Function $\lim\limits_{x\to a}f(x) = L$ If $f$ is a function then
$$\lim_{x\to a}f(x) = L$$
implies that $f(x)$ gets closer to $L$ as $x$ gets closer to $a$. 
Is this right?
 A: That is what it means, yes.
And even better: $f(x)$ can come arbitrarily close to $L$ as $x$ gets closer to $a$. This is a subtle, but important difference: $\cos(x)$ gets closer and closer to $2$ as $x$ gets closer to $0$, but you can never actually get really close to $2$. So $\lim_{x\to0}\cos(x)$ isn't $2$.
Also important: as $x$ gets closer to $a$, $f(x)$ not only gets close to $L$, but it gets close to $L$ for all values of $x$ close to $a$. For instance, $\sin(1/x)$ gets close to $0$ as $x$ gets close to $0$, but it is also far away for some $x$ close to $0$ as well. So $\lim_{x\to0}\sin(1/x)$ isn't $0$.
A: Yes it is the naive idea for the definition of limit for which get closer means "get eventually and arbitrarly closer" with respect to any prescribed bound but not necessarly equal to the limit and not necessarly monotonically, that is formally
$$\left(\lim_{x\rightarrow a} f(x) = L\right) \iff \Big(\forall \varepsilon >0\, \exists \delta > 0: \big(0<\vert x-a\vert <\delta \implies \vert f(x)-L\vert <\varepsilon\big)\Big)$$
A: No. Take$$f(x)=\begin{cases}x\sin\left(\frac1x\right)&\text{ if }x\neq0\\0&\text{ if }x=0\end{cases}$$Then $\lim_{x\to0}f(x)=0$. However, no matter how close you are to $0$, there are points $x$ at which $f(x)=0$ and there are points $x$ at which $f(x)\neq0$.
