$\limsup$ and $\liminf$ of real-valued function (Exercise 9.3.4 in Analysis by Tao ) Exercise 9.3.4. Propose a definition for limit superior $\limsup_{x \to x_0; x \in E} f(x)$ and $\liminf_{x \to x_0; x\in E} f(x)$, and then propose an analogue of Proposition 9.3.9 for your definition. (For an additional challenge: prove that analogue.)
Proposition 9.3.9. Let $X$ be a subset of $\mathbb{R}$, let $f : X \to \mathbb{R}$ be a function, let $E$ be a subset of $X$, let $x_0$ be an adherent point of $E$, and let $L$ be a real number. Then the following two statements are logically equivalent:
(a) $f$ converges to $L$ at $x_0$ in $E$. 
(b) For every sequence $(a_n)_{n=0}^\infty$ which consists entirely of elements of $E$ and converges to $x_0$, the sequence $(f(a_n))_{n=0}^\infty$ converges to $L$. 
Attempt: $\limsup_{x \to x_0; x \in E} f(x) = \lim_{\delta \to 0} \sup \{f(x): \text{$x \in E$ s.t. $|x- x_0| < \delta$}\}$. Similarly, $\liminf_{x \to x_0; x \in E} f(x) = \lim_{\delta \to 0} \inf \{f(x): \text{$x \in E$ s.t. $|x- x_0| < \delta$}\}$. Is this correct? Then, what is the analogue of Proposition 9.3.9?
 A: As you can see on the Wikipedia page, just as in the case of limits, you have to disregard the value at $x_0$ itself, so use $0 < |x-x_0| < \delta$ to exclude it. This is for both definitions, of course. 
For 9.3.9. analogues, try:
Equivalent are 


*

*$\limsup_{x \to x_0; x \in E} f(x)=L$. 

*There exists a sequence $(x_n)$ with all $x_n \in E$ and $x_n \neq x_0$, such that $\lim_{n \to \infty} f(x_n)=L$, and for all such sequences $(y_n)$ , $\lim_{n \to \infty} f(y_n) \le L$ if that limit exists.
I think you should be able to show that. What would the analogue for $\liminf$ be? 
A: The purpose of Proposition 9.3.9 is to convert limits of continuous functions to limits of sequences (so we can reuse all our propositions of the latter).
Your attempt uses $\lim_{\delta \to 0} \sup \{f(x): \text{$x \in E$ s.t. $|x- x_0| < \delta$}\}$. This is of the form $\lim_{x \to 0} g(x)$ where $x \in \mathbb{R}$, so it is a limit of a continuous function, and as a result, it won't help us reformulate limits of functions as limits of sequences. 
My suggestion:
First, set the scene exactly like Proposition 9.3.9 and Definition 9.3.6 do. Copied directly:

Let $X$ be a subset of $\mathbb{R}$, let $f : X \to \mathbb{R}$ be a function, let $E$ > be a subset of $X$, let $x_0$ be an adherent point of $E$, and let $L$ be a real number. 

Define $\limsup_{x \to x_0; x \in E}f(x)$ and $\liminf_{x \to x_0; x \in E}f(x)$
Define the set $A_{\delta} = \{f(k) : k \in E, |x_0 - k| < \delta\}$ (chosen to match Definition 9.3.3).
Let $n > 1$ be an integer, and define $ S_n:= \sup(A_{\frac{1}{n}})$ and $I_n := \inf(A_{\frac{1}{n}})$ to be the supremum of $A_{\delta}$ and infimum of $A_{\delta}$ respectively for when $\delta = \frac{1}{n}$.
Consider the sequence of supremums $S_1, S_2, S_3, ...$ = $(S_n)_{n=1}^{\infty}$. If convergent, this sequence will have a limit as per Definition 6.1.8. Make the definition:  
$$\limsup_{x\to x_0; x\in E}f(x) := \lim_{n \to \infty} S_n$$
Similarly, consider the sequence $I_1, I_2, I_3,... = (I_n)_{n=1}^{\infty}$ and make the definition:
$$\liminf_{x\to x_0; x\in E}f(x) := \lim_{n \to \infty} I_n$$
Finally, we are in a position to propose an analogy to Proposition 9.3.9.
Analogous Proposition
The following two statements are logically equivalent:


*

*$f$ convergest to $L$ at $x_0$ in $E$. 

*$\limsup_{x\to x_0; x\in E}f(x)$ and $\liminf_{x\to x_0; x\in E}f(x)$ both exist and are equal to $L$.


Note: the construction of $\limsup$ and $\liminf$ makes the proof of this proposition smooth as they both relate closely to the definition of local ε-closeness.
