How "sharp" does a cusp have to be in order for the equation to be nondifferentiable? From a mathematical standpoint, I understand the concept of cusps: for example, a cusp exists at the origin of $y=|x|$ because one cannot take the limit from both sides, and therefore the derivative does not exist. However, I have always wondered: how "sharp" does a cusp have to be in order to make the region nondifferentiable? 
To extend the previous example, intuitively I would assume that if the slopes from both sides of $y = |x|$ where to decrease (eg $y = |\frac{1}{2}x|$, etc) the point at (0,0) would stay nondifferentiable until the slopes from both sides became 0, at which point $f'(x) = 0$. Is this correct, and if so, why? Does it have something to do with the "abruptness" of the change between values, or is there a more basic underlying concept that I am missing?
 A: You're correct. If $f$ is differentiable at a point $x = a$, then we know that the limit:
$$
\lim_{x \to a} \frac{f(x) - f(a)}{x - a}
$$
must exist (as a finite number). So from what we know about limits, we in particular we require that:
$$
\lim_{x \to a^-} \frac{f(x) - f(a)}{x - a}
= \lim_{x \to a^+} \frac{f(x) - f(a)}{x - a}
$$
In other words, the slope from the left must match the slope from the right. If there is any difference in slopes (no matter how small), then there is a cusp (actually, I call it a kink) and $f$ is not differentiable at that point.
A: In the calculus book from which I teach (Calculus: Graphical, Numerical, Algebraic, by Finney et al.), $y=|x|$ is not considered to have a cusp, but rather a corner at $x=0$.
A corner is where the two one-sided derivatives exist, are finite, and are not equal to each other. In the case $y=|x|$ the left derivative is $-1$ and the right one is $1$ at $x=0$.
A cusp is where the one-sided derivatives tend to opposite infinities. Either the left derivative is $-\infty$ and the right is $+\infty$, as with $y=x^{2/3}$ at $x=0$, or the reverse.  (Note that I am speaking informally here of a derivative equaling infinity, but I think the meaning is clear.)
For completeness, a vertical tangent is where the one-sided derivatives equal the same infinity, such as both equaling $+\infty$ for $y=\sqrt[3]{x}$ at $x=0$.
Any of these prevent the derivative from existing, since the derivative is a two-sided limit at an interior point of the domain of the function. Any two-sided limit does not exist if either one-side limit is undefined or is infinite, or is the one-sided limits are not equal.
A: Two points:

First of all, there is a nice way of considering curves that have a limited "sharpness" to them.  In particular, when (continuous) curves have a maximum "sharpness" that they achieve, we say that they are Lipschitz continuous.  Moreover, their sharpness is bounded by something called a "Lipschitz constant".
Note, the definition of "sharpness" that I'm using here is not exactly what you might expect.  In particular, there are differentiable functions that aren't Lipschitz.

The notion that the derivative at $x = 0$ of $|x|$ should be $0$ is not entirely unreasonable.  In fact, there are many circumstances when it is useful to consider the limits 
$$
f'_+(x) = \lim_{h \to 0^+} \frac{f(x + h) - f(x)}{h}\\
f'_-(x) = \lim_{h \to 0^+} \frac{f(x) - f(x- h)}{2h}\\
\frac{f'_+(x) + f'_-(x)}{2} = \lim_{h \to 0^+} \frac{f(x + h) - f(x- h)}{2h}
$$
which is equal to the derivative whenever the derivative exists, but exists when the derivative might not.
However, the definition of the derivative implies that the derivative only exists if the "slope" at the point is the same, no matter how you approach it.  That is, the derivative only exists if coming just from the right, just from the left, or "averaging" the two sides all gives you the same answer.

A possibly helpful experiment:
Note that for any value of $a>0$, the function
$$
f(x) = \sqrt{x^2 + a^2}
$$
is differentiable when $x = 0$.  What happens to the graph as we make $a$ smaller?
