. lim ln x = lim xp = ∞. 59. If x → 0, then x2 → 0, so if lim
f (x)
= L,
x→∞ x→∞
ln x ∞ x→0 g(x)
lim is of type f (x2 )
x→∞ xp ∞ then lim = L (but not conversely). If
we use L’Hˆpital’s Rule to get
o x→0 g(x2 )
1
x 1 f (x)
lim p−1
= lim = 0 (since p > 0). a = 0 or 1, then lim involves the be-
x→∞ px x→∞ pxp x→a g(x)
p
Therefore, x dominates ln x. f (x2 )
havior of the quotient near a, while lim
t t
x→a g(x2 )
55. lim e 2 − t3 Since e 2 dominates t3 . So involves the behavior of the quotient near the
t→∞
t diﬀerent point a2 .
lim e − t3 = ∞
2
t→∞
60. Functions f (x) = |x| and g(x) = x work.
√ f (x)
x − ln x ∞ lim does not exist as it approaches −1
56. lim √ is type . x→0 g(x)
x→∞ x ∞
from the left and it approaches 1 from the
o
√ − 1
1 √ f (x2 )
2 x x x−2 x right, but lim = 1.
lim = lim x→0 g(x2 )
x→∞ 1
√ x→∞ x
2 x
2 2.5(4ωt − sin 4ωt)
= lim 1 − √ = 1. 61. lim
x→∞ x ω→0 4ω 2
2.5(4t − 4t cos 4ωt)
= lim
ln x3 + 2x + 1
ω→0 8ω
57. lim 2.5(16t2 sin 4ωt)
x→∞ ln (x2 + x + 2) = lim =0
ω→0 8
we apply L’Hˆpital’s Rule
o
d
dx ln x3 + 2x + 1 π
lim 2.5 − 2.5 sin(4ωt + )
d 2 2 is type 0 ;
dx (ln (x + x + 2))
x→∞
62. lim 0
3x2 +2
ω→0 4ω 2
x3 +2x+1 we apply L’Hˆpital’s Rule to get
o
= lim 2x+1 −10t cos(4ωt + π )
x→∞ 2
x2 +x+2 lim
3x + 3x + 8x2 + 2x + 4
4 3
3 ω→0 8ω
= lim = 40t2 sin(4ωt + π )
x→∞ 2x 4 + x3 + 4x2 + 4x + 1 2 = lim 2
= 5t2 .
In general, for numerator and denominator the
ω→0 8
highest degee of polynomials p and q, such that
p(x) > 0 and q(x) > 0 for x > 0, 2

should be the lim ln(p(x)) .
ln(q(x))
x→∞
1.5

3x
ln e + x ∞
58. lim 2x + 4)
is ; 1
x→∞ ln (e ∞
we apply L’Hˆpital’s Rule
o 0.5
d 3x
dx ln e +x
lim d
x→∞
dx (ln (e2x + 4)) 0
0 0.1 0.2 0.3 0.4 0.5 0.6
3e3x +1 t
e3x +x
= lim 2e2x
x→∞
e2x +4
5x
3e + 12e3x + e2x + 4 3 63. The area of triangular region 1 is
= lim = (1/2)(base)(height)
x→∞ 2e5x + 2xe2x 2

Ism et chapter_3

More Related Content

Ism et chapter_3