Practice Quiz IC for Calculus I, MATH 1501
(I)
Let
f
(
x
) be given by
f
(
x
) =
1

x
1 +
x
for
x
6
= 0.
(a) Find all
x
such that

f
(
x
)

f
(1)

<
1.
(b) Find a
δ
( )
>
0 such that

x

1

< δ
( )
⇒ 
f
(
x
)

f
(1)

<
.
and justify your answer.
ANSWERS
For (a),
x >
0. for (b), here are two ways:

f
(1 +
h
)

f
(1)

=

h


2 +
h

First Way
Apply

a
+
b
 ≥ 
a
  
b

with
a
= 2,
b
=
h
to get

2 +
h
 ≤
2
 
h

.
Hence

f
(1 +
h
)

f
(1)
 ≤

h

2
 
h

Now for

h

<
2, 2
 
h

>
0, so

h

2
 
h

<
⇐⇒ 
h

<
2
 
h

⇐⇒ 
h

<
2
1 +
This gives
δ
( ) = min
2
,
2
1 +
.
Second Way
Inpose the condition

h

<
1, under which
1
<
2 +
h <
3
and hence

2 +
h

>
1. Decreasing the denominator gives

f
(1 +
h
)

f
(1)
 ≤

h

2
 
h

<

h

and so if

h

<
, we have what we want. Therefore,
δ
( ) = min
{
1
,
}
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
(II)
Which of the following limits exist? If they don’t exist, explain why not. If they do,
compute the value of the limit.
(a)
lim
x
→
0
1
x
2
√
2 +
x

√
2
(b)
lim
x
→
1
cos
(
x

1)
2
(
x
2

1)
2
(c)
lim
x
→
2
x
2
+ 5
x
+ 6
(
x
+ 2)
2
(d)
lim
x
→
2
(
x
2
+ 5
x
+ 6)
2
(
x
+ 2)
2
ANSWERS
(a):

1, (b) cos(1
/
4), (c) does not exist, (d) 1.
(III):
Consider the following function defined for
x
6
= 0:
f
(
x
) =
1
x
1
x

1
sin(
x
)
.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '08
 N/A
 Calculus, Practice Quiz IC, lim x→−2

Click to edit the document details