Limit of a function involving infinity What would be the limit of $\frac{1}{\alpha} \left(x+\frac{\alpha}{2}\right)$ as $\alpha$ approaches 0, where $-\frac{\alpha}{2} \leq x \leq \frac{\alpha}{2},\  $  and $\alpha > 0$? 
I have tried the following, but i think it may not be correct.
$$\lim\limits_{\alpha \to 0}  \frac{x}{\alpha} +\frac{1}{2}$$
Taking the limit 
$$=\frac{x}{0} + \frac{1}{2}
 = (\infty + \frac{1}{2})
 \to\infty$$
 A: If $x=0$, the limit is $1/2$. Suppose $x>0$; then
$$
\lim_{x\to0^-}\frac{x}{\alpha}=-\infty,
\qquad
\lim_{x\to0^+}\frac{x}{\alpha}=\infty
$$
Similarly (with signs swapped) if $x<0$. Adding $1/2$ doesn't change the limits.
A: Break the problem down into cases:
Case 1: $x=-\frac{a}{2}$
$$\lim_{a\to0}\frac{1}{a}\left(\frac{a}{2}-\frac{a}{2}\right)=\lim_{a\to0}\frac{1}{a}(0)=0$$
This follows from $\forall a>0.\frac{1}{a}0=0$ for arbitrarily small $a$
Case 2: $x=\frac{a}{2}$
$$\lim_{a\to0}\frac{1}{a}\frac{2a}{2}=\lim_{a\to0}\frac{1}{a}a=1$$
This follows from $\forall a>0.\frac{1}{a}a=1$ for arbitrarily small $a$
This shows that the limit is not independent of choice of $x$, and therefore does not exist (this can be shown more explicitly by considering the function $f(x,y)=\frac{1}{y}\left(x+\frac{y}{2}\right)$. Different limits are obtained for $(x,y)\to(0,0)$ depending on the path you take to get to $(0,0)$.

While this shows that the limit does not exist, for the sake of completion, you may consider the final case.
Case 3: $x\in\left(-\frac{a}{2},\frac{a}{2}\right)\setminus\left\{0\right\}$
$$\lim_{a\to0}\frac{1}{a}\left(x+\frac{a}{2}\right)=\lim_{a\to0}\frac{x}{a}+\frac{1}{2}=\infty$$
Correction: As per Eevee Trainer's comment, $\infty$ is not a [real] number, so it is an abuse of notation to write $\lim_{a\to0}\frac{x}{a}+\frac{1}{2}=\infty$ assuming that $x$ and $a$ are real. However, this is exactly the case given a one-point compactification of the real number line.
In the absence of a point at infinity, if it is decided that the expression must have a limit, then that limit must be infinite. This can be proven using nonstandard techniques
$$\lim_{a\to0}\frac{1}{a}\left(x+\frac{a}{2}\right)\approx\lim_{a\to\epsilon}\frac{1}{a}\left(x+\frac{a}{2}\right)=\epsilon^{-1}\left(x+\frac{\epsilon}{2}\right)$$
Where $\epsilon$ is infinitesimal. It follows that $x+\frac{\epsilon}{2}\in\text{monad}(x)$, thus $x\approx x+\frac{\epsilon}{2}$. Since the reciprocal of an infinitesimal is nonfinite and $x$ is finite by the previous, the product of $\epsilon^{-1}$ and any $y$ s.t. $y\in\text{monad}(x)$ must be infinite. Therefore the limit, evaluated this way, is infinite.
STILL the limit does not exist due to the contradictions in cases 1 and 2.
You can also look at a graph of the function $f(x,y)$.

Edit: as per J.G.'s comment, the limit does exist for fixed $x=0$. This is, in fact, the only fixed value of $x$ for which the limit exists. However, whether or not this is "the" limit depends closely on how you evaluate the expression.
Edit: In response to KaviRamaMurthy's comment, it is worth noting that if $x$ is evaluated according to:
$$x=\lim_{a\to0}\left\{y\in\mathbb{R}\mid\frac{a}{2}\leq y\leq\frac{a}{2}\right\}$$
Then the result is indeterminate, whereas if $x$ is evaluated as a [single-valued] function of $a$ as in cases 1 and 2 the result is either 0, 1, or infinite.
If $x$ is evaluated as a constant, then the limit is either $\infty$ (using one-point compactification of the real-number line), $\pm\infty$ depending on whether $x$ is greater than or less than zero, or $\frac{1}{2}$ if $x$ is equal to zero.
Hopefully this covers every possible interpretation of the expression.
Yet another correction
From jnxd:

My question was transpired by the need to verify a statement $u(x)=\lim_{\alpha\to0}u_\alpha(x)$ mentioned in an
  article here. The statement is marked 4.9.

This is a different question. The statement given is
$$\hspace{50pt} \delta_{\alpha}(x)=\frac{d}{dx} u_{\alpha}(x), \hspace{15pt}  u(x)=\lim_{\alpha \rightarrow 0} u_{\alpha}(x) \hspace{50pt} (4.9)$$
Where
$$u_\alpha(x)=\begin{cases}1&x>\frac{\alpha}{2}\\\frac{1}{\alpha}\left(x+\frac{\alpha}{2}\right)&-\frac{\alpha}{2}\leq x\leq\frac{\alpha}{2}\\0&x<\frac{\alpha}{2}\end{cases}$$
In this context, the limit refers to the function $u(x)$, not its value at 0.
$$u(x)=\begin{cases}1&x>0\\0&x<0\end{cases}\quad\text{or}\quad u(x)=\begin{cases}1&x\geq0\\0&x<0\end{cases}$$
You will notice that smaller values of $\alpha$ lead to larger values of the derivative of $u_\alpha(x)$ in the region $[-\frac{\alpha}{2},\frac{\alpha}{2}]$. Because this piece of $u_\alpha(x)$ is linear for all $\alpha\neq0$, its derivative is a constant. When $\alpha=0$, $u_\alpha(x)$ becomes discontinuous at $0$ - it jumps from a value of $0$ to a value of $1$. The derivative of $u(x)\vert_{x=0}$ is thus "infinite". Ergo,
$$\delta(x)=\begin{cases}\infty&x=0\\0&\text{otherwise}\end{cases}$$
This is not the formal definition of the Dirac Delta, but it is the right idea. I think the confusion arises from the fact that the value of the Dirac Delta at $0$ is not well defined in terms of real numbers. The key takeaway is that the Dirac Delta can be approximated to arbitrary precision as the derivative of $u(x)$ - I would not consider the value of $u(0)$ itself.
If it is absolutely necessary to evaluate the function $u(x)$ at zero, then I would say $u(0)=1/2$. This can be determined either by evaluating the limit whilst fixing $x=0$ or by considering:
$$\int \delta(x)\ dx=\frac{\text{sgn}(x)+1}{2}$$
It is possible to justify this analytically (I definitely wouldn't say prove), but not without going far off topic. I would be very careful when using this, as it could easily lead to errors. Actually, if the application is probability, I would not use this at all.
