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MATH- 102 / Spring2005, MT- 1 Solutions
1- a)
lim 5 3
lim 2
lim 2
2
2 2
2 2
2
2
2
→− →− →− x x
x
x
x
x
x
x
x
x x x
lim
2 5 3
lim 2 2 2
2
2
→ − →− x
x
x x
x
x x
Another Solution:
Since lim ( 5 3 ) 0
2 2
→−
x x
, lim( 2 ) 0
2
→ −
x x
2 x + − and x + 2 are differentiable
functions at x =− 2 , we have
lim 2
lim 2 2 '
2
2
→ − ↑ →− x
x
x
x
x LH
x
b) x + 1 = 0 ⇒ x =− 1.
→ −+^ →−+ 1
lim lim (^1 1) x
x y x x
and (^) =+∞
−
→ −−^ →−+ 1
2 6 lim lim (^1 1) x
x y x x
Hence x =− 1 is the vertical asymptote.
Since 2 1 1
lim 1
lim lim =
→∞ →∞ →∞
x
x
x
x y x x x
, y = 2 is the horizontal asymptote.
c- i) ( )
2
2
2
2
0
2
2 0 2
2
0 0
cos sin
sin lim
cos
sin
sin sin lim tan
sin lim lim x x
x
x
x
x
x
x x
x
x f x x x x x
→ → → →
cos 1111 1. sin
sin sin lim
2 2
2
0
→
x x
x
x
x
x
x
x
Another way for finding this limit;
Since lim sin 0
2 0
→
x x
, lim tan 0
2 0
→
x x
, x
2 sin and
2 tan x are differentiable functions at
x = 0 , we have
( ) cos cos 111 1
sin lim sec 2
2 sin cos lim tan
sin lim lim
2 2 0 2 2 0 '
2
2
0 0
→ → ↑ → →
x x x
x
x x
x x
x
x f x x x LH
x x
ii) f is continuous at x = 0 if and only if lim ( ) ( ) 0
0
f x f x
→
Since f ( ) x
x 0
lim →
=1 and ( )
2 f 0 = c , we have 1
2 c =.
2 c = ⇒ c =±.
2-a) ( )
2 1 cos
cos 1 cos sin 1 sin
1 cos
1 sin 2
r ′ = ⋅
⇒ r ′^ ( ) 0 =
2 1 cos 0
cos 0 1 cos 0 sin 0 1 sin 0
1 cos 0
1 sin 0 2
⇒ r ′ ( ) 0 =
b) 2 y y 2 y 2 x y 2 x 4 ( x y x ) ( y 3 x y y 2 x )
3 23 3 2 ⋅ ′+ + ⋅ ′+ = ⋅ ⋅ + ⋅ + ⋅ ⋅ ⋅ ′+
When x = 1 and y = 0 :
3 ⋅ y ′ + = ⋅ ⋅ = ⇒ y ′= = So the slpoe of the line tangent to this
curve at ( 1 , 0 )is 3.
c) t
t
dt
dx
dt
dy
dx
dy
sec
2
= =. When 4
t = :
( )
π π π
π
sec 2
2
dx
dy .
2
x
= x ⋅ h ⇒ x ⋅ h = ⇒ h =
c^ (^ x )^ = 10 ⋅( 3 x + 3 x )^ + 6 ⋅(^2 x ⋅ h + 6 x ⋅ h )
2 2
⇒ c ( x )
2
x
= x + x ⋅
x
c x x
2 ⇒ = +.
x ∈ ( 0 ,∞).
3 2
′ (^) = − = ⇒ x = ⇒ x = x
c x x m.
( ) ( )
( ) ( ) 2
2 3 2 2 2
3
2
x
x x x x x x x
x
x
c x x
So c ′( )^ x > 0 when x > 2 and c ′ ( )^ x < 0 when 0 < x < 2. Therefore at x = 2 , c ( ) x has its
minimum value. Dimensions are 2 × 6 × 5.
Alternative way of determining that c ( x ) is minimum when x = 2 ;
3
x
c ′′^ x = + ⋅. So
3 c ′ ′ = + ⋅ > ( i.e. we have minimum at x = 2 ).
