SOLUTION OF HW2
MINGFENG ZHAO
September 15, 2012
1. [5 Points, Exercises 4, on Page 49] Find lim f (x), and justify your answers with − δ proof.
x→x0
a. x2 + 2x + 1,
x0 = 1.
x3 − 8
, x0 = 2.
x−2
1
, x0 = 0.
c. 2
x −1
√
d. x, x0 = 4.
b.
e.
x3 − 1
+ x,
(x − 1)(x − 2)
x0 = 1.
Proof. a. Claim: lim (x2 + 2x + 1) = 4.
x→1
For any 0 < |x − 1| < 1, then −1 < x − 1 < 1, that is, 0 < x < 2. Hence we have
|(x2 + 2x + 1) − 4| =
|x2 + 2x − 3|
=
|(x − 1)(x + 3)|
=
|x + 3||x − 1|
≤
(|x| + 3)|x − 1|
≤
5|x − 1|,
∀0 < |x − 1| < 1.
n o
> 0, then for any 0 < |x − 1| < δ, we have
For any > 0, let δ = min 1,
5
|(x2 + 2x + 1) − 4| < 5 · δ
≤ 5·
= .
1
5
2
MINGFENG ZHAO
By the definition of limit, then lim (x2 + 2x + 1) = 4.
x→1
3
b. Claim: lim
x→2
x −8
= 12.
x−2
For any 0 < |x − 2| < 1, then −1 < x − 2 < 1, that is, 1 < x < 3. Hence we have
3
x − 8
x − 2 − 12
= |(x2 + 2x + 4) − 12|
= |x2 + 2x − 8|
= |(x − 2)(x + 4)|
= |x + 4||x − 2|
≤ (|x| + 4)|x − 2|
≤ 7|x − 2|,
∀0 < |x − 2| < 1.
n o
Now for any > 0, let δ = min 1,
> 0, then for any 0 < |x − 2| < δ, we have
7
3
x − 8
x − 2 − 12 <
7·δ
≤
7·
=
.
7
x3 − 8
= 12.
x→2 x − 2
By the definition of limit, then lim
1
= −1.
−1
1
For any 0 < |x| < , then we have
2
c. Claim: lim
x→0 x2
1
−
(−1)
x2 − 1
1
= 2
+ 1
x −1
1 + x2 − 1 = x2 − 1 =
x2
|x2 − 1|
=
x2
1 − x2
Since 0 < |x| <
1
2
SOLUTION OF HW2
<
1
For any > 0, let δ = min
1 3
, 2 2
1
2 |x|
2
− 12
=
1 4
· · |x|
2 3
=
2
|x|,
3
3
Since 0 < |x| <
∀0 < |x| <
1
2
1
.
2
> 0, then for any 0 < |x| < δ, we have
1
<
−
(−1)
x2 − 1
2
·δ
3
2 3
· 3 2
≤
= .
By the definition of limit, then lim
x→0
d. Claim: lim
x→4
√
1
= −1.
x2 − 1
x = 2.
For any 0 < |x − 4| < 1, then −1 < x − 4 < 1, that is, 3 < x < 5. Hence we have
√
√
( x − 2)( x + 2) √
= x+2
√
| x − 2|
=
|x − 4|
√
x+2
<
|x − 4|
,
2
∀0 < |x − 4| < 1.
For any > 0, let δ = min {1, 2} > 0, then for any 0 < |x − 4| < δ, we have
√
| x − 2| <
2
2
= .
By the definition of limit, then lim
√
x→4
x3 − 1
+ x = −2.
x→1 (x − 1)(x − 2)
e. Claim: lim
x = 2.
4
MINGFENG ZHAO
For any 0 < |x − 1| <
1
2,
then − 12 < x − 1 <
1
2,
that is,
1
2
< x <
3
2.
Which implies that
− 32 < x − 2 < − 12 and
x3 − 1
+
x
−
(−2)
(x − 1)(x − 2)
x3 − 1
= + x + 2
(x − 1)(x − 2)
2
x + x + 1
= + x + 2
x−2
2
x + x + 1 + (x + 2)(x − 2) = x−2
2
x + x + 1 + x2 − 4 = x−2
2
2x + x − 3 = x−2 (x − 1)(2x + 3) = x−2
=
|2x + 3|
· |x − 1|
|x − 2|
≤
2|x| + 3
· |x − 1|
2−x
≤
=
For any > 0, let δ = min
1 ,
2 4
2·
3
2 +
3
2
3
· |x − 1|
4|x − 1|.
> 0, then for any 0 < |x − 1| < δ, we have
x3 − 1
(x − 1)(x − 2) + x − (−2)
< 4δ
≤ 4·
4
= .
x3 − 1
+ x = −2.
x→1 (x − 1)(x − 2)
By the definition of limit, then lim
2. [5 Points, Exercises 15, on Page 50] Find lim f (x) if it exists, and justify your answer directly
x→∞
from Definition 2.1.7.
SOLUTION OF HW2
5
1
.
x2 + 1
sin x
(α > 0).
b.
|x|α
sin x
c.
(α ≤ 0).
|x|α
a.
d. e−x sin x.
e. tan x.
2
f. e−x e2x .
1
= 0.
x2 + 1
1
> 0, then for any x ≥ G, we have
For any > 0, let G = max 1,
Proof. a. Claim: lim
x→∞
1
=
−
0
x2 + 1
<
1
x2
≤
1
x
≤
=
By the Definition 2.1.7, on Page 40, then lim
x→∞
b. Claim: If α > 0, then lim
x→∞
For any > 0, let G =
1
1
α
1
x2 + 1
Since x > 1
1
1
.
1
= 0.
x2 + 1
sin x
= 0.
|x|α
> 0, then for any x ≥ G, since α > 0, then |x|α ≥
sin x
−
0
|x|α
sin x = α |x|
=
| sin x|
|x|α
≤
1
|x|α
≤
1
1
= .
1
and
6
MINGFENG ZHAO
By the Definition 2.1.7, on Page 40, then lim
x→∞
c. Claim: If α ≤ 0, then lim
x→∞
sin x
= 0.
|x|α
sin x
does not exists.
|x|α
Otherwise, by the Definition 2.1.7, on Page 40, then there exists some A ∈ R such that lim
x→∞
sin x
=
|x|α
A, which implies that there exists some G0 > 0 such that for all x ≥ G0 , we have
sin x
1
< .
−
A
|x|α
4
Hence for all x, y ≥ G0 , then we can get
sin x sin y −
|x|α
|y|α sin x
sin y = α − A + A −
|x|
|y|α sin x
sin y ≤ α − A + A −
|x|
|y|α <
1 1
+
4 4
=
1
.
2
Hence, we get
sin x sin y 1
|x|α − |y|α < 2 ,
(1)
∀x, y ≥ G0 .
When taking y = G0 , then we know that for all x ≥ G0 , we have
−
1
sin x sin G0
1
<
−
< .
α
α
2
|x|
|G0 |
2
In particular, there exists some B > 0 such that
sin x |x|α ≤ B,
(2)
Case I: If α = 0, then
2([G0 ] + 1)π +
3π
2
for all x ≥ G0 .
sin x
= sin x for all x > 0. Now taking x0 = 2([G0 ] + 1)π +
|x|α
> G0 , but we have
sin x0
sin y0 |x0 |α − |y0 |α = | sin x0 − sin y0 |
π
2 , y0
=
SOLUTION OF HW2
7
π
3π = sin 2([G0 ] + 1)π +
− sin 2([G0 ] + 1)π +
2
2 = |1 − (−1)|
=
2
>
1
.
2
Which contradicts with (??).
Case II: If α < 0, for any k ∈ N, taking xk = 2k([G0 ] + 1)π +
B
π
2
> G0 , by (??), then we know that
sin xk |xk |α sin 2k([G ] + 1)π + π 0
2 2k([G0 ] + 1)π + π2 α ≥
=
1
α .
2k([G0 ] + 1)π + π2
=
Let C = [2([G0 ] + 1) + π2 ]α > 0, then we get
B≥
1
,
Ck α
∀k ∈ N.
kα ≥
1
,
BC
∀k ∈ N.
Which implies that
By taking ln on both sides, then
α ln k ≥ ln
1
= − ln(BC),
BC
∀k ∈ N.
Since α < 0, then
ln k ≤ −
ln(BC)
,
α
∀k ∈ N.
That is, we have
k ≤ e−
ln(BC)
α
,
Since k is arbitrary, then we get a contradiction.
∀k ∈ N.
8
MINGFENG ZHAO
In summary, we get a contradiction. Therefore, we can conclude that if α ≤ 0, then lim
x→∞
sin x
|x|α
does not exists.
d. Claim: lim e−x sin x = 0.
x→∞
For any > 0, let G = max{1, − ln } > 0, then for any x ≥ G, we have
|e−x sin x − 0| =
|e−x sin x|
=
| sin x|
ex
≤
1
ex
≤
e−x
≤
e−(− ln )
=
eln =
.
By the Definition 2.1.7, on Page 40, then lim e−x sin x = 0.
x→∞
e. Claim: lim tan x does not exists.
x→∞
Since for any k ∈ N, the function tan (2k+1)π
is not well defined. By the Definition 2.1.7, on Page
2
40, then lim tan x does not exists.
x→∞
2
f. Claim: lim e−x e2x = 0.
x→∞
r
For any > 0, let G = max 1, ln
+1
e
> 0, then for any x ≥ G, then we have
2
|e−x e2x − 0| =
e−x
2
+2x
2
=
e−(x−1)
=
e · e−(x−1)
≤
e · eln e
+1
2
SOLUTION OF HW2
=
e·
=
.
9
e
2
By the Definition 2.1.7, on Page 40, then lim e−x e2x = 0.
x→∞
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu