I'm reviewing a previous test, and this is bothering me:
An example function: $f(x)=x-[x]$ at $a=3$, where $[]$ is the ceiling function.
I'm supposed to find the left and right derivatives of functions like this, and determine if they are continuous or not at the given point $a$.
I'm not quite sure where to start on problems like these.
I know that the ceiling function's derivative doesn't exist at integer values, and at non-integer values its derivative is 0. Is this useful information for this case?
Edit
So evaluating $$\lim_{x\to3^+}\frac{f(x)-f(3)}{x-3}=\lim_{x\to3^+}\frac{x-[x]-3-3}{x-3}=\lim_{x\to3^+}\frac{x-[x]}{x-3}$$
I note that since $x\to3$ from the right, $[x]=x+1$ as the difference gets closer. Plugging that in, the limit ends up being $\lim_{x\to3^+}\frac{1}{x-3}=\infty$.
Doing the same with $x\to3^-$, by the same reasoning, $[x]=x$, and the limit evaluates to 0.
Since the limits don't match, the function is discontinuous at $a=3$.
Is this correct?
Lets look at the left derivative. We wish to evaluate $$ \lim_{x \to 3^+}\frac{f(x)-f(3)}{x-3}$$ We may assume that $x = 3-h$ for some $h<1$ and evaluate $$ \lim_{h \to 0} \frac{f(3-h)-f(3)}{3-h-3} = \lim_{h \to 0}\frac{(3-h)-[3-h] - (3-[3])}{-h} = \lim_{h\to 0}\frac{3-h-3}{-h} = 1$$