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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