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The solutions to exam 3 of math 112 fall 2006. It includes the calculation of limits using l'hopital's rule, the evaluation of improper integrals, and the analysis of trigonometric functions. The document also covers the comparison theorem and the determination of the area under a curve.
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Math 112 Fall 2006 Exam 3 Solutions Some solutions include the details; others have just the final answer.
ex^ − 1
x
= lim x → 0 +
x − ex^ + 1 x ( ex^ − 1 )
L’H = lim x → 0 +
1 − ex xex^ + ex^ − 1
L’H = lim x → 0 +
− ex ( x + 2 ) ex^
(b) (^) x lim→∞
x − 2 x
) x has the indefinite form “1∞”. Let y =
( x − 2 x
) x , so ln y = x ln
( x − 2 x
. Then
x lim→∞ ln^ y^ =^ x lim→∞ x^ ln
x − 2 x
= (^) x lim→∞
ln
( x − 2 x
1 x
L’H = lim x →∞
( (^) x x − 2
x^2
− 1 x^2
= (^) x lim→∞
− 2 x x − 2
Then
x lim→∞
x − 2 x
) x = (^) x lim→∞ y = (^) x lim→∞ e ln^ y^ = e −^2
(c) lim θ → 0
sin θ 1 + cos θ
∫ (^) ∞ 2
dx ( x − 7 )^4 = lim t → 7 −
∫ (^) t 2
dx ( x − 7 )^4
∫ (^8) t
dx ( x − 7 )^4
∫ (^) t 8
dx ( x − 7 )^4 and
lim t → 7 −
∫ (^) t
2
dx ( x − 7 )^4 = lim t → 7 −
( x − 7 )−^3
t
2
= lim t → 7 −
3 ( t − 7 )^3
That is, the limit is not a finite number, so the improper integral is divergent.
∫ (^) ∞ 4
dx x
x − 1
= (^) t lim→∞
∫ (^) t 4
dx x
x − 1
Let u =
x − 1 = ( x − 1 )^1 /^2. Then du = 12 ( x − 1 )−^1 /^2 , u^2 + 1 = x , and 2 udu = dx. So ∫ (^) t 4
dx x
x − 1
∫ √ t − 1 √ 3
u^2 + 1 du = 2 (arctan(
t − 1 ) − arctan
arctan(
t − 1 ) − π 3
So (^) ∫ (^) ∞
4
dx x
x − 1
= lim t →∞
arctan(
t − 1 ) − π 3
(π 2
π 3
π 3
3 x^2 + 2 sin( 2 x ) + 2
3 x^2
Since
∫ (^) ∞
1
dx 3 x^2
∫ (^) ∞
1
dx x^2 is convergent, by the comparison theorem we conclude that ∫ (^) ∞ 1
3 x^2 + 2 sin( 2 x ) + 2 dx
is convergent.
1
(∫ (^) π/ 6
0
(sin θ )^2 d θ +
∫ (^) π/ 2 π/ 6
( 1 − sin θ )^2 d θ
We will need dx dt
= 15 t^4 + 4 and dy dt
= 6 t + 2. Then dy dx
dy dt dx dt
6 t + 2 15 t^4 + 4
so at t = 0, dy dx
For d
(^2) y dx^2 , we will need d dt
dy dx
d dt
6 t + 2 15 t^4 + 4
( 15 t^4 + 4 )( 6 ) − ( 6 t + 2 )( 60 t^3 ) ( 15 t^4 + 4 )^2 Then d^2 y dx^2
d dt
dy dx
dx dt
( 15 t^4 + 4 )( 6 ) − ( 6 t + 2 )( 60 t^3 ) ( 15 t^4 + 4 )^3 and at t = 0, we find d^2 y dx^2
sin θ θ
L’H = lim θ → 0
cos θ 1
= 1 so we must choose A = 1. (b) (The sketch will be shown in class.)
r =
1 + cos θ r + r cos θ = 1 r = 1 − r cos θ r^2 = ( 1 − r cos θ )^2 x^2 + y^2 = ( 1 − x )^2 = 1 − 2 x + x^2 y^2 = 1 − 2 x
This last equation shows that the curve is a parabola.
∫ (^2) π
0
r^2 +
dr d θ
d θ =
∫ (^2) π
0
e θ^ /^2
e θ^ /^2 2
d θ =
∫ (^2) π
0
e θ^ d θ =
∫ (^2) π 0
e θ^ /^2 =
5 e θ^ /^2
∣∣^2 π 0 =
5 ( e π^ − 1 )