If $f$ is equal to an affine function up to $1$-th order at $a$, then $f$ is differentiable at $a$, proof more subtle then it appears?

Question

I came across the following exercise:

Two functions $f, g : \mathbb R \to \mathbb R$ are equal up to $n$th order at $a$ if $$ \lim_{h \to 0} \frac{f(a + h) - g(a + h)}{h^n} = 0. $$ Show that $f$ is differentiable at $a$ if and only if there is a function $g$ of the form $g(x) = a_0 + a_1(x - a)$ such that $f$ and $g$ are equal up to first order at $a$.

For the proof, if $f$ is differentiable at $a$, then $a_0 := f(a)$ and $a_1 = f'(a)$ fulfill the requirements, this is the easy direction. For the other direction, suppose there exists such an $g(x) = a_0 + a_1(x-a)$, then I found [precisely I found it here, it is Problem 2-9] the following solution, on $$ \lim_{h \to 0} \frac{f(a+h) - (a_0 + a_1h)}{h} $$ add $\lim a_1 = a_1$ to get $$ \lim_{h\to 0} \frac{f(a+h) - a_0}{h} = a_1. $$ For $a_0 \ne f(a)$, the limit diverges, so we must have $a_0 = f(a)$, then we get the limit for the derivative, and so $a_1 = f'(a)$. $\square$

But I doubt that $$ \lim_{h\to 0} \frac{f(a+h) - a_0}{h} < \infty \qquad (*) $$ implies $a_0 = f(a)$ (if nothing is assumed about $f : \mathbb R\to \mathbb R$).

I think everything that follows is that:

1) $\lim_{h\to 0} f(a+h) = a_0$

2) $f(a) \to a_0$ faster then any linear term (otherwise the $h$ in the denominator would not be "compensated" and it still diverges)

And so $a_0 = f(a)$ just follows for continuous $f$, where we have $\lim_{h\to 0} f(a+h) = f(a)$?

Am I right? And by the way, conclusion 1) and 2) I just reached by an intuitive feeling, any ideas how to make this precise?

Answer 1

As Daniel Fischer said, the assumptions say nothing about $f(a)$. E.g., the function $$f(x)=\begin{cases}1,\quad & x =0\\ 0,\quad & x\ne 0\end{cases}$$ agrees with $g(x)\equiv 0$ up to any order, according to the definition. Yet, it's not differentiable at $0$, since it's not continuous there.

There are three ways to fix this:

1. Add the requirement $f(a)=g(a)$ in the definition of "agree up to order $n$" (this is what I would do).
2. Change the statement to "Show that $f$ is differentiable at $a$ if and only if it is continuous at $a$ and there is a function $g$..."
3. Change the statement to "Show that $f$ can be made differentiable at $a$ (by redefining $f(a)$ if needed) if and only if there is a function $g$..."

Once the statement is corrected, the exercise should not present any difficulty.