Homework 2: solutions
Section 1.2
39. f ◦ f (x) = sin(sin(x)),
Dom(f ◦ f ) = R.
√
f ◦ g(x) = sin(1√− x), Dom(f ◦ g) = Dom(g) = [0, +∞).
g ◦ f (x) = 1 − sin x, Dom(g ◦ f ) = {x : sin(x) > 0} =
= · · · ∪ [−2π, −π] ∪ [0, π] ∪ [2π, 3π] ∪ · · · = ∪n∈Z [2nπ, 2(n + 1)π].
56. With v = 350mi/h and
√ h = 1mi: d = v · t = d(t); s =
s(t) = s ◦ d(t) = s(d(t)) = h2 + v 2 t2 .
√
h2 + d2 = s(d);
60. A(n) (x) = notation for composing n times = A ◦ A ◦ · · · ◦ A(x) = (1.04)n x.
Although c = 1.04 is very near 1, at one point the powers of c will get very large.
Example: c20 ≈ 2.19, c70 ≈ 7.1, c100 ≈ 50.5, c200 ≈ 2550.7, c400 ≈ 6, 506, 324 etc.
61. a) h(x) = 4x2 + 4x + 1 + 6 = (2x + 1)2 + 6 = g(x)2 + 6, so f (x) = x2 + 1.
Section 1.3
3. c) limx→1− f (x) = 2, limx→1+ f (x) = 3, so limx→1 f (x) does not exist.
11. If x = 2 + ² is a point near 2 (with ² ' 0), then
(This is not the complete answer to this exercise.)
x2 −2x
x2 −x−2
=
2² + ²2
2+²
2
=
' .
2
3² + ²
3+²
3
Section 1.4
2. b) limx→1− [f (x) + g(x)] = limx→1− f (x) + limx→1− g(x) = 1 + 2 = 3.
limx→1+ [f (x) + g(x)] = limx→1+ f (x) + limx→1+ g(x) = 1 + 1 = 2.
Since limx→1− 6= limx→1+ , limx→1 [f (x) + g(x)] does not exist.
4. limt→−1 (t2 + 1)3 (t + 3)5 = [limt→−1 (t2 + 1)]3 · [limt→−1 (t + 3)]5 .
Since limt→−1 (t2 + 1) = (limt→−1 t)2 + limt→−1 1 = 1 + 1 = 2,
and limt→−1 t + 3 = limt→−1 t + limt→−1 3 = −1 + 3 = 2,
the answer is 23 · 25 = 28 = 256.
10. a) The identity holds only for x 6= 2 [the left-hand side doesn’t make sense at
x = 2].
b) The equation is correct since the expression limx→2 means that x approaches 2
without equalling 2.
√
(x + 2) − 32
x−7
21. x + 2 − 3 = √
=√
, hence
x+2+3
x+2+3
√
x+2−3
1
=√
(this holds for x 6= 7.
x√
−7
x+2+3
1
x+2−3
1
1
lim
= lim √
=√
= .
x→7
x→7
x−7
6
x+2+3
7+2+3
24. We first simplify the expression: for t 6= 0,
1
1
(t2 + 1) − 1
t2
t
− 2
=
=
=
.
2
2
t
t +1
t(t + 1)
t(t + 1)
t+1
1
1
t
limt→0 t
0
lim ( − 2
) = lim
=
= = 0.
t→0 t
t→0 t + 1
t +1
limt→0 t + 1
1
√
√
√
32. For x >√0, x ≤ x√· [1 + sin2 (2π/x)] ≤ 2 x.
Since lim x = lim 2 x = 0, we can apply the squeeze theorem.
x→0+
x→0+
1