Why those division by zero are formalized? Easy example first:
$f(x) = nx$
$f'(x) = (f(x+0)-f(x))/0 = (nx+0n-nx)/0 = (0n)/0 = n$
Hard one:
$f(x) = a^x$
$f'(x) = (f(x+0)-f(x))/0 = (a^{x+0}-a^x)/0 = (a^x(a^0-1))/0 = (a^x(e^{\ln(a^0)}-1))/0 = (a^x(e^{0\ln(a)}-1))/0$
Place $e^x = 1 + x + x^2/2 + ... + x^k/!k$ in $f'(x)$.
$f'(x) = (a^x(1 + 0\ln(a) + ... - 1))/0 = (a^x(0\ln(a) + ...))/0 = a^x(\ln(a) + (...)/0)$
Discard values that are $0$ that we can't rescue them with that $/0$.
$f'(x) = a^x(\ln(a)) = a^x\ln(a) $
Why is this working? Is there a name for this type of equations? (Or type of a way of solving equations) This way of solving involves expanding $x$ of $x/0$ until some $*0$ to elimnate the outside $/0$.
But as you see this is not exactly limit, but I tagged it because it's similar.

The definition of a derivative involves a limit:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$
Which is why there is no division by zero.
For your first example:
$$f'(x)= \lim_{h \to 0} \frac{(nx+nh)-(nx)}{h}=\lim_{h \to 0} \frac{nh}{h}=\lim_{h \to 0} n=n$$
In response to the edit: You can't find a derivative without taking the limit or using s shortcut - and all the shortcuts are based on limits!

Let me see if I can try to improve this answer (as per Meelo's suggestions).
When defining a derivative, we use the form
$$\lim_{h \to 0}$$
This is the primary difference between your method and taking the derivative by its definition. This avoids the issues with forms like $\frac{0}{0}$. For example, take the function
$$f(x)=x^2$$
If you used your method, we would get
$$f'(x)=\frac{f(x+0)-f(x)}{0}=\frac{(x+0)^2-x^2}{0}=\frac{x^2-(2)(0)x-x^2}{0}=\frac{0}{0}$$
which is unhelpful. However, if we use the limit definition, we get
$$f'(x)=\lim_{h \to x}\frac{f(x+h)-f(x)}{h}=\lim_{h \to 0}\frac{(x+h)^2-x^2}{h}=\lim_{h \to 0}\frac{x^2 + 2xh + h^2-x^2}{h}$$
$$=\lim_{h \to 0}\frac{2xh +h^2}{h}=\lim_{h \to 0}2x+h$$
When $h \to 0$, the last term disappears, and we get
$$f'(x)=2x$$
The reason this is so convenient is that $h$ always disappears when we take this final limit. This goes for all higher powers of $h$.
