HOMEWORK #15 – SOLUTIONS Page 305 48) 49) cos x ⎞
⎛ 1
⎛ 1 − cos x ⎞ H
lim ( csc x − cot x ) = lim ⎜
−
=
lim
⎟
⎜
⎟=
x→0
x→0 ⎝ sin x
sin x ⎠ x→0 ⎝ sin x ⎠
sin x
lim
=0
x→0 cos x
lim
x→∞
(
⎛ 2
x2 + x + x ⎞
x + x − x = lim ⎜ x + x − x ⋅
⎟=
2
x→∞
x + x + x⎠
⎝
)
2
⎛ x2 + x − x ⎞
x
lim ⎜
=
lim
= lim
⎟⎠ x→∞ 2
2
x→∞ ⎝
x +x+x
x + x + x x→∞
1
1
=
2
1
1+ +1
x
54) lim+ ( tan 2x ) ⇒ y = ( tan 2x ) ⇒ ln y = x ln tan 2x =
x
x
x→0
ln tan 2x
1
x
1
⋅ sec 2 2x ⋅ 2
ln tan 2x
lim+ ln y = lim+
= lim+ tan 2x
=
1
−1 2
x→0
x→0
x→0
x
x
−2x 2 cos 2x
2x
−x
lim+
=
lim
⋅
lim
= 1⋅ 0 = 0 ⇒
2
x→0 sin 2x cos 2x
x→0 + sin 2x x→0 + cos 2x
H
lim+ ( tan 2x ) = e0 = 1
x
x→0
56) a⎞
⎛
lim ⎜ 1 + ⎟
x→∞ ⎝
x⎠
62) bx
a⎞
⎛
⇒ y = ⎜1 + ⎟
⎝
x⎠
a⎞
⎛
b ln ⎜ 1 + ⎟
⎝
x⎠
lim ln y = lim
x→∞
x→∞
1
x
bx
⎛
⎞
⎜ 1 ⎟ ⎛ −a ⎞
b⎜
a ⎟ ⎜⎝ x 2 ⎟⎠
1
+
⎜⎝
⎟⎠
H
x
= lim
=
x→∞
−1 2
x
ab
a⎞
⎛
lim
= ab ⇒ lim ⎜ 1 + ⎟
x→∞
x→∞ ⎝
a
x⎠
1+
x
lim ( 2 − x )
⎛ πx⎞
tan ⎜ ⎟
⎝ 2 ⎠
x→1
a⎞
⎛
⇒ ln y = bx ln ⎜ 1 + ⎟
⎝
x⎠
bx
⇒ y = (2 − x)
= eab
⎛ πx⎞
tan ⎜ ⎟
⎝ 2 ⎠
⇒
⎛ πx⎞
ln y = tan ⎜ ⎟ ln ( 2 − x ) ⇒
⎝ 2 ⎠
ln ( 2 − x ) H
⎛ πx⎞
lim ln y = lim tan ⎜ ⎟ ln ( 2 − x ) = lim
=
x→1
x→1
x→1
⎝ 2 ⎠
⎛ πx⎞
cot ⎜ ⎟
⎝ 2 ⎠
⎛ πx⎞
1
sin 2 ⎜ ⎟
⋅ ( −1)
⎝ 2 ⎠ 2 1 2
2
lim 2 − x
= lim
= ⋅ = ⇒
x→1
π x→1 2 − x
π 1 π
2 ⎛ πx⎞ π
− csc ⎜ ⎟ ⋅
⎝ 2 ⎠ 2
lim ( 2 − x )
x→1
64) ⎛ πx⎞
tan ⎜ ⎟
⎝ 2 ⎠
=e
2
π
⎛ 2x − 3 ⎞
lim ⎜
⎟
x→∞ ⎝ 2x + 5 ⎠
2 x+1
⎛ 2x − 3 ⎞
⇒y=⎜
⎝ 2x + 5 ⎟⎠
2 x+1
⎛ 2x − 3 ⎞
⇒ ln y = ( 2x + 1) ln ⎜
=
⎝ 2x + 5 ⎟⎠
( 2x + 1) ⎡⎣ ln ( 2x − 3) − ln ( 2x + 5 )⎤⎦ ⇒
ln ( 2x − 3) − ln ( 2x + 5 ) H
lim ln y = lim
x→∞
1
x→∞
( 2x + 1)
=
1
1
2
⋅2 −
⋅2
−8 ( 2x + 1)
2x
−
3
2x
+
5
lim
= lim
=
x→∞
x→∞ ( 2x − 3) ( 2x + 5 )
−2
( 2x + 1)2
2
1⎞
⎛
−8 ⎜ 2 + ⎟
⎝
−8 ⋅ 2 2
x⎠
lim
=
= −8 ⇒
x→∞ ⎛
3⎞ ⎛
5⎞
2⋅2
⎜⎝ 2 − ⎟⎠ ⎜⎝ 2 + ⎟⎠
x
x
⎛ 2x − 3 ⎞
lim ⎜
⎟
x→∞ ⎝ 2x + 5 ⎠
2 x+1
= e−8
80) ⎛ sin 2x + ax 3 + bx ⎞ H
b⎞
⎛ sin 2x
lim ⎜ 3 + a + 2 ⎟ = lim ⎜
⎟⎠ =
x→0 ⎝
x
x ⎠ x→0 ⎝
x3
2 cos 2x + 3ax 2 + b
lim
⇒ 2 + 0 + b = 0 ⇒ b = −2
x→0
3x 2
2 cos 2x + 3ax 2 − 2 H
−4 sin 2x + 6ax H
−8 cos 2x + 6a
lim
=
lim
=
lim
⇒
x→0
x→0
x→0
3x 2
6x
6
4
−8 + 6a = 0 ⇒ a =
3
Page 314 38) sin x
⇒D=
2 + cos x
x − intercepts: sin x = 0 ⇒ x = nπ
y=
f ( x ) = − f ( x ) ⇒ symmetric about the origin
No asymptotes.
2 cos x + 1
1
f '( x) =  =
>
0
⇒
cos
x
>
−
⇒ 2
( 2 + cos x )2
⎛ 2π ⎞ ⎛ 4π
⎞
f is increasing ⎜ 0, ⎟ , ⎜
, 2π ⎟
⎝ 3⎠ ⎝ 3
⎠
⎛ 2π 4π ⎞
f is decreasing ⎜
,
⎝ 3 3 ⎟⎠
⎛ 2π 3 ⎞
local max ⎜
,
⎟
⎝ 3 3 ⎠
⎛ 4π
3⎞
local min ⎜
,−
3 ⎟⎠
⎝ 3
CU : (π , 2π )
CD : ( 0, π )
POI : ( 0, 0 ) , (π , 0 ) , ( 2π , 0 )
Page 320 2) f ( x ) = x 6 − 15x 5 + 75x 4 − 125x 3 − x
D=
x − intercepts: x = 0, x ≈ 5.33
local min: ( 2.50, −246.6 ) , ( 5.05, −5.03)
( 4.95, −4.965 )
CU : ( −∞, 0 ) , (1.38, 3.62 ) , ( 5, ∞ )
CD : ( 0,1.38 ) , ( 3.62, 5 )
POI : ( 0, 0 ) , ( 5, −5 ) , (1.38, −126.38 ) , ( 3.62, −128.62 )
local max: