Does a limit exist at a cusp or sharp point Does a limit exist at a sharp curve or point? The limit as $x \to c+$ and $x \to c-$ won't equal at a cusp right? (or would it in some cases? So depending on that, does a limit always or never or sometimes exists at a cusp.

Thanks guys
Yes there exists a limit at a sharp point.
According to the definition of limit. Limit $L$ exists if $$\lim _{x\rightarrow n^+}f\left( x\right) =\lim _{x\rightarrow n^-}f\left( x\right) $$
The function is of course still continuous at the cusp so the limit exists and is evaluated as
$$\lim _{x\rightarrow n^+}f\left( x\right) = L =\lim _{x\rightarrow n^-}f\left( x\right) $$
There is a mathematical way to deal with sharp edges or cusps,
if $f(x)$ is the function with sharp edge at $x=c$ then shift the origin at $(c,0)$ and then obtain fourier series for $f(x)$,
Let $f(x)=a_0+a_1\cos{x}+a_2\cos{2x}...+b_1\sin{x}+b_2\sin{2x}...$
Then, $f'(0)=\sum_{1}^{\infty}nb_n$
If you can evaluate it you could obtain the derivative at the sharp edge.