In the text that I'm following, the construction of the definition of differentiation is given as follows:
My problem : In the second paragraph of point-2, we see that if the condition $\lim_{x \to 0}\frac{E(x)}{x}=0$ holds, then it is easy to see that $a=f(0)$. I'm encountering a problem here. The limit in the condition has absolutely nothing to do with what happens to $f$ at the point $0$.
If $f$ is continuous at $0$, then we can enforce the claim easily from the definition of continuity. But we're not given that $f$ is continuous (of course, differentiability implies continuity, but we cannot use that fact before defining differentiation, otherwise circularity creeps in).
What I thought about it : Well one possibility is that I'm missing something obvious and I'll greatly appreciate if someone points it out. Otherwise, I think we need an initial condition that $E(0)=0$, which enforces $a$ to be $f(0)$. Now we can impose the previously stated limit condition to try to compute $b$. So in total, we need two conditions: \begin{align} &1.\,\, E(0)=0\\ &2. \,\, \lim_{x \to 0}\frac{E(x)}{x}=0 \end{align}
Is this extra initial condition really needed? Or the limit condition alone does the job? Any help would be greatly appreciated. Thank you.
I think you are right, and the definition you copied in point 2 is flawed.
For example, simply take $$f(x) = \begin{cases} 0 & \text{if $x \ne 0$} \\ 1 & \text{if $x=0$} \end{cases} $$ Taking $a=b=0$ and hence $E(x)=0$ if $x \ne 0$, it follows that $$\lim_{x \to 0} \frac{E(x)}{x} = 0 $$
One bare minimum way to correct the discussion in point 2, without actually specifying the value of $E(0)$, is to require that $E(0)$ be defined and that $\lim_{x \to 0} E(x)$ exists. The only possible value of that limit is $0$, because $$\lim_{x \to 0} E(x) = \lim_{x \to 0} \left(\frac{E(x)}{x} \cdot x\right) = \lim_{x \to 0} \frac{E(x)}{x} \, \cdot \, \lim_{x \to 0} x = 0 \cdot 0 = 0 $$ (in this calculation we are taking "deleted limits", since the fraction $E(x)/x$ is undefined at $0$).
Together with existence of $E(0)$ this forces $E(0)=0$.