6c) For
ϵ
>
0
∧
δ
>
0,
where x
3
8=x2x
2
+3x+4<
ϵ
,
for every 0<x2<
δ
Let S
1
=1. Then 0<x2<
δ
1
=1, then 1<x<3
Then x2(3^2)+2(3)+4=19x2<
ϵ
and
δ
=
ϵ
19
.
9a) By Theorem 5.1.10, There exists a sequence S
n
in D with each S
n
≠
c such that S
n
converges to c, but f(s
n
) is not convergent in
R
. Let f(x)=
1
x
and S
n
=
1
n
.
lim
n→∞
S
n
=
0
∀
n
but f(s
n
)= Lim n=+
∞
so
f(s
n
) is divergent and the limit does not exist.
9b) By Theorem 5.1.10,
∀
k
∈
N
Sin(2
πk
+
nπ
2
¿
= {0, if n=0
1, if n=1
0, if n=2
1, if n=3}
S
n
=
n
2
π
, for all n
∈
N
, Then the limit as n approaches positive infinity of S
n
=0
But f(s
n
) is the sequence {0,1,0,1,0, 1, ….} which does not converge.
9c)
x→
0
+
¿
xsin
1
x
lim
¿
¿
. Let
ϵ
>
0
∧
δ
>
0
where
ϵ
=
δ
. Then 
xsin
1
x
0<x as 
sin
1
x

≤
1
∀
x
>
0.

xsin
1
x
∨
≤x ,
∀
x
>
0.

xsin
1
x
∨
≤
ϵ
where 0 <x <
δ
x→
0
+
¿
xsin
1
x
=
0
lim
¿
¿
13) From Theorem 4.1.13, 4.4.11, and 5.1.8 we can see that
a)
Let S
n
∈
D
that converges to c with S
n
≠c
b)
Knowing
lim
x→ c
f
(
x
)
=
L
∧
lim
x →c
h
(
x
)
=
L,
by Theorem 5.1.8,
lim
n→∞
f
(
s
n
)
=
L
∧
lim
n→ ∞
h
(
s
n
)
=
M
c)
Since f(x)
≤
g(x)
≤
h(x) for every x
∈
D
f(s
n
)
≤
g(s
n
)
≤
h(s
n
)
∀
n
∈
N
d)
Given
ϵ
>
0,
∃
N
1
, such that h(s
n
)M<
ϵ
∀
n≥ N
1
e)
Given
ϵ
>
0,
∃
N
2
, such that f(s
n
)L<
ϵ
∀
n≥ N
2
f(s
n
)L<
ϵ
and h(s
n
)M<
ϵ
∀
n≥ N
N
1
∧
N
2
aremaximums
f)
Therefore –
ϵ
< f(s
n
)L<
ϵ
and –
ϵ
< g(s
n
)M<
ϵ
g)
From line c, f(s
n
)
−
L≤
g(s
n
)
−
L≤
h(s
n
)
−
L
∀
n
∈
N
–
ϵ
< g(s
n
)L<
ϵ
∀
n
∈
N
lim
n→∞
g
(
s
n
)
=
L
∀
n
∈
N
****Similar can be said that
lim
n→∞
g
(
s
n
)
=
M
16)By theorem 5.1.2, let f:d>
ℝ
and c be an accumulation point of D. Then lim as x
approaches c of f(x)=L iff for each neighborhood V of L, there exists a deleted
neighborhood U of c such that f(U
∩
D)
⊆
V .
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 Fall '08
 Staff
 Limits, Limit of a sequence, Limit superior and limit inferior, lim sup, subsequence, Lim Sup Vn