MAT 137Y Solutions to Term Test 1, 2001-2002

1. Evaluate the following limits.

(5%) (i) lim

x

0

x

9

x

3.

Multiplying numerator and denominator by the conjugate,

lim

x

0

x

9

x

3

lim

x

0

x

9

x

3

9

x

3

9

x

3

lim

x

0

x

9

x

3

9

x

9

lim

x

0

x

9

x

3

x

lim

x

0

9

x

3

6

(5%) (ii) lim

x

1

x1

3

1

x1

2

1.

Let x

u6. Then

lim

x

1

x1

3

1

x1

2

1

lim

u

1

u2

1

u3

1

lim

u

1

u

1

u

1

u

1

u2

u

1

lim

u

1

u

1

u2

u

1

2

3

(5%) (iii) lim

x

1

cos

π

2x

x

1.

Let u

x

1. Then

lim

x

1

cos

π

2x

x

1

lim

u

0

cos

π

2

u

1

u

lim

u

0

cos

π

2u

π

2

u

lim

u

0

cos π

2ucos π

2

sin π

2usin π

2

u

lim

u

0

sin π

2u

u

π

2lim

u

0

sin π

2u

π

2u

π

2

2. Solve the following inequalities.

(8%) (i) x3

x2

1

x3

x2

x

1

1.

Isolating one side, we get

x3

x2

1

x3

x2

x

1

x3

x2

x

1

x3

x2

x

1

0

x

x3

x2

x

1

0

x

x

1

2

x

1

0

Now we consider the sign of each factor:

x

1

1

x

0 0¡x¡1 x

1

x

x

1

2

x

1

x

x

1

2

x

1

At x

0 we have equality, and at x

1 the rational function is undeﬁned. Therefore the set of

all solutions which satisfy the inequality are x

∞

1

0

1

1

∞

.

(7%) (ii)

x

1

2

x

3

.

One can take cases on the values of x

1 and x

3, but it is simpler to note that both sides are

positive, so we are allowed to square both sides, giving us

x

1

2

x

3

!

x2

2x

1

4

x2

6x

9

"#

x2

2x

1

4x2

24x

36

#

3x2

26x

35

0

1