Why do we say in limits that $x$ must approach a but not be equal to a Having a little bit of unrigorous foresight as to what will come after limits, I'm wondering why in the definition of the limit we require $x$ to approach a but not be equal to $a$. What is the mathematical reason to this; what situations does it help avoid? To me it seems arbitrary. 
 A: Because the notion of the limit is used to investigate the behaviour of a function in the neighborhood of a point. The value of the function on that point is irrelevant because: 


*

*a). the function may not even be defined on that point (the point maybe excluded from the domain). Think about the behaviour of $\frac{1}{x}$ in a neighborhood of $0$.

*b). the function may be defined on some point but its behaviour close to that point maybe completely different. Think for example points where the function is not continuous.

A: Usually there is some kind of "impossible" calculation if you were actually at the limit.
The classic case is trying to find the gradient of a curve.
You can estimate the gradient of the curve $y=f(x)$ at the point $(x,f(x))$ by choosing a point a little bit away - say $(x+h,f(x+h))$ and calculating the gradient of the line joining those points.
The estimated gradient would be $\frac {f(x+h)-f(x)}{h}$.
To get a better estimate, we can try making $h$ very very small.
So maybe the best estimate would be making $h=0$. But that would mean dividing by zero and that causes us some difficulties...
So instead we look at the limit as $h \rightarrow 0$
A: because the limit is a more widely applicable concept if $x$ itself is excluded: For real or complex functions, $\lim_{t \rightarrow x}(f(t))$ equals (among other equivalent definitions) the COMMON limit of $f(t_n)$ for  ALL sequences $(t_n)$ that converge to $x$ as $n \rightarrow \infty$, IF that common limit exists. Since $f$ may be undefined or unpleasant at $x$, the sequence $(t_n)$ that is constantly $t_n = x$ for all $n$ (and therefore cvgt. to $x$) would in many cases yield a trivial limit of $f(t_n)$ that is different from all the other $lim_{n \rightarrow \infty}(f(t_n))$ (even if these are consistent), and therefore $\lim_{t \rightarrow x}(f(t))$ would not exist. In particular, making sequences that include $x$ itself admissible to the limit would lead to there being no limiting value at discontinuities of $f$. So excluding $x$ allows an analysis of $f$'s behaviour in a neighbourhood of $x$ in terms of the limit, even if for some reason $f(x)$ itself does not "fit in".
A: If a function $f(x)$ is nice enough to have a limit as $x$ approaches $a$, that limit tells you, loosely speaking, what $f(a)$ ‘ought’ to be in order to fit in with nearby values. Consider the function 
$$f(x)=\begin{cases}
?,&\text{if }x=0\\
0,&\text{otherwise}\;,
\end{cases}$$
where I’m not going to tell you what $f(0)$ is. If I start at $x=1$, say, or $x=-2$, and move towards $0$, keeping an eye on $f(x)$ as I go, what am I going to see? The value of $f(x)$ will be steadily $0$ as long as I’ve not yet reached $x=0$. If I had to predict $f(0)$ on the basis of what I was seeing, the only sensible prediction would be that $f(0)=0$, no matter what value the function actually has at $x=0$, if any.
For a slightly less trivial example, let
$$f(x)=\begin{cases}
?,&\text{if }x=0\\
x+1,&\text{otherwise}\;.
\end{cases}$$
If I start at any non-zero value of $x$ and move towards $0$, keeping an eye on $f(x)$ as I go, I’ll see $f(x)$ steadily approaching $1$. If I had to predict $f(0)$ on the basis of what I was seeing, the only sensible prediction would be that $f(0)=1$, and again this would be true no matter what value the function actually has at $x=0$, if any.
If I want to know what value $f$ does have at $x=a$, I compute $f(a)$. If I want to know what value $f$ should have at $a$ if $f(a)$ is to be consistent with the general tendency of the values of $f(x)$ for $x$ that are near but not equal to $a$, I need to ignore the actual value of $f(a)$ and see how the values of $f$ at nearby points are actually behaving: the actual value of $f(a)$ is irrelevant.
