2002 Test 1 solution

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10 Apr 2012
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MAT 137Y 2002-2003, Test 1 Solutions
(12%) 1. Solve the inequality x2
x
1
x2
2x
1
Express your answer in interval notation.
Note that
x2
x
1
x2
2x
1

x2
x
1
x2
2x
1
0

x2
x
1

x2
2x
x2
2x
0

x
1
x
x
2
0
Therefore
x
1

x
x
2

is undefined when x
0
2, and equality holds when x
1. A table of
signs is created as follows:
x
1
1
x
0 0
x
2x
2
x
1
 
1
x
 
 
1
x
2
 
x
1

x
x
2

So the inequality is true if and only if x

1
0

2
.
2. Evaluate the following limits. (No proofs are required.)
(6%) (i) lim
x
2
x3
8
x2
4.
Factoring the numerator and denominator,
lim
x
2
x3
8
x2
4
lim
x
2
x
2
x2
2x
4
x
2
x
2
lim
x
2
x2
2x
4
x
2
3
(6%) (ii) lim
x
1
x2

x
1
x.
Let u
x. Since x
1, then u
1, so we have
lim
u
1
u4
u
1
u2
lim
u
1
u
u3
1
1
u
1
u
lim
u
1
u
u
1
u2
u
1
1
u
1
u
lim
u
1
u
u2
u
1
1
u
3
2
Alternate solution: Multiply top and bottom by x2
x. Then
lim
x
1
x4
x
1
x
x2
x
lim
x
1
x
x
1
x2
x
1
1
x
x2
x
lim
x
1
x
x2
x
1
x2
x
3
2
(6%) (iii) lim
x
0
sin3x
tan5x.
Rewriting tan5xusing sines and cosines,
lim
x
0
sin3x
tan5x
lim
x
0
sin3x
cos5x
sin5x
lim
x
0
3
5
sin3x
3x
cos5x
5x
sin5x
3
5
since lim
t
0
sint
t
1.
1
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