Problem with differentiation as a concept. I don't understand quiet good something here, for example if we want to find the derivative of the function $\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(h)}{h} $ and if we compute it from the function: $ f(x) = 12 + 7x $
We get that the derivative of $f(x)$ is equal to $$\lim_{h \to 0} \frac{7h}{h}$$
But I thought that we can't divide by zero (here we cancel 0 over 0), I'm I wrong or $\displaystyle \frac{0}{0}$ equals 1? 

If you're familiar with the $\varepsilon-\delta$ definition of a limit, then you can prove that if two functions $f$ and $g$ that satisfy $f(x) = g(x)$ for all $x \neq c$ in an open interval containing $c$, and $\lim_{x \to c}g(x)$ exists, then
$$\lim_{x \to c}g(x) = \lim_{x \to c}f(x)$$
To prove this (feel free to skip this part if you don't care), call $\lim_{x \to c} g(x) = L$, and then by definition for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $f(x) = g(x)$ in the intervals $(c - \delta, c)$ and $(c, c+\delta)$, and
$$0 < |x - c| < \delta \implies |g(x) - L| < \varepsilon$$
But then because $f(x) = g(x)$ for all $x$ in the interval other than $x = c$, you know that
$$0 < |x - c| < \delta \implies |f(x) - L| < \varepsilon$$
so you can say that $$\lim_{x \to c}f(x) = L$$

This relates to your question because the functions $f(x) = \frac{7h}{h}$ and $g(x) = 7$ only differ at $h = 0$, which combined with the theorem above shows that:
$$\lim_{h \to 0}7=7 \implies \lim_{h \to 0} \frac{7h}{h} = 7$$
In general this theorem is what justifies being able to cancel off variables in a limit.
