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Quick Review 3.1
1.
lim
h
→
0
}
(2
1
h
4
)
2
2
4
}
5
lim
h
→
0
}
(4
1
4
h
1
h
h
2
)
2
4
}
5
lim
h
→
0
4
1
h
5
4
1
0
5
4
2.
lim
x
→
2
+
}
x
1
2
3
} 5 }
2
1
2
3
} 5 }
5
2
}
3.
Since
}
)
y
y
)
}52
1 for
y
,
0, lim
y
→
0
2
}
)
y
y
)
}52
1.
4.
lim
x
→
4
}
ˇ
2
x
x
w
2
2
8
2
} 5
lim
x
→
4
5
lim
h
→
4
2(
ˇ
x
w
1
2)
5
2(
ˇ
4
w
1
2)
5
8
5.
The vertex of the parabola is at (0, 1). The slope of the line
through (0, 1) and another point (
h
,
h
2
1
1) on the parabola
is
}
(
h
2
h
1
2
1)
0
2
1
}5
h
. Since lim
h
→
0
h
5
0, the slope of the line
tangent to the parabola at its vertex is 0.
6.
Use the graph of
f
in the window [
2
6, 6] by [
2
4, 4] to find
that (0, 2) is the coordinate of the high point and (2,
2
2) is
the coordinate of the low point. Therefore,
f
is increasing
on (
2‘
, 0] and [2,
‘
).
7.
lim
x
→
1
1
f
(
x
)
5
lim
x
→
1
1
(
x
2
1)
2
5
(1
2
1)
2
5
0
lim
x
→
1
2
f
(
x
)
5
lim
x
→
1
2
(
x
1
2)
5
1
1
2
5
3
8.
lim
h
→
0
1
f
(1
1
h
)
5
lim
x
→
1
1
f
(
x
)
5
0
9.
No, the two onesided limits are different (see Exercise 7).
10.
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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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