HOMEWORK #2 – SOLUTIONS Page 106 2) a) 2 + 0 = 2 c) 0(1.5)=0 4) e) b) d) 2 3 ⋅ 2 = 16 f) 1 + DNE = DNE ‐1 / 0 = DNE 3+1 = 2 2x 2 + 1
9 3
lim 2
=
= x→2 x + 6x − 4
12 4
x (x − 4)
x 2 − 4x
x
4
= lim
= lim
= 14) lim 2
x→4 x − 3x − 4
x→4 ( x − 4 ) ( x + 1)
x→4 x + 1
5
(
)
( x − 1) x 2 + x + 1
x3 − 1
x2 + x + 1 3
= lim
= lim
= 18) lim 2
x→1 x − 1
x→1
x→1
x +1
2
( x − 1) ( x + 1)
26) ⎛ t2 + t − t ⎞
1 ⎞
t2
⎛1
lim ⎜ − 2 ⎟ = lim ⎜ 2
=
⎟ = lim
t→0 ⎝ t
t + t ⎠ t→0 ⎝ t t + t ⎠ t→0 t ( t + 1)
(
lim
t→0
30) 1
=1
t +1
)
x2 + 9 − 5
= lim
x→−4
x+4
lim
x→−4
x 2 + 9 − 25
(
lim
(x + 4) x + 9 + 5
( x − 4 )( x + 4 )
lim
x→−4
( x + 4 ) x2 + 9 + 5
x→−4
x2 + 9 − 5
(x + 4)
)
= lim
)
= lim
x→−4
x→−4
(
)(
x2 + 9 + 5
x2 + 9 + 5
lim xe
π
sin
x
)
(x + 4)
(
x +9 +5
2
x−4
x2 + 9 + 5
4
5
=−
= 0 by the Squeeze Theorem.
2 ( x + 6)
2x + 12
2x + 12
= lim+
= lim+
=2
x→−6 x + 6
x→−6
x→−6
x+6
x+6
lim
lim−
x→−6
2 ( x + 6)
2x + 12
= lim−
= −2
− ( x + 6 ) x→−6 − ( x + 6 )
2x + 12
DNE
x→−6 x + 6
∴ lim
52) )=
x 2 − 16
π
sin
π
−1
−1 ≤ sin ≤ 1 ⇒ e ≤ e x ≤ e1 ⇒
x
π
sin
1
x ⋅ ≤ x ⋅e x ≤ x ⋅e
e
x
lim+
= 0 = lim+ x ⋅ e ⇒
x→0
x→0
e
x→0 +
40) 2
(
38) (
)
= v2
lim L0 1− 2 = L0 1− 1 = 0
v→c −
c
v must be less than c because velocity cannot exceed the speed of light.
56) lim
x→0
f ( x)
= 5 ⇒ f ( x ) = 5x 2 2
x
a) lim f ( x ) = lim 5x 2 = 5 ⋅ 0 = 0 b) f ( x)
5x 2
lim
= lim
= lim 5x = 5 ⋅ 0 = 0 x→0
x→0
x→0
x
x
lim
6−x −2
=
3− x −1
60) x→2
(
lim
(
x→0
x→0
) ( 6 − x + 2 ) ( 3 − x + 1) =
6 − x + 2 ) ( 3 − x + 1) ( 3 − x − 1)
( 6 − x − 4 ) ( 3 − x + 1)
( 2 − x ) ( 3 − x + 1)
lim
= lim
=
( 6 − x + 2 ) ( 3 − x − 1) ( 2 − x ) ( 6 − x + 2 )
6−x −2
x→2
x→2
2 1
=
4 2
x→2