Page 93 - Schaum's Outline of Theory and Problems of Advanced Calculus

84 DERIVATIVES [CHAP. 4

2
1=x . Then ln FðxÞ¼ðln cos xÞ=x to which L’Hospital’s rule can be applied. We
2
Let FðxÞ¼ ðcos xÞ
have
ln cos x sin x cos x 1
lim ¼ lim ð sin xÞ=ðcos xÞ ¼ lim ¼ lim ¼
x!0 x 2 x!0 2x x!0 2x cos x x!0 2x sin x þ 2cos x 2
1
Thus, lim ln FðxÞ¼ . But since the logarithm is a continuous function, lim ln FðxÞ¼ lnðlim FðxÞÞ. Then
x!0 2 x!0 x!0
1 or 1=x 2 ¼ e 1=2
lnðlim FðxÞÞ ¼ 2 lim FðxÞ¼ limðcos xÞ
x!0 x!0 x!0

4.31. If FðxÞ¼ðe 3x 5xÞ 1=x , find (a) lim FðxÞ and (b) lim FðxÞ.
x!0 x!0
0
The respective indeterminate forms in (a) and (b)are 1 and 1 .
1
lnðe 3x 5xÞ
x x!1 x!0
Let GðxÞ¼ ln FðxÞ¼ . Then lim GðxÞ and lim GðxÞ assume the indeterminate forms 1=1
and 0/0 respectively, and L’Hospital’s rule applies. We have
lnðe 3x 3e 3x 5 9e 3x 27e 3x
lim 5xÞ ¼ lim ¼ lim ¼ lim ¼ 3
x x!1 e 5x x!0 3e 5 x!1 9e
ðaÞ 3x 3x 3x
x!1
3
Then, as in Problem 4.30, lim ðe 3x 5xÞ 1=x ¼ e .
x!1
lnðe 3x 3e 3x 5 2
lim 5xÞ ¼ lim ¼ 2 and limðe 3x 1=x ¼ e
ðbÞ 3x 5xÞ
x!0 x x!0 e 5x x!0

4.32. Suppose the equation of motion of a particle is x ¼ sinðc 1 t þ c 2 Þ, where c 1 and c 2 are constants.
(Simple harmonic motion.) (a) Show that the acceleration of the particle is proportional to its
distance from the origin. (b)If c 1 ¼ 1, c 2 ¼ , and t 0, determine the velocity and acceleration
at the end points and at the midpoint of the motion.

2
dx d x 2 2
¼ c 1 cosðc 1 t þ c 2 Þ; ¼ c 1 sinðc 1 t þ c 2 Þ¼ c 1 x:
dt dt
ðaÞ 2
This relation demonstrates the proportionality of acceleration and distance.
(b) The motion starts at 0 and moves to 1. Then it oscillates between this value and 1. The absolute value
of the velocity is zero at the end points, and that of the acceleration is maximum there. The particle
coasts through the origin (zero acceleration), while the absolute value of the velocity is maximum there.

p ffiffiffi
4.33. Use Newton’s method to determine 3 to three decimal points of accuracy.
p ffiffiffi 2 2
3 is a solution of x 3 ¼ 0, which lies between 1 and 2. Consider f ðxÞ¼ x 3then f ðxÞ¼ 2x.
0
The graph of f crosses the x-axis between 1 and 2. Let x 0 ¼ 2. Then f ðx 0 Þ¼ 1 and f ðx 0 Þ¼ 1:75.
0
f ðx 0 Þ
¼ 2 :25 ¼ 1:75.
According to the Newton formula, x 1 ¼ x 0
0
f ðx 0 Þ
f ðx 1 Þ 2
¼ 1:732. To verify the three decimal point accuracy, note that ð1:732Þ ¼ 2:9998
Then x 2 ¼ x 1
0
f ðx 1 Þ
2
and ð1:7333Þ ¼ 3:0033.