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L is at the point (x , 0), where x is the solution of
negligible mass. (Here, m ≪ m means that a is much
2
B
2
A
smaller than m .) Assume that object B orbits in a circular
A
4
2
5
(1 + k)x − (3k + 2)x + (3k + 1)x − (2k + 1)x + 2x − 1 = 0
path around the common center of mass. There are five cir-
cular orbits for object C that maintain constant relative posi-
and L is at the point (−x , 0), where x is the solution of
tions of the three objects. These are called Lagrange points
3
3
L , L , L , L and L , as shown in the figure.
3
4
1
2
5
3
5
2
(1 + k)x + (3k + 2)x + (3k + 1)x − x − 2x − 1 = 0,
m
B
where k =
. Use Newton’s method to find approximate
m
A
solutions of the following.
(a) Find L for the Earth-Sun system with k = 0.000002.
1
This point has an uninterrupted view of the Sun and
is the location of the solar observatory SOHO.
B
(b) Find L for the Earth-Sun system with k = 0.000002.
A
2
This is the location of NASA’s Microwave Anisotropy
L
L
L
3
2
1
Probe.
(c) Find L for the Earth-Sun system with k = 0.000002.
3
This point is invisible from the Earth and is the loca-
tion of Planet X in many science fiction stories.
(d) Find L for the Moon-Earth system with k = 0.01229.
1
This point has been suggested as a good location for a
L
space station to help colonize the moon.
(e) The points L and L form equilateral triangles with ob-
To derive equations for the Lagrange points, set up a coor-
4
dinate system with object A at the origin and object B at the
jects A and B. Explain why this means that polar coor-
4 - 2 L 4 5 1 Indeterminate Forms 4 5 3 ( 3 2 6 ) . Find (x, y)-coordinates
point (1, 0). Then L is at the point (x , 0), where x is the
dinates for L are (r, ) = 1,
1
1
4
solution of
for L and L . In the Jupiter-Sun system, these are loca-
4
5
3
5
4
tions of numerous Trojan asteroids.
2
(1 + k)x − (3k + 2)x + (3k + 1)x − x + 2x − 1 = 0;

Indeterminate Forms
s Rule
and L
’Hôpital’
and L’Hôpital’s Rule
4.2
INDETERMINATE FORMS AND l’HÔPITAL’S RULE
In this section, we reconsider the problem of computing limits. You have frequently
seen limits of the form
f(x)
lim ,
x→a g(x)
where lim f(x) = lim g(x) = 0 or where lim f(x) = lim g(x) =∞ (or −∞). Recall that from
x→a x→a x→a x→a
0
either of these forms ( or ∞ , called indeterminate forms), we cannot determine the
∞
0
value of the limit, or even whether the limit exists, without additional work. For in-
stance, note that
2
lim x − 1 = lim (x − 1)(x + 1) = lim x + 1 = 2 = 2,
x→1 x − 1 x→1 x − 1 x→1 1 1
lim x − 1 = lim x − 1 = lim 1 = 1
2
x→1 x − 1 x→1 (x − 1)(x + 1) x→1 x + 1 2
and lim x − 1 = lim x − 1 = lim 1 , which does not exist,
2
x→1 x − 2x + 1 x→1 (x − 1) 2 x→1 x − 1
0
even though all three limits initially have the form . The lesson here is that the ex-
0
pression 0 is mathematically meaningless. It indicates only that both the numerator
0
and denominator tend to zero and that we’ll need to dig deeper to find the value of the
limit or whether the limit even exists.
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240 | Lesson 4-2 | Indeterminate Forms and L’Hôpital’s Rule