Not sure how these two options are different in interpreting $\lim_{x\to a} f(x) = L$ I came across a multiple choice question asking "What do we mean when we write $\lim\nolimits_{x \to a} f(x) = L$?" The only two of the four answers that made sense are


*

*As $x$ gets closer to $a$, $f(x)$ gets closer to $L$.

*We can make $f(x)$ as close as we like to $L$, by choosing $x$ closer to $a$.


Using the precise definition of a limit, I knew that the answer was 2., but I am unsure how these two answers are different. 
The only thing that I could come up with is that option 1. has to do with continuity, for $x$ getting closer to $a$ is not sufficient to conclude that $f(x)$ gets closer to $L$, unless $L=f(a)$, but I'm not sure.
Edit: I added $L=f(a)$ in the last sentence of the preceding paragraph.
 A: They are different. Because the first statement only says we're getting closer, but not as close as we want (see Example 1). Moreover we can get as close as we want without constantly getting closer (see Example 2, in which $L=f(a)$).
Let me explain with an example:
Example 1. Let $f:[0,1]\rightarrow\mathbb{R}$ be the function $f(x)=x$. Now if $L=1$, $a=1$ both statements hold.
However you can now take $L=2$. I claim that the first statement hold, but the second doesn't.
Indeed, the first holds. Since $f(x)$ is monotonically increasing, it gets larger as $x$ gets closer to $1$. In particular as $x$ gets closer to $1$, $f(x)$ get closer to $L=2$. Formally, $|2-f(x)| = 2-x$ decreases as $x$ gets closer to $1$.
On the other hand, the second property doesn't hold. Because no matter how close you get to $1$, $f(x)$ will always be at least $1$ point far from $L=2$.
If you take $L=f(a)$ the statements still don't mean the same thing (the first one would imply continuity but it is much stronger).
Example 2. For instance consider
$$f(x) = \begin{cases} x & \text{there is no } n \text{ s.t. } x=1-\frac{1}{n}\\ 1 & \text{if }  x=1-\frac{1}{n} \end{cases}$$
Take $L=f(1)=1$. Then the second statement holds, but the first doesn't. Because $f(\frac{1}{2})=1$, and if $x=\frac{98}{100}$ then $f(x)=\frac{98}{100}\not = 1$. Clearly the value $f$ gives to $\frac{1}{2}$ is closer to $1$ then the value it gives $\frac{98}{100}$ even though $\frac{98}{100}$ is closer to $a=1$.
A: Both of the statements are defective. In 1.) you do not need to get arbitrarily close to $L$, as is necessary for a limit, and in 2.) one can not exclude that there are multiple values for $L$ that $f(x)$ gets close to for $x$ approaching $a$. For instance in $f(x)=\sin(1/x)$  you can chose $x$ arbitrarily close to $a=0$ so that $f(x)$ is equal to any value in $[−1,1]$.

Under the assumption that the "dynamics" of "getting closer" includes that the distance has to go to zero, the first variant seems to be more correct. A more formal formulation of the dynamical statement would be that the sets of function values $f((a-δ,a+δ)\cap D_f)$ gets more concentrated around $L$ the smaller $δ$ gets, the radius shrinking to zero.
The second variant in my view describes a limit value, that is, the sets $f((a-δ,a+δ)\cap D_f)$ need not concentrate around $L$, they just need to all contain $L$ in their closure, the radius of the sets can remain above some positive bound.

Or in still other words, in the sequential interpretation of the limit, the first condition demands that $f(x_k)\to L$ for all $x_k\to a$, while the second condition only demands that $f(x_k)\to L$ for at least one sequence $x_k\to a$.
A: The reason the second definition is better is primarily because of sets other than $\mathbb{R}$ limits can exist in any sets that have distance measures. If you think about for example $\mathbb{R^2}$ getting closer to a number doesn't mean the same thing as in $\mathbb{R}$ In $\mathbb{R}$ it is very simple to think about following a point along a function to infinity to approach a point but this is more complicated in other dimensions as there is not necessarily a set path to follow. 
Another issue with the first definitions is discontinuities that aren't related to the limit. For example $\lim_{x\to\infty}\frac{1}{x}=0$ but if you look at the left side of the y-axis and try to follow the function you would be justified saying that the limit is $-\infty$ since as you move x closer to $\infty$ it gets closer to $-\infty$
The third reason is that it doesn't say anything about how close, like in my last issue while $\frac1x$ might approach 0 as it goes to $\infty$ it also gets closer to $-1$ and really any negative number. 
A: The second answer would be the more correct of the two.
The reason is because of the notion of "arbitrary closeness."
The limit of a function $f(x)$ as $x \to a$ exists and is equal to $L$ if and only if given any $\epsilon > 0$, we can find another number $\delta$ for which whenever $x$ is within a $\delta$-neighborhood of $a$, it necessarily follows that $f(x)$ is within the associated $\epsilon$-neighborhood of $L$.
The $\delta$, however, is not a universal $\delta$. If you change $\epsilon$ we may need to recalculate $\delta$, but we are assured that such a $\delta$ exists by the underlying definition of limit. 
