Where does the limit go in the "little o" definition of $f'(x)$? \begin{equation}
f'(x)=\lim_{a\to x} \frac{f(a)-f(x)}{a-x}
\end{equation}
is the definition of a limit that I've known forever, but my professor was saying that we can rearrange this "ignoring the limit" to get
\begin{equation}
f'(x)(a-x)=f(a)-f(x)\tag{1}
\end{equation}
\begin{equation}
f'(x)(a-x)+f(x)=f(a)\tag{2}
\end{equation}
But then we have to add a "little o" term so 
\begin{equation}
f(a)=f(x)+f'(x)(a-x)+o(a-x)
\end{equation}
where $\lim_{a\to x} o(a-x)/(a-x)=0$
I just don't understand how the "little o" term makes up for the original limit that we ignored until the last step. If we rearranged this formula again, it wouldn't equal the original limit because only one term has a limit on it...
 A: We say that $f(x)=o(g(x))$ near $a$ if $\lim_{x\to a}\frac{f(x)}{g(x)}=0$.
This is really an extreme shorthand. We should really define $o_a(g(x))$ as the set of all function $h$ with the property that $h(x)/g(x)\to 0$ as $x\to a$, and then write $f(x)\in o_a(g(x)).$
But the shorthand is quite convenient, as long as you take care to not abuse it.
Now, when we say: $f(x)=f_1(x)+o(g(x))$ near $a,$ we mean $f(x)-f_1(x)\in o_a(g(x))$.
So when we say: $$f(x)=f(a)+f'(a)(x-a)+o(x-a)$$ we really mean: $$f(x)-f(a)-f'(a)(x-a)=o(x-a)$$
Which means that $$\lim_{x\to a}\frac{f(x)-f(a) - f'(a)(x-a)}{x-a}=0$$
But this is equivalent to saying:
$$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)$$

If you recognize $f_1(x)=f(a)+f'(a)(x-a)$ as the line through $(a,f(a))$ that is tangent to the graph of $f$, what this statement is saying is that, close to $a$, that error between the linear estimate $f_1(x)$ and $f(x)$ is a "whole lot smaller than" the difference between $x$ and $a$.
A: The definition of differential
$$f(x)=f(a)+f'(a)(x-a)+o(x-a) \tag 1$$
where $$\lim_{x\to a} \frac{o(x-a)}{x-a}=0$$
is equivalent to the definition of limit.
Indeed if (1) holds
$$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{f'(a)(x-a)+o(x-a)}{x-a}=\lim_{x\to a}f'(a)+\frac{o(x-a)}{x-a}=f'(a)$$
And if limit exists we can show that (1) holds, indeed we have to show that
$$f(x)=f(a)+f'(a)(x-a)+o(x-a)$$
thus
$$f(x)-f(a)-f'(a)(x-a)=o(x-a) \quad x\to a$$
and
$$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}-f'(a)=0$$
A: You cant ignore the limit. The identities of $(1)$ and $(2)$ in your question are not true. From the $\epsilon-\delta$ definition of limit we find that for each $\epsilon>0$ there is a $\delta>0$ such that
$$0<|x-a|<\delta\implies\left|\frac{f(x)-f(a)}{x-a}-f'(a)\right|<\epsilon\tag1$$
In other words: $\lim_{x\to a}\left|\frac{f(x)-f(a)}{x-a}-f'(a)\right|=0$. Then from the RHS of $(1)$ we have that
$$\left|\frac{f(x)-f(a)}{x-a}-f'(a)\right|<\epsilon\implies \left|f(x)-f(a)-f'(a)(x-a)\right|<\epsilon|x-a|\tag2$$
Now observe that the RHS of $(2)$ is the definition of little-o, that is $$f(x)-f(a)-f'(a)(x-a)=o(x-a),\quad\text{as }x-a\to 0\tag3$$
Then from $(3)$ we find that
$$f(x)=f(a)+f'(a)(x-a)+o(x-a),\quad\text{as }x\to a\tag4$$
