When does $\lim\limits_{x\to 0}(p(x)-q(x))/x = 0$  imply $p = q$? It’s obviously not true in general, but is it true if, for example, there exists a function $f$ such that $x$ appears in $(p(x) - q(x))$ only within an arithmetic argument of $f$?
Edit:
Background:
I was trying to establish that the rule for multiplying complex numbers be obtained from that fact that the derivative of the sine is the cosine. The significance would be that this would mean that all of trigonometry is implied by the fact that sin’ = cos, because if you have the complex number system, you have trigonometry. (The formula for the ADDITION of complex numbers is self-evident; it is only the formula for the MULTIPLICATION of complex numbers that needs justification.) However, I ran into a snag. Here is my attempt:
sin’ = cos
implies (by l’Hopital’s Rule)
lim as x goes to 0 of sin(x)/x = 1
implies (by changing the variable name from ‘x’ to ‘h’)
lim as h goes to 0 of sin(h)/h = 1
implies
lim as h/2 goes to 0 of sin(h/2)/(h/2) = 1
implies
cos(x) = (lim as h/2 goes to 0 of sin(h/2)/(h/2))(lim as h/2 goes to 0 of cos(x + h/2))
implies
cos(x) = lim as h/2 goes to 0 of ((sin(h/2)/(h/2))cos(x + h/2)
implies
cos(x) = lim as h/2 goes to 0 of 2(sin(h/2)/h)cos(x + h/2)
implies
cos(x) = lim as h goes to 0 of 2(sin(h/2)/h)cos(x + h/2)
implies
cos(x) = lim as h goes to 0 of (2sin(h/2)cos(x + h/2))/h
implies (since sin' = cos)
lim as h goes to 0 of (2sin(h/2)cos(x + h/2))/h
= lim as h goes to 0 of (sin(x + h) – sin(x))/h
At this point I want to conclude that for all x and h
2sin(h/2)cos(x + h/2) = sin(x + h) – sin(x),
from which one could then deduce the formula for the sine of a difference. One could also then deduce the formula for the cosine of a difference. From these would follow the multiplication rule of complex numbers simply by associativity.
It seems to me that the fact that the two functions in question, namely,
2sin(h/2)cos(x + h/2), and sin(x + h)- sin(x) are as “lean” as possible might allow for the conclusion of their equality, based on their common limiting behavior. Anyway, that’s where I was coming from with my question. I didn’t give the backround at first because I wanted the cast this net as wide as possible first, without anyone being influenced by the particular functions I had in mind.
 A: I can't understand parts of the body of your post ( what does it mean to be within an arithmetic argument of f?) however, to answer the question in the title: Rarely. 
In fact, there are many many functions that have the property $ \lim_{x\to 0} \frac{f(x)}{x} = 0$. There is no way to characterize $f$ is any simpler way than that property right there. I'm sure you can think of many types of functions that have that property. 
With that in consideration, we realize that we can pick any 2 functions with the above property, and we will have $\lim_{x\to 0} \frac{p(x) - q(x)}{x}=0$. Moreover, for this new limit, we will have an even larger family of functions p,q which satisfies it. Functions having the first property only form a subset of this more general family. So, since $p,q$ may be a large variety of different functions yet still satisfy your limit condition, having the limit in itself does not give us any information about equality. You will need more specific information on what $p,q$ actually are. 
Another way to answer this question is to think like this: The limit condition is a local property, it only encodes some information about the value of $p(x)-q(x)$ near $0$. However, equality of 2 functions is a global property, you need information about every point. So, only knowing something in a small region, can not possibly tell you the behavior in the entire domain (unless perhaps you have some special condition on your functions which tells us that the behavior locally is linked to it's global behavior). 
A: One sufficient condition for $\lim_{x\to0}\frac{p(x)-q(x)}x = 0$ to imply $p = q$ is for $p$ and $q$ to be linear, i.e.
$$p(x) = a+bx, \qquad q(x) = c+dx$$
for some constants $a$, $b$, $c$ and $d$.  Then
$$\lim_{x\to0}\frac{p(x)-q(x)}x = \lim_{x\to0}\frac{a-c}x+b-d,$$
which is $0$ if and only if $a = c$ and $b = d$.

On the other hand, if $p$ or $q$ are higher-order polynomials, the implication won't always hold; for example,
$$\lim_{x\to0}\frac{ax^2-bx^2}x = \lim_{x\to0}\,(a-b)x = 0$$
for any constants $a$ and $b$.
A: Generalizing from Ilmari Karonen's example, if $p(x)$ and $q(x)$ are analytic in a disk about $0$ then $\lim\limits_{x\to 0}(p(x)-q(x))/x = 0$ means $p(0) = q(0)$ and $p'(0) = q'(0)$ but nothing more.
$$0 = \lim_{x \to 0} \frac{p(x) - q(x)}{x} = \lim_{x \to 0} \frac{1}{x} \left(\sum_{k=0}^\infty \frac{p^{(k)}(0) x^k}{k!} - \frac{q^{(k)}(0) x^k}{k!}\right)  = \lim_{x \to 0} \frac{p(0) - q(0)}{x}  + p'(0)- q'(0)$$
$$ \Rightarrow p(0) = q(0), p'(0) = q'(0).$$
In other words, for $p(x)$ and $q(x)$ analytic, $\lim_{x \to 0} \frac{p(x) - q(x)}{x} = 0$ simply means that $p(x)$ and $q(x)$ have the same linearization near $0$. 
This can be used to construct all kinds of functions for which $\lim_{x \to 0} \frac{p(x) - q(x)}{x} = 0$ but $p(x) \neq q(x)$: Just let $p(x)$ be any function that admits a first-order Taylor polynomial approximation near $0$, and let $q(x)$ be that first-order Taylor approximation.  For example, $p(x) = \sin x, q(x) = x$.  Or $p(x) = \cos x, q(x) = 1$.  Etc.
