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MAT 137Y, Solutions to Term Test 2, 2001-2002

1. Evaluate the following expressions.

(5%) (i) d

dx sin2

1

x

cosx

x

. (Do NOT simplify your answer.)

2sin

1

x

cosx

x

cos

1

x

cosx

x

x

1

sinx

2

x

cosx

1

x

cosx

x2.

(5%) (ii) lim

x

π

1

cosx

π

x

2.

Let Lbe the limit. Applying L’Hˆopital’s Rule twice, L

lim

x

π

sinx

2

π

x

lim

x

π

cosx

2

1

2

(5%) (iii) lim

x

∞

x2

x2

sinx.

Dividing both sides by x2, we have lim

x

∞

1

1

sinx

x2

. Since sinx

x2

0 by the squeeze theorem, it follows that

the limit is 1.

Note that if you apply L’Hˆopital’s Rule twice, we get lim

x

∞

2

2

sinx, which does not exist. However, you

cannot apply L’Hˆopital’s Rule when the resulting limit does not exist.

(5%) (iv) lim

x

0

sinx

x2

cosx

x

.

Let Lbe the limit. Obtaining a common denominator and applying L’Hˆopital’s Rule once,

L

lim

x

0

sinx

xcosx

x2

lim

x

0

cosx

xsinx

cosx

2x

lim

x

0

sinx

2

0

For the rest of this question, suppose gand hare functions such that

h

0

2

h

0

5

g

0

3

g

0

2

g

2

4

g

2

1

h

1

2

h

1

3

h

2

6

h

2

3

(3%) (v)

g

h

0

g

h

0

h

0

g

2

5

4

5

20

(3%) (vi)

h

g

2

h

g

2

g

2

h

1

4

3

4

12

(15%) 2. Find the equation of the line through the point

3

5

which cuts off the least area from the ﬁrst quadrant.

Let

x

0

and

0

y

be the points on the line which intersect the point

3

5

. Then the slope mcan be calculated

in two ways, giving us

m

5

y

3

5

3

x

5

y

15

3

x

1

y

5

15

3

x

1

Then the area of the triangle is A

x

1

2xy

1

2x

5

15

3

x

1

5

2x

15

2x

3

x, for x

3

∞

. Differentiating,

A

x

5

2

15

2

3

3

x

2

5

3

x

2

15

2

3

x

2

0

3

x

2

3

x

0 or x

6

1