Limit of $0/x$ as x goes to 0 What is the limit of $0/x$ as x goes to $0$, without using L'Hopital's rule? Clearly it should be $0$, but I'm not sure how it is any different from something like the divergence of $1/r^2$ which yields a dirac delta.
 A: You don't need L'Hopital's rule. Every term of the sequence $\frac{0}{x_n}$ is $0$ where $x_n \to 0$, so the limit  is $0$.
More Detail
You can picture the two functions you've mentioned:
$
  f(x)
= \frac{0}{x}
$
 is just the $x$-axis minus the point $(0,0)$ where the function is not defined, whereas
$$
  g(r)
= \frac{1}{r^2}
$$
is a function which escapes to $\infty$ quite rapidly as $r \to 0$. Picturing these two functions should help clarify why they have different limits at $0$ (I've included a picture). 

A: Since $\frac0x = 0$ for $x\neq 0$, the limit of $\frac0x$ as $x$ goes to $0$ is the same as the limit of $0$ as $x$ goes to $0$:
$$\lim_{x\to0}\frac0x = \lim_{x\to 0 } 0$$
A: You could prove it formally by applying the squeeze theorem:

Let I be an interval having the point a as a limit point.
Let g, f, and h be functions defined on I, except possibly at a
itself.
Suppose that for every x in I not equal to a, we have
$$g(x)\leq f(x)\leq h(x)$$ and also suppose that
$$ \lim _{x\to a}g(x)=\lim _{x\to a}h(x)=L.$$ Then
$$\lim_{x \to a}f(x) = L.$$

Setting $a=0$, $f(x)=\frac0x$ and $L=0$, let's use this theorem to prove that $\lim_{x \to 0}\frac0x = 0$:
Let $I$ be an interval around $0$, e.g. $(−1,1)$. I is now an interval having the point a as a limit point. Furthermore, let $g(x)=h(x)=0$, which satisfies $g(x)\leq f(x)\leq h(x)=0$ for all x in I except a (since $f(x)=\frac0x=0$ for all x≠0). Since $ \lim _{x\to a}g(x)=\lim _{x\to a}h(x)=\lim _{x\to 0}0=0$, then $\lim_{x \to 0}\frac0x = 0$.  QED.
