陶哲轩实分析（上）9.5及习题-Analysis I 9.5

左极限和右极限，相对简单的一节。

Exercise 9.5.1

Define $\lim_{x→x_0;x∈E}f(x)=+∞$ iff $∀M>0,∃δ>0$, s.t. $x∈E,|x-x_0 |<δ$ would lead to $f(x)>M$. $\lim_{x→x_0;x∈E}f(x)=-∞$ iff $∀M>0,∃δ>0$, s.t. $x∈E,|x-x_0 |<δ$ would lead to $f(x)<-M$.
For $f(x)=1/x$, we consider $f(0+)$ and $f(0-)$:
For $∀M>0,∃δ=1/(M+1)>0$, s.t. $x∈E∩(0,+∞),|x-0|<δ$, we have
$f(x)=1/x>1/δ=M+1>M$
Thus $f(0+)=+∞$.
For $f(0-)$, let $∀M>0,∃δ=-1/(M+1)<0$, s.t. $x∈E∩(-∞,0),|x-0|<δ$, we have
$f(x)=1/x<1/δ=-M-1<-M$
Thus $f(0-)=-∞$.

An analogue of Proposition 9.3.9: the following two are equivalent:
(a) $\lim_{x→x_0;x∈E)}f(x)=+∞$(or $-∞$)
(b) $∀(a_n )_{n=0}^∞∈E, \lim_{n→∞}a_n=x_0$, we have $\lim_{n→∞}⁡f(a_n )=+∞$(or $-∞$)
Proof :
(a) implies (b):
$∀M>0$, since (a) is true, $∃δ>0$, s.t. $x∈E,|x-x_0 |<δ$ would lead to $f(x)>M$. We have $\lim_{n→∞}a_n=x_0$, so $∃N∈\mathbf N$, s.t.
$|a_n-x_0 |<δ,\quad∀n>N$
Then we have $f(a_n )>M,∀n>N$, this means $\lim_{n→∞}⁡f(a_n )=+∞$.
(b) implies (a):
Let (b) be true, assume $\lim_{x→x_0;x∈E)}f(x)≠+∞$, then we can find a $M_0$, s.t. for any $δ>0$, the set $\{x∈(x_0-δ,x_0+δ) ∶f(x)≤M_0\}$ is non-empty. We use axiom of choice to select
$a_n∈\{x∈(x_0-1/n,x_0+1/n) ∶f(x)≤M_0\}$
Then $\lim_{n→∞}a_n=x_0$, but $\lim_{n→∞}⁡f(a_n )≤M_0<+∞$, this contradicts (b).
The case of $-∞$ can be similarly proved.