Download Math 112 Fall 2006 Exam 3 Solutions: Limits, Integrals, and Trigonometric Functions and more Exams Calculus in PDF only on Docsity!
Math 112 Fall 2006 Exam 3 Solutions Some solutions include the details; others have just the final answer.
- (a) lim x → 0 +
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
) (^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
3 ( 2 − 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
- We know − 1 ≤ sin( 2 x ) ≤ 1 so − 2 ≤ 2 sin( 2 x ) ≤ 2 and therefore 0 ≤ 2 sin( 2 x ) + 2 ≤ 4. (We really only need the result that 2 sin( 2 x ) + 2 ≥ 0.) Then 3 x^2 + 2 sin( 2 x ) + 2 ≥ 3 x^2 so 0 ≤
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
- ( 0 , 0 ) and ( 1 / 2 , 1 / 2 ).
- The area is A = 2
(∫ (^) π/ 6
0
(sin θ )^2 d θ +
∫ (^) π/ 2 π/ 6
( 1 − sin θ )^2 d θ
- First, we find that the curve is at ( 5 , 3 ) when t = 0.
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
- (a) lim θ → 0
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.)
- (a) From the relation tan θ = y x , we conclude y x = 5, or y = 5 x. This is a straight line through the origin with slope 5. (b) We will use the r^2 = x^2 + y^2 and x = r cos θ in the following:
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 )
