Why is $\lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f(x) - f(0)}{g(x)-g(0)}$ ? In my lecture notes: 

Why is $\lim_{x\to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f(x) - f(0)}{g(x)-g(0)}$ and so on. I know its trying to get to "$\frac{\text{change in y}}{\text{change in x}}$" but can I actually add stuff like that? 
 A: Since $f(0)=0$, it follows that $f(x) = f(x)-f(0)$. Similarly for $g(x)$.
Also, for any number $a$, and any number $b \ne 0$, we have that $a = a\cdot \frac{b}{b}$. Hence, 
$ \frac{f(x) - f(0)}{g(x)-g(0)} =  \frac{f(x) - f(0)}{g(x)-g(0)} \frac{\frac{1}{x-0}}{\frac{1}{x-0}} = \frac{\frac{f(x)-f(0)}{x-0}}{\frac{g(x)-g(0)}{x-0}}
$
A: This is not strictly an application of L'Hopital's rule, but rather (I think) a warm up to the proof of the rule for a specific case.
As anonymous points out above, given any expression, you may:
$\ \ \ $1) add 0 to it, without changing it
$\ \ \ $2) multiply it by 1 without changing it
In your problem, starting with the expression
$$\tag{1}
{1-\cos x\over x-x^2}
$$
and setting $f(x)=1-\cos x$ and $g(x)=x-x^2$, we have (as noted in your notes) 
$f(0)=0=g(0)$.
So, using 1), the expression $(1)$ can be written
$$
{(1-\cos x) - (1-\cos 0)\over (x-x^2)- 0}
$$
and then using 2)
$$\tag{2}
{(1-\cos x) - (1-\cos 0)\over (x-x^2)- 0}\cdot{x-0\over x-0}
={ {(1-\cos x) - (1-\cos 0)\over x-0 }\over{ (x-x^2)- 0 \over x-0}}
$$
In terms of $f$ ang $g$,  the right hand side of $(2)$ can be written as
$$\tag {3}
{f(x)-f(0)\over x-0}\over{g(x)-g(0)\over x-0}
$$
(note the second to last line of the first bullet point in your notes has an typo)
So, since $(1)$ and $(3)$ are the same
$$
\lim_{x\rightarrow0}{1-\cos x\over x-x^2}
=\lim_{x\rightarrow0}{{f(x)-f(0)\over x-0}\over{g(x)-g(0)\over x-0}}
={\lim\limits_{x\rightarrow0}{f(x)-f(0)\over x-0}\over\lim\limits_{x\rightarrow0}{g(x)-g(0)\over x-0}}
$$
Now here's the important observation: the limits on the right hand side of the above 
are by definition derivatives.  The whole point of doing the above manipulations was to arrive at this stage.
