How to interpret limit notation $\lim\limits_{x \to a} f(x)= L$ is by most; intuitively thought of "as $x$ gets close to $a$, $f(x)$ gets close to $L$", however my lecturer said this is not correct. She told me to go away and somehow find out why, by formal definition, the intuition "$f(x)$ is close to $L$, for all $x$ sufficiently close to $a$" is correct, not the former.
I went on to find examples; Simply consider; $f(x) = x/|x|$ when $x$ tends to some number.
and to recall an emphasise; “As $x$ gets close to $a$, $f(x)$ gets close to $L$”
The emphasise on gets is important as it suggests some change towards $L$, however when investigating, as $x$ tends to some number (like $0$), $f(x) = L$, no matter where on the domain you fly. There ceases to be a case in this function where $f(x)$ moves/gets close to $L$ anywhere.
“$f(x)$ is close to $L$, for all $x$ sufficiently close to $a$” includes the idea of ‘there exists some interval’ where $f(x)$ is close to $L$.
Is that a sufficient answer to the question? I can't find anything online.
 A: The epsilon-delta definition is pretty straight-forward:
$$\lim _{{x\to c}}f(x)=L\iff (\forall \varepsilon >0)(\exists \ \delta >0)(\forall x\in D)(0<|x-c|<\delta \ \Rightarrow \ |f(x)-L|<\varepsilon )$$

What does this mean?  Well, we break it down, part by part:


*

*$(\forall\varepsilon>0)\dots(\dots|f(x)-L|<\varepsilon)$


This means that $f(x)$ can get arbitrarily close to $L$.


*$(\exists \ \delta >0)(\forall x\in D)(0<|x-c|<\delta\dots)$


This means that the previous statement is true for every $x$ in the domain that is a certain distance from $c$, the value $x$ is approaching.

This is different from your definition in that it requires $f(x)$ to be close to $L$ with some maximum error $\varepsilon$ for all values $x$ close to $c$.  $x$ does not merely approach $c$, but instead, we must have
$$|f(x)-L|<\varepsilon$$
for every $x$ close to $c$.  The next requirement would then be that the distance between $f(x)$ and $L$ can keep getting smaller and smaller, and that it would still hold for every $x$ values a certain distance from $c$.
There is no such "$x$ approaches $c$" here.

So yes, $f(x)$ is close to $L$ for all $x$ sufficiently close to $a$ is the accurate statement.

In the example $x/|x|$, notice that no matter what $x$ value you choose, either the result will be $1$ or $-1$.  Let's imagine taking the limit to $10$.
$$\lim_{x\to10}\frac x{|x|}\stackrel?=1$$
We then make a table of values:
$$\begin{array}{c|c}x&f(x)\\\hline9&1\\9.9&1\\9.99&1\\\vdots&\vdots\\10.01&1\\10.1&1\\11&1\end{array}$$
Notice that $f(x)$ does not approach anything.  It is simply constant.  I suppose you could then try to fix your statement with "as $x$ approaches $10$, $f(x)$ is close to $1$ within some amount of error that tends to zero."
But that misses the intuition you can get with the epsilon-delta definition:
Notice that $f(x)-L=0\forall x>0$.  It thus follows that $|f(x)-L|=0<\varepsilon$, which holds when $\delta=10$.  $\delta$ needn't get smaller.  It simply needs to be small enough.
A: It is wise to tread carefully when semantically wording some intuition of a precise definition. 
However; in this particular case there exists at least one function that smashes the first and is consistent with the latter. That is taking the limit as $x$ tends to $0$ of;
$f(x) = xsin(1/x)$.
The statement "as x gets close to a, f(x) gets close to L" simply does not hold because no matter how close to $a$ you get, your function continues to oscillate towards and away from L. However by definition the limit does exist. 
A: I think that it is easier to understand the negation.
$f (x) $  could never be closer to $L $ $(|f (x)-L|>\epsilon) $
even if $x $ is closer to $a  \;(|x-a|<\delta) $
