MATH 118 SAMPLE FINAL SOLUTIONS
FINAL EXAM GIVEN MAY 4, 2004
Problem 1.
Let
f
(
x
)
=
1
x
+
1
,
∞
<
x
<

1
1
+
x
1

x
,

1
≤
x
<
1
2
x
2

1
x
2
+
1
1
≤
x
<
+∞
a)
(5 points) List all the values of x for which f
(
x
)
is not continuous.
Calculate the following limits. If the limit is infinite, indicate whether it is
+∞
or
∞
.
b)
(3 points)
lim
x
→
1
+
f
(
x
)
c)
(3 points)
lim
x
→
1

f
(
x
)
d)
(3 points)
lim
x
→
1
+
f
(
x
)
e)
(3 points)
lim
x
→
1

f
(
x
)
f)
(3 points)
lim
x
→+∞
f
(
x
)
g)
(3 points)
lim
x
→∞
f
(
x
)
Solution.
a) You really need to complete the rest of the parts of this problem be
fore you do this part. Go ahead and do that now, then come back and
read this solution.
OK, you’re back? The function is not continuous at
x
= 
1 and
at
x
=
1. This is because the limits from the left and from the right
aren’t equal at these two points. (Part b) and c): 1
/
2
= +∞
; Part d)
and e): 0
= ∞
.) At every other point,
f
(
x
)
is a rational function
nearby
x
, whose denominator is never 0, and is therefore continuous
at
x
.
b) As
x
→
1
+
, i.e. as
x
approaches 1 from the right, the values of
x
are
(
2
x
2

1
)/(
x
2
+
1
)
, and by the limit theorems this approaches
(
2

1
)/(
1
+
1
)
=
1
/
2:
lim
x
→
1
+
f
(
x
)
=
1
2
.
c) As
x
→
1

, i.e. as
x
approaches 1 from the left, the value of
f
(
x
)
will be
(
1
+
x
)/(
1

x
)
as soon as
x
is close enough to 1.
The
numerator of this approaches 2 and the denominator approaches 0;
this increases without bound in absolute value. Since the quotient is
always positive for
x
in this range, therefore
lim
x
→
1

f
(
x
)
= +∞
.
d) As
x
→ 
1
+
, i.e. as
x
approaches

1 from the right, the values of
f
(
x
)
are governed by the middle condition,
f
(
x
)
=
1
+
x
1

x
.
The numerator approaches 0 and the denominator approaches 2, there
fore by the limit theorems the quotient approaches 0:
lim
x
→
1
+
f
(
x
)
=
0
.
e) As
x
→ 
1

, i.e. as
x
approaches

1 from the left, the values of
f
(
x
)
are govered by the first condition,
f
(
x
)
=
1
1
+
x
.
The numerator is constantly 1 and the denominator approaches 0, so
the quotient “blows up” (becomes infinite in absolute value). Since
the quotient is always negative for
x
<

1, therefore
lim
x
→
1

f
(
x
)
= ∞
.
f) As
x
→ +∞
, the value of
f
(
x
)
is
(
2
x
2

1
)/(
x
2
+
1
)
. We rewrite
this by dividing numerator and denominator by
x
2
,
f
(
x
)
=
2

1
x
2
1
+
1
x
2
.
In this form it’s clear that 1
/
x
2
→
0, and thus
lim
x
→∞
f
(
x
)
=
2
1
=
2
.
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SOLUTIONS
g) As
x
→ ∞
, the value of
f
(
x
)
is 1
/(
1

x
)
, and therefore
lim
x
→∞
f
(
x
)
=
0
.
(Divide the numerator and denominator by
x
and use the fact that
1
/
x
→
0 as
x
→ ∞
.)
Problem 2.
Find an equation of the tangent line to the curve x
3

3
y
3
+
2
x
+
y
=
1
at the point x
=
1
, y
=
1
.
Solution.
To find the tangent line we need to find
dy
/
dx
, which we do by
implicit differentiation:
d
dx
x
3

3
y
3
+
2
x
+
y
=
d
dx
1
,
i.e.
3
x
2

9
y
2
dy
dx
+
2
+
dy
dx
=
0
which we solve for
dy
/
dx
:
dy
dx
=

3
x
2

2

9
y
2
+
1
.
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 Fall '07
 Vorel
 Math, Calculus, Critical Point

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