HOMEWORK #4 – SOLUTIONS Page 141 +∞ d) −∞ e) 4) a) 2 b) ‐2 c) f) HA: y = 2, y = ‐2 VA: x = 0, x = 3, x = ‐2 8) (presented in class). 3x + 5
= 3 Numerator and denominator have same degree x→∞ x − 4
16) lim
t2 + 2
20) lim 3
= 0 t→−∞ t + t 2 − 1
deg (N) < deg (D) x 3 − 2x + 3
32) lim
= −∞ deg (N) > deg(D) x→∞
5 − 2x 2
34) (
)
(
(
lim tan −1 x 2 − x 4 = tan −1 lim x 2 − x 4
x→∞
−π
tan ( −∞ ) =
2
x→∞
−1
arctan is continuous. 40) 43) −∞ x2 + 1
x2 + 1
y= 2
=
⇒
2x − 3x − 2 ( 2x + 1) ( x − 2 )
1
HA : y =
2
1
VA : x = − , x = 2
2
)) =
(
VA : x = 5
48) )
x x2 − 1
x ( x − 1) ( x + 1) x ( x + 1)
x3 − x
y= 2
=
=
=
x − 6x + 5 ( x − 5 ) ( x − 1) ( x − 5 ) ( x − 1)
x−5 HA : none SA : y = x + 6
VA : x = 1, x = 3 ⇒ D = ( x − 1) ( x − 3) = x 2 − 4x + 3
HA : y = 1 ⇒ N has same degree as D and is x 2
x2
∴ f ( x) = 2
x − 4x + 3
58) a) After t minutes, 25t liters of brine with 30g of salt per liter has been pumped into the tank. Therefore it contains ( 5000 + 25t ) liters of water and 25t ⋅ 30 = 750t grams of salt. 750t
30t g
=
5000 + 25t 200 + t L
30t
g
lim
= 30 t→∞ t + 200
L
∴C ( t ) =
b) which says that the salt concentration approaches that of the brine being pumped in.