derivative of $f(n)=(-1)^n$ can this function $f(n)=(-1)^n$ be differentiated?
When trying to find the limit, I could not apply L'Hopital's rule because $\ln (-1)$ is undefined.
Also looking at the range/ outputs of this function, it alternates between -1 and 1. I cannot relate this to the slope/ derivative of the function.
 A: It only makes sense to talk about the derivative of a function $f$ at a point $x \in \mathbb{R}$ if $f$ is defined in a neighborhood (i.e. an interval) around $x$. Since $f(n) = (-1)^n$ only makes sense for integer values of $n$ (without going into complex numbers), the domain of $f$ must be a discrete set (I.e. a subset of $\mathbb{Z}$). Thus, it doesn't make sense to take the derivative of $f$ at any point in its domain.
A: While there are already answers in the classical sense, i would like to point out that sometimes when you can't do something directly, you may find ways to overcome difficulties, one of such being a derivative of a function defined on $\mathbb{Z}$ or even more ridiculously on $\{q^{n}| n\in\mathbb{Z} \}$ for some $ q\in \mathbb{R}$\ $ \{ \pm 1, 0 \} $.
In the first case there is a discrete derivative: $\Delta f (n) = f(n + 1) - f(n)$ and in the second one there is $q$-derivative: $\Delta_{q}f = \frac{f(qx) - f(x)}{qx - x}$.
Both satisfying usual properties, e.g. linearity, analogues of Leibniz rule(which you can find yourself), derivative is $0$ iff function is constant.
Reasonable question is "and what are they good for?". Answer is: combinatorics, reccurent sequences, $q$-stuff, modern number theory and much more.   
A: The derivative of a function will exists if you can draw a tangent line to the curve at every point.
If you think on $(-1)^n$, its graph has two set of alternate points. Try to draw any imaginary tangent line to the graph and you will realize that there exists a lot of "tangent" lines. Since there is no an unique tangent, you can't define a derivative.
