MAT 137Y, 2008-2009 Winter Session, Term Test 2 Solutions

1. Evaluate the following limits, or show the limit does not exist.

(8%) (i) lim

x→0cotx−1

x.

Let Lbe the limit. Then

L=lim

h→0

cosx

sinx−1

x=lim

h→0

xcosx−sinx

xsinx

H

=lim

h→0

cosx−xsinx−cosx

sinx+xcosx

=lim

h→0−xsinx

sinx+xcosx

H

=lim

h→0−sinx−xcosx

cosx+cosx−xsinx=0

2=0.

(8%) (ii) lim

x→∞

x2+cosx

x2+2009.

Since −1≤cosx≤1, we have x2−1

x2+2009 ≤x2+cosx

x2+2009 ≤x2+1

x2+2009,

so by the squeeze theorem, lim

x→∞

x2+cosx

x2+2009 =1.

(Note, applying L’Hˆ

opital’s Rule twice yields lim

x→∞

2−cosx

2, which does not exist, but

L’Hˆ

opital’s Rule does not apply in this situation.)

(15%) 2. Recall that the volume of a right circular cone is V=1

3πr2h, where ris the radius of the

(circular) base, and his the height of the cone. Find the largest possible volume of a right

circular cone that is inscribed in a sphere of radius R. Make sure that your answer yields an

absolute maximum.

This question is posed in SHE 4.5 #43, which provides a diagram. We maximize V=1

3πr2h.

Since r2+ (h−R)2=R2by the Pythagorean Theorem, we solve for r2to get

V(h) = 1

3πR2−(h−R)2)h=1

3πR2h−1

3π(h−R)2h,

where h∈(0,2R). We solve for the absolute maximum; differentiating,

V0(h) = 1

3πR2−1

3h(h−R)2+2h(h−R)i=0=⇒R2−(h−R)2−2h(h−R) = 0

=⇒R2−h2+2hR −R2−2h2+2hR =0=⇒ −3h2+4hR =0=⇒h=4R

3.

Since lim

h→0V(h) = lim

h→2RV(h) = 0, the critical point h=4R

3must yield maximum volume.

Therefore the maximum volume is

V(4R

3) = 1

3πR2−R2

9·4R

3=32πR3

81 .

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