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Math 112 Fall 2006 Exam 3 Solutions: Limits, Integrals, and Trigonometric Functions, Exams of Calculus

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.

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

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Math 112 Fall 2006 Exam 3 Solutions
Some solutions include the details; others have just the final answer.
1. (a) lim
x0+
1
ex11
x=lim
x0+
xex+1
x(ex1)
L’H
=lim
x0+
1ex
xex+ex1
L’H
=lim
x0+ex
(x+2)ex=1
2
(b) lim
xx2
xxhas the indefinite form “1”. Let y=x2
xx, so lny=xlnx2
x. Then
lim
xlny=lim
xxlnx2
x=lim
x
lnx2
x
1
x
L’H
=lim
xx
x22
x2
1
x2
=lim
x2x
x2=2
Then
lim
xx2
xx
=lim
xy=lim
xelny=e2
(c) lim
θ0
sinθ
1+cosθ=0
2. Z
2
dx
(x7)4=lim
t7Zt
2
dx
(x7)4+lim
t7+Z8
t
dx
(x7)4+lim
tZt
8
dx
(x7)4
and
lim
t7Zt
2
dx
(x7)4=lim
t71
3(x7)3
t
2=lim
t71
3(t7)3+1
3(27)3=
That is, the limit is not a finite number, so the improper integral is divergent.
3. Z
4
dx
xx1=lim
tZt
4
dx
xx1.
Let u=x1= (x1)1/2. Then du =1
2(x1)1/2,u2+1=x, and 2udu =dx. So
Zt
4
dx
xx1=Zt1
3
2
u2+1du =2(arctan(t1)arctan3) = 2arctan(t1)
π
3
So Z
4
dx
xx1=lim
t2arctan(t1)
π
3=2π
2
π
3=π
3
4. We know
1sin(2x)1
so
22sin(2x)2 and therefore 0 2sin(2x) + 24.
(We really only need the result that 2sin(2x) + 20.) Then
3x2+2sin(2x) + 23x2
so
01
3x2+2sin(2x) + 21
3x2
Since Z
1
dx
3x2=1
3Z
1
dx
x2is convergent, by the comparison theorem we conclude that
Z
1
1
3x2+2sin(2x) + 2dx
is convergent.
1
pf2

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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.

  1. (a) lim x → 0 +

ex^ − 1

x

= lim x → 0 +

xex^ + 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

  • lim t → 7 +

∫ (^8) t

dx ( x − 7 )^4

  • (^) t lim→∞

∫ (^) 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

  1. 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

  1. ( 0 , 0 ) and ( 1 / 2 , 1 / 2 ).
  2. The area is A = 2

(∫ (^) π/ 6

0

(sin θ )^2 d θ +

∫ (^) π/ 2 π/ 6

( 1 − sin θ )^2 d θ

  1. 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

  1. (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.)

  1. (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 )