Math 137
Assignment No. 8  Solutions
Fall 2010
1.
(a) lim
x
→
0
parenleftbigg
cot
x

1
x
parenrightbigg
. This is an indeterminate form of the type
∞  ∞
.
If we put the two functions over a common denominator, i.e.,
lim
x
→
0
parenleftbigg
cot
x

1
x
parenrightbigg
= lim
x
→
0
x
cos
x

sin
x
x
sin
x
= lim
x
→
0
f
(
x
)
g
(
x
)
:
a
0
0
indeterminate form.
If we now differentiate numerator and denominator, we obtain
lim
x
→
0
f
′
(
x
)
g
′
(
x
)
= lim
x
→
0

x
sin
x
sin
x
+
x
cos
x
,
another
0
0
indeterminate form.
Differentiating again (with faith),
lim
x
→
0
f
′′
(
x
)
g
′′
(
x
)
= lim
x
→
0

sin
x

x
cos
x
cos
x
+ cos
x

x
sin
x
= 0.
We can now work backward and conclude that
lim
x
→
0
parenleftbigg
cot
x

1
x
parenrightbigg
= lim
x
→
0
f
(
x
)
g
(
x
)
= lim
x
→
0
f
′
(
x
)
g
′
(
x
)
= lim
x
→
0
f
′′
(
x
)
g
′′
(
x
)
= 0.
(b)
lim
x
→∞
(
xe
1
/x

x
). This is an indeterminate form of the type
∞  ∞
.
If we factor the expression and then rewrite it as a product, i.e.,
(
xe
1
/x

x
) =
x
(
e
1
/x

1) =
e
1
/x

1
1
/x
,
we have produced a
0
0
indeterminate form since
e
1
/x
→
e
0
= 1 as
x
→ ∞
. It is convenient to make
the change of variable
y
=
1
x
so that
lim
x
→∞
e
1
/x

1
1
/x
= lim
y
→
0
e
y

1
y
.
Now differentiate numerator and denominator to find that
lim
y
→
0
e
y

1
y
= lim
y
→
0
e
y
1
= 1.
Therefore
lim
x
→∞
(
xe
1
/x

x
) = 1.
2. Required to find
a
such that
lim
x
→∞
parenleftbigg
x
+
a
x

a
parenrightbigg
x
=
e.
(1)
•
Method No. 1: Using the logarithm.
If we take the natural logarithm of the LHS of (1),
ln
parenleftbigg
x
+
a
x

a
parenrightbigg
x
=
x
ln
parenleftbigg
x
+
a
x

a
parenrightbigg
=
ln
bracketleftbig
(
x
+
a
)(
x

a
)
−
1
bracketrightbig
x
−
1
.
From (1), we would like the limit as
x
→ ∞
of this expression to be ln
e
= 1.
In the limit
x
→ ∞
, the rightmost term is a
0
0
indeterminate form. Rewrite this term as follows,
f
(
x
)
g
(
x
)
=
ln(
x
+
a
)

ln(
x

a
)
x
−
1
,
1
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which implies that
f
′
(
x
)
g
′
(
x
)
=
(
x
+
a
)
−
1

(
x

a
)
−
1

x
−
2
=
· · ·
=
2
ax
2
(
x
+
a
)(
x

a
)
.
Then
lim
x
→∞
f
′
(
x
)
g
′
(
x
)
= lim
x
→∞
2
ax
2
(
x
+
a
)(
x

a
)
= lim
x
→∞
2
a
(1 +
ax
−
1
)(1

ax
−
1
)
=
· · ·
= 2
a.
Recalling that this limit must be equal to ln
e
= 1, we have 2
a
= 1, implying that
a
=
1
2
.
•
Method No. 2: Using the limit definition for
e
.
The appearance of
x
both inside the brackets
as well as in the exponent suggests that the limit definition for
e
may possibly be involved, i.e.,
lim
n
→∞
parenleftbigg
1 +
1
n
parenrightbigg
n
=
e
(2)
We’ll return to this equation. In the original expression, let
x

a
=
y
which implies that
x
=
y
+
a
, so
that
x
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 Math, Derivative, lim, Mathematical analysis, Limit of a function, Concave function

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