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MATH- 102/Spring 2005, Final Exam Solutions
1) Each piece of this piecewise defined function f is continuous, so f is continuous if and only
if f is continuous at x = 0. f is continuous at x = 0 if and only if ( ) =
→+
f x x 0
lim lim ( ) ( ) 0
0
f x f x
→−
→+
f x x 0
lim (^) = =
→ + 2
limarctan 0
x (^) x 2
lim 0
→−
x a a a x
2
2 3
(^02)
3
0 2
3 1 1 1 2
2 2
= = =
x x
x
x
x
t dt x x dx
d
dx
df x f x t dt
2 3 ⋅ ⋅ = + ⋅ ⋅ = ⋅
So, the equation of the tangent line at x = 2 is y − f ( 2 ) = 6 ⋅ 2 ⋅( x − 2 ).
3) If ( )
v ( ) x
ux
f x = then ( )
2 v x
u x vx v x ux f x
x
u x x u x x
= sin ln ⇒ ′ =cosln ⋅ and ( ) ( ) ( ) ( )
x x x v x = cos e ⇒ v ′ x =−sin e ⋅ e. So
2 cos
cos sin sinln
cosln
x
x x x
e
e e e x x
x
f x
′ (^) = and
( ) e ( ) e
e
e
e e f cos
cos
cos
cos
cosln 1 1 cos sin sinln 1 1 1 2 2
4) Since
( ) (^)
+ ⋅
+
= =
+^ n
n n n
n n
e e n
n
n 3 2 ln 3 ln 2 2 3
and ( )
x f x = e is a continuous function, we have
( ) ( )
+ ⋅
+ ⋅
→∞ →∞
→∞ = =
+^ n
n n n
n n
n
n
n
n e e n
n
3 lim^2 ln
3 2 ln 2
lim
lim.
( ) 6 3
lim
lim
ln
lim
lim 2 ln
2
2
'
→∞ →∞ ↑ →∞ →∞ n
n
n
n n
n
n
n
n
n
n n n n LH
n n
. So,
6
2 3 lim e n
n
n
n
→∞
Another solution:
Since
k
n
n
e n
k =
→∞
lim 1 , ( ).
lim 1
lim 1
lim
3 2 6
2 2 2
e e n n n
n
n
n
n
n
n
n
→∞ →∞ →∞
5) In order to evaluate this integral, we will use Integration by Parts;
x u = x , dv = e
2
x du = 2 x ⋅ dx , v = e and
( ) ( ) ∫ ∫ ∫ ∫
1
0
0
2 1
1
0
2 u dv u v v du x e dx x e e 0 2 x e dx
x x
b
a
x a
b
b
a
To evaluate ∫
1
0
2 x e dx
x , we will use integration by parts again;
u x dv e dx
x = 2 , = ⋅
x du = 2 ⋅ dx , v = e and
2 ( 2 ) 2 2 0 ( 2 ) 2 ( 2 2 ) 2 0
1
1
0
0
1
1
0
∫ ∫
x e dx x e e dx e e e e
x x x x .
Therefore 2
1
0
2 ⋅ ⋅ = − ∫ x e dx e
x
2
2 2
x x if x
x x if x
y x
y x
2 2 0 (^2 ) (^1 )^012
2 2 x − =− x ⇒ x + x − = ⇒ x + ⋅ x − = ⇒ x = or x =−.
2 2 0 ( 2 ) ( 1 ) 0 1 2
2 2 x − = x ⇒ x − x − = ⇒ x − ⋅ x + = ⇒ x =− or x =.
8-a) Since the Harmonic Series (^) ∑
∞
= 1
n n^
is divergent and = <∞
lim
3
2
n
n n
n n
n
, by The Limit
Comparison Test, this series is divergent.
b) Since cos 0 1 0
lim cos = = ≠
n → ∞ (^) n
, by the n-th Term Test, this series is divergent.
c)
( ) ( )
( ( ))
( )
( ) ( )
( ( ))
( )
n n
n
n
n n
n
nn
n
n n
a
a n
n
n
n
n
n n
n n (^) ⋅ ⋅
→∞
→∞
→ ∞ 2!
lim
lim lim
1
1
1
( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( ) ( )
( ) ( )
lim 2 2 2 1
lim 2!
lim = < ⋅ + ⋅ +
→∞ →∞ →∞ n n
n n
n n
n n
nn
n
n n n
n n n n n n n
n
n
So, by The Ratio Test , this series is convergent.
9) Let n
n
n n
x a ⋅ 3
=. By the Ratio Test, we have
lim 1 3 3
lim
lim lim 1
1 1
1
1 x
n
x n
x n
x n
n
x
n
x
a
a
n n n
n n
n
n
n
n
n
n n
n n
→∞
→∞
→∞
ρ and
< 1 ⇒ < ⇔ x < ⇔− < x <
x
ρ. So the radius of convergence is 3.
When n
x an
= 3 ; = and (^) ∑
∞
= 1
n n^
diverges, since this is a p-series with 1 2
p = <.
When
( ) ( )
n n
x a
n
n
n
n
= − = and
( ) ∑
∞
=
1
n
n
n
converges by the Alternating Series Test.
∴ (^) The interval of convergence is [− 3 , 3 ).
