HOMEWORK 3
SHUANGLIN SHAO
1. P 74. Ex. 3.1.2
Proof. a). The limit does not exist. Since tan x has period π, we choose
1
1
and yn = nπ+π/4
. Both xn and yn converges to zero as n goes
xn = nπ+π/6
to infinity. However,
tan
π √
1
π
= tan(nπ + ) = tan = 3/2,
xn
6
6
and
tan
√
π
1
= tan(nπ + ) = 2/2.
yn
4
These two output sequences converge to different numbers; so the limit does
not exist.
x2 +1 b). Since cos x3 ≤ 1, we see that
|x cos
x2 + 1
| ≤ |x|.
x3
So
lim x cos
x→0
x2 + 1
= 0.
x3
c). The limit does not exist. If we choose xn approaches to 1 from the left,
as n goes to infinity, we have
1
= −∞.
x→1 log x
lim
While if yn approaches to 1 from the right, as n goes to infinity,
1
= −∞.
x→1 log x
lim
1
2. P75. Ex. 3.1.6
Proof. By definition, if limx→a f (x) = L, we have for any ε > 0, ∃δ > 0 such
that for any 0 < |x − a| < δ,
|f (x) − L| < ε.
By the triangle inequality, ||f (x)| − |L|| ≤ |f (x) − L|. So
lim |f (x)| = |L|.
x→a
b). We choose the following function,
(
1
f (x) =
−1
if x ≥ 0
if x < 0.
Then
lim |f (x)| = 1.
x→0
But limx→0 f (x) does not exist.
3. P75. Ex. 3.1.7
Proof. a). This part follows from the definition of f + and f − .
b). This part follows from Ex. 3.1.6.
4. P 76. Ex. 3.1.8
Proof. For x ∈ Dom(f ) ∩ Dom(g), if f (x) ≥ g(x), then
(f ∨ g)(x) = max{f (x), g(x)} = f (x),
while
f (x) + g(x) + f (x) − g(x)
(f + g)(x) + |(f − g)(x)|
=
= f (x).
2
2
On the other hand,
(f ∧ g)(x) = min{f (x), g(x)} = g(x).
And
(f + g)(x) − |(f − g)(x)|
f (x) + g(x) − f (x) + g(x)
=
= g(x).
2
2
This proves the identities if f (x) ≥ (x). The proof is similar if f (x) ≤ g(x).
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b). If limx→a f (x) = L and limx→a g(x) = M , we have
lim (f ± g)(x) = L ± M,
x→a
and
lim |(f ± g)(x)| = |L ± M |.
x→a
By definition, as x → a,
(f ∨ g)(x) → L ∨ M, and (f ∧ g)(x) → L ∧ M.
5. P81. Ex. 3.2.1
√
x2 = |x| = −x. So
√
−x
x2
=
= −1.
x
x
This is a constant function. For any ε > 0, we choose δ = ε; then for
−δ < x < 0,
√
x2
|
− (−1)| = | − 1 − (−1)| = 0 < ε.
x
Hence
√
x2
lim
= −1.
x→1− x
Proof. a). When x < 0,
b).
sin x
= 0.
x→∞ x2
This is because, for any ε > 0, we choose M =
lim
|f (x) − 0| = |
√1 ;
ε
thus for any x > M ,
sin x
1
1
| ≤ 2 ≤ 2 ≤ ε.
2
x
x
M
Hence
lim
x→∞
sin x
= 0.
x2
c).
1
= ∞.
−1
1
For any M > 0, we take 0 < δ < min{2, 4M
}, then for 1 < x < 1 + δ,
lim
x→1+ x2
0 < x2 − 1 < 2δ + δ 2 ≤ 2δ + 2δ = 4δ ≤
Hence
x2
1
> M.
−1
3
1
.
M
This shows that limx→1+
1
x2 −1
= ∞.
d).
x−3
= ∞.
x→1+ 3 − x − 2x2
lim
We write
x−3
3−x
= 2
.
2
3 − x − 2x
2x + x − 3
1
For any M > 0, we choose 0 < δ < min{1, 7M
}. Then for 1 < x < 1 + δ,
3 − x > 1, and 2x2 + x − 3 = (2x + 3)(x − 1) < 7δ <
Hence
1
.
M
3−x
1
>
= M.
+x−3
1/M
2x2
Thus
x−3
= ∞.
x→1+ 3 − x − 2x2
lim
e).
cos tan x
= 0.
x→−∞ x + 1
For ε > 0, take M < −1 − 1ε . for x < M ,
lim
1
1
x + 1 < M + 1 < − , ⇒ |x + 1| > .
ε
ε
Hence
|
cos tan x
1
− 0| ≤
< ε.
x+1
|x + 1|
This implies that
lim
x→−∞
cos tan x
= 0.
x+1
6. P82. Ex. 3.2.4
Proof. a). Since limx→a g(x) = ∞, for any M > 0, there exists δ > 0 such
that for 0 < |x − a| < δ,
g(x) > M.
Together with f (x) ≥ g(x), f (x) > M . Thus we have limx→a f (x) = ∞.
b). Case 1. |L| < ∞. The limits limx→∞ f (x) = limx→∞ h(x) = L: for any
ε > 0, there exists M > 0 such that for x > M ,
L − ε < f (x) < L + ε, and L − ε < h(x) < L + ε.
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Together with f (x) ≤ g(x) ≤ h(x),
L − ε < g(x) < L + ε.
So
lim g(x) = L.
x→∞
Case 2. L = ∞. The limit limx→∞ f (x) = ∞: for any N > 0, there exists
M > 0 such that for x > M ,
f (x) > N.
Together with f (x) ≤ g(x),
g(x) > N.
So
lim g(x) = ∞.
x→∞
The proof for L = −∞ is similar.
7. P82. Ex. 3.2.6
Proof. If f (x) = 0 for all x ∈ [0, 1], q ∈ Q ∩ [0, 1] ⊂ [0, 1],
f (q) = 0.
Conversely, for x ∈ [0, 1], we choose a sequence of rational numbers rn ∈ [0, 1]
such that limn→∞ rn = x. By f (a) = limx→a f (x), we see that
f (x) = lim f (rn ) = 0.
n→∞
8. P82. Ex. 3.2.7
Proof. We know that P (a) > 0. Since P (x) is continuous at x = a, for
ε = P (a)/2, there exists δ0 > 0 such that for 0 < |x − a| < δ0 , P (x) >
P (a) − ε = P (a)/2.
(a)
Thus for any M > 0, take δ such that 0 < δ < min{δ0 , P2M
}. In this case,
for a < x < a + δ,
P (a)
0<x−a<δ <
;
2M
thus
P (a)/2
P (x)
>
> M.
x−a
δ
(x)
This implies that limx→a+ Px−a
= ∞.
5
P (a)
On the other hand, for M < 0, take δ such that 0 < δ < min{δ0 , −2M
}; for
a − δ < x < a,
P (a)
;
0 > x − a > −δ >
2M
thus
P (x)
P (a)/2
P (a)/2
<
<
= M.
x−a
x−a
P (a)/2M
This implies that limx→a−
P (x)
x−a
= −∞.
Combining these together, we see that limx→a
P (x)
x−a
does not exist.
9. P83. Ex. 3.2.8
Proof. We assume that |L| < ∞. Let an = f (n). We have limn→∞ an+1 −
an = L. For any ε > 0, ∃N ∈ N such that for any n ≥ N ,
L − ε < an+1 − an < L + ε.
In particular,
L − ε < aN +1 − aN < L + ε,
L − ε < aN +2 − aN +1 < L + ε,
······
L − ε < an − an−1 < L + ε.
This implies that
(n − N )(L − ε) < an − aN < (n − N )(L + ε).
Diving it by n,
an aN
N
N
)(L − ε) <
−
< (1 + )(L + ε).
n
n
n
n
By taking n → ∞, we have
an
an
≤ lim sup
≤ L + ε.
L − ε ≤ lim inf
n
n
n
n
Since ε > 0 is arbitrary, we have
an
an
lim inf
= lim sup
= L.
n
n
n
n
(1 −
Thus limn→∞
f (n)
n
= L.
The case where L = ∞ or L = −∞ is similar. We omit the details.
Department of Mathematics, KU, Lawrence, KS 66045
E-mail address: [email protected]
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