How are the epsilon delta definitions equivalent? 
The epsilon–delta limit definition $1$:
A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ has limit $L$ at point $x_0 \in \mathbf{R}$ if:
  $\bbox[yellow] {\text{for every $\epsilon > 0$ there exists a $\delta > 0$ such that whenever $|x-x_0| < \delta$ then}\ |f(x)-L| < \epsilon}$
Intuitive definition $2$:
A function $f(x)$ from $\mathbf{R}$ to $\mathbf{R}$ has limit $L$ at point $x_0 \in \mathbf{R}$ if:
  $\bbox[yellow] {\text{I can get $f(x)$ as close as I want to $L$ by choosing $x$ close enough to $x_0$}}$

Can anybody explain in a step-by-step manner how the definition $1$ implies definition $2$?
 A: Unfortunately, both definitions are incorrect.
The first definition says, rather, that $f$ is continuous at $x_0$ and $f(x_0)=L.$ Instead, it should read

if for all  $\epsilon>0,$ there exists $\delta>0$ such that whenever $0<\left\lvert x-x_0\right\rvert<\delta$ then $\left\lvert f(x)-L\right\rvert<\epsilon.$

Consider, for example, the function $$f(x)=\begin{cases}1 & x=0\\0 & x\ne 0.\end{cases}$$
Readily, $f$ has limit $0$ at $x_0=0,$ but doesn't satisfy the given definition, since for $0<\epsilon<1,$ no appropriate $\delta$ can be found.
Moreover, the second definition is inaccurate, as well, as any constant function $f(x)=L$ demonstrates. There is no way for $f(x)$ to get closer to $L$ as $x$ gets closer to $x_0,$ since $f(x)$ is always equal to $f(x_0)$!
Instead, I tend to think about it this way. Saying that $f$ has limit $L$ at $x_0$ says that we're guaranteed to be able to get $f(x)$ as close to $L$ as we like (that is, within $\epsilon$ for any positive $\epsilon$ we choose), so long as we get $x$ sufficiently close to (within some positive $\delta$ of), but not equal to, $x_0.$
