10.
lim
y
→2
}
y
2
1
y
1
5
y
2
1
6
}
5
}
2
2
1
2
5
1
(2
2
)
1
6
}
5
}
2
4
0
}
5
5
Graphical support:
[
2
3, 3] by [
2
5, 10]
11.
lim
y
→
2
3
}
y
2
1
y
2
4
2
y
3
1
3
}
55
}
0
6
}
5
0
Graphical support:
[
2
5, 5] by [
2
5, 5]
12.
lim
x
→
1/2
int
x
5
int
}
1
2
}
5
0
Note that substitution cannot always be used to find limits
of the int function. Its use here can be justified by the
Sandwich Theorem, using
g
(
x
)
5
h
(
x
)
5
0 on the interval
(0, 1).
Graphical support:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
13.
lim
x
→
2
2
(
x
2
6)
2/3
5
(
2
2
2
6)
2/3
5
ˇ
3
(
2
w
8
w
)
2
w
5
ˇ
3
6
w
4
w
5
4
Graphical support:
[
2
10, 10] by [
2
10, 10]
14.
lim
x
→
2
ˇ
x
w
1
w
3
w
5
ˇ
2
w
1
w
3
w
5
ˇ
5
w
Graphical support:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
15.
lim
x
→
0
(
e
x
cos
x
)
5
e
0
cos 0
5
1
?
1
5
1
Graphical support:
[
2
4.7, 4.7] by [
2
3.1, 3.1]
16.
lim
x
→
p
/2
ln (sin
x
)
5
ln
1
sin
}
p
2
}
2
5
ln 1
5
0
Graphical support:
[
2
p
,
p
] by [
2
3, 1]
17.
You cannot use substitution because the expression
ˇ
x
w
2
w
2
w
is not defined at
x
52
2. Since the expression is not
defined at points near
x
2, the limit does not exist.
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.
 Spring '08
 GERMAN

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