Show that this definition of derivative implies the other one. I have two definitions of derivatives. The first one is the one provided in Rudin:

Let $f: [a,b] \to \mathbb{R}$ be a function. Let $x \in [a,b]$. 
Let $\phi: (a,b) \setminus \{x\}\to\mathbb{R}: t \mapsto \frac{f(t)-
 f(x)}{t-x}$
Then we define $f'(x) = \lim_{t \to x} \phi(t)$, provided that the
  limit exists.

Here's the second one:

Let $f: E \subseteq \mathbb{R} \to \mathbb{R}$ be a function, $x \in
 E$ and $x$ a limit point of $E$. Put 
$\phi: E \setminus \{x\} \to \mathbb{R}:  t \mapsto \frac{f(t)- 
 f(x)}{t-x} $.
Then we define $f'(x) = \lim_{x \to t} \phi (t)$, provided that the limit exists.

I want to show that the second definition implies the first, when we put $E = [a,b]$.
So, in particular, I want to show that for every $x \in [a,b]$
$\lim_{t \to x, t \in (a,b) \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$ exists $\iff$ $\lim_{t \to x, t \in [a,b] \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$
and if one of the two exists, both are equal.
Clearly, $\Leftarrow$ is satisfied (immediately from the limit definition).
For the other direction, put $q:= \lim_{t \to x, t \in (a,b) \setminus \{x\}} \frac{f(t)-f(x)}{t-x}$. If $x \in (a,b)$, then it is an interior point and the limit is equal to the other one.
WLOG, assume $x = a$. Let $\epsilon > 0$. Choose $\delta > 0$ such that $|q - \frac{f(t)-f(a)}{t-a}| < \epsilon$ for all $t \in (a,b)$ with $0 < |t-x|< \delta$.
Then, if $t \in [a,b] \setminus \{a\}$ with $0 < |t-a| < \min \{\delta, |b-a|\}$, then $t \neq b, t \neq a$ and we can make the quantity smaller than $\epsilon$. 

Does this seem correct? Are there any arguments against using the
  second definition over the first one, as it seems more general?

 A: Let us generalize the second definition. For $x \in \mathbb{R}$ we denote by $\mathfrak{N}(x)$ the set of all neighborhoods of $x$ in $\mathbb{R}$ (a neighborhood of $x$ is a set $N \subset \mathbb{R}$ such that $(x -\varepsilon, x +\varepsilon ) \subset N$ for some $\varepsilon > 0$).
Let $x \in E \subset  \mathbb{R}$. Then obviously the following are equivalent:
(1) $x$ is a limit point of $E$.
(2) There exists $N \in \mathfrak{N}(x)$ such that $x$ is a limit point of $N \cap E$.
(3) For all $N \in \mathfrak{N}(x)$, $x$ is a limit point of $N \cap E$.
Now let $f: E \to \mathbb{R}$ be a function and $x \in  E$ be a limit point of $E$. For each $N \in \mathfrak{N}(x)$ define
$$\phi_N: N \cap E \setminus \{x\} \to \mathbb{R}:  t \mapsto \frac{f(t)- 
f(x)}{t-x} .$$
The map $\phi$ from the second definition is given as  $\phi = \phi_{\mathbb{R}}$. If $\lim_{x \to t} \phi_N (t)$ exists, we denote it by $f'_N(x)$. In case $N = \mathbb{R}$ we simply write $f'(x)$.
The following are obvious:
(1) If $\lim_{x \to t} \phi (t)$ exists, then $\lim_{x \to t} \phi_N (t)$ exists for all $N \in \mathfrak{N}(x)$ and $f'_N(x) = f'(x)$.
(2) If $\lim_{x \to t} \phi_N (t)$ exists for some $N \in \mathfrak{N}(x)$, then $\lim_{x \to t} \phi (t)$ exists and $f'(x) = f'_N(x)$.
Now let $E = [a,b]$. To avoid confusion the function $\phi$ from the first definition will be denoted by $\Phi$.  
For $x \in (a,b)$ we have $\Phi = \phi_{(a,b)}$, for $x = a$ we have $\Phi = \phi_{(a-1,b)}$ and for  $x = b$ we have $\Phi = \phi_{(a,b+1)}$.
This shows that the first and the second definition are equivalent. 
