Math UA 325: Analysis
New York University
Homework 5
Due March 2 at 11:59pm
1) [Exercise 1.3.2] Given two real numbers x, y show that
(a)
max(x, y) =
x+ y + |x y|
2
(b)
min(x, y) =
x+ y |x y|
2
2) Let
A =
⇢
1
n
n 2 N>0 [⇢1 nn
n 2 N>0 ⇢ R
Find supA and inf A.
3) For n > 1 let
xn =
p
n+ (1)n
n
.
Find the limit limn!+1 xn, or prove that it does not exist.
4) For n > 1 let
xn =
n3 + 3n+ 1
2n3 2n2 + n+ 1 .
Find the limit limn!+1 xn, or prove that it does not exist.
1

Q⽐⽐ that x.ge to real numbers 从 ⼓130
Firstly if Xy the max of Kylie X
the min of lxyl is yand Ky1 x y
Thus we have
maxlx.glxtyt x y I X which goals to 02
minlxyjxtylx D_y.ph to
2Thus we have a
If xcy the max of Kyl is y
min of Kyl is x 40and k y l y x
Thus we have
max X7
⼼ y equals to
win x y
州 4 D__x yds to
2
Given that n E Mo nil 2 in
According to example 2.1.4 on the textbook
in no and 䘕 is convergent
And according to def 219 that
a sequence n is moone decreasingifxnTXntl.siis monotone decreasi g Gwen ⼤了 is
convergent it is also bounded Thus according
to def 21.10 签ㄨn 讨xninENJ.intto his bounded above as 廿 ⼆ 1
for all n E Mo The 1 is an upper band for t
take a umber b st b is also an upper bad
for 式 bit for all u.and b4
Since n EN n 1，2 n
古 ⼆ ⼗ Ì I But we assume b ⼆⽓
which is a contradiction This l is the last
upper bound for ⻓ Sup 䘕 1
Gwen AH nt No U 1 In EM CR
we have All i I 去 u H.it.it in B
ni is a unit left more front According to
prop ⼈26 sup仪 AxtsupA.inflxtAkxthftus.mehave 5up 0 Inf 1
This sup A ⼆ sup Gk l
Inf At Inf ⼆ 1
According to Ex in prop 217 we know that
⼀ is divergent
Since rn we have
⽆⼆ ⼆ i fun
Gwen Go we must find most
0 Ymc E then for all n M
we have 1舁 ⼀ 1型 it ⼆⻰式
since 啙去 ⼆ 0 we have
Lemma227 OEÈEÈ faunal
Thus 啙吉⼆ 0 and 啙 不 i 0
according to pop 225It is convergent
As for 不 1
n
Gwen Go we must find most
0 Ymc G then for all n MI
we have 1丁兴 ⼀ 1型 ⼆ it ⼆⻰⼗式
器 后 i 0 CanveyThus we here
coding to pop 225
1 1元 anti til for un
sie ne hae
nf字型 ⼆ 品 型 ⼆ 0
Then lion In ⼗ 州
anges andn n
0
Q4
since nsl n_n 1
ni
Xu
P 3n 1
zn3_3in 1
⼆
zrE3ritntln3
lt iz f
tiwehaneqo'FEifraunEN
0ET3EE
We know him É 0 Thus use Lemma 221
we ouhdfgf see him an
mo
啙⼗⼆0 c Lim Xu
hto3F3 no 点名 T 3 0 0
ㄨn 1
0 0
2 0 0 o Ì