calculus - Finding a function based on its Derivative without Integrating

My question revolves around finding a function based on its derivative of the type below :

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Problem :
The limit below represents the derivative of some real-valued function $f$ at some real-number $a$. State such an $f$ and $a$ in this case.

$$\lim_{h \ \to \ 0} \frac{\sqrt{9+h}-3}{h}$$

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Now this type of problem is slightly unique. In general we can always find a function based on it's derivative by taking the indefinite integral of the derivative, however in this case we don't have the derivative in a general form, we only have the value for the derivative function at some point $a$, and there are a large number of $f$'s which can produce the value for the derivative at that point. Am I correct in saying this?

This problem above is easily solvable, anyone can see already that the function is obviously going to be $f(x) = \sqrt{x}$, but the way I seem to solve it is a very heuristic method, which bothers me greatly (i.e. similar to guessing a function and working from there). I'm trying to find a methodical way of solving problems of this type, as the way I solve it (shown below) will definitely break down for harder examples.

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This is my solution :

By the definition of a derivative :

$$f'(x) = \lim_{h \to\ 0} \frac{f(x+h)-f(x)}{h}$$

In the above case we can see that $\ f(a+h) = \sqrt{9+h}\ $ and $\ f(a)=3 = \sqrt{9}$

This implies that $a = 9$ and $f(x) = \sqrt{x}$ or written more formally ($f : x \to \sqrt{x}, \ \forall x\in\mathbb{R^{+}}$)

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As you can see that is a very hap-hazard solution, and I would like to find a better way to solve problems of these types, but it eludes me.

Is there a more methodical approach (or formal approach) to solving problems of this type, where we are given a limit representing the derivative of some real-valued function $f$ at some real-number $a$, and asked to find $f$ and $a$?

If you have any other suggestions for solving these types of problems, please comment below.

Answer

You are correct in saying there is a large number of maps $f$ and points $a$ satisfying this. Take any real number $a$, and any map $f$ differentiable at $a$, then this equals $g'(a)$ where $\forall x \in \mathbb{R}, g(x) = f(x) +(\frac{1}{6} - f'(a))x$. This exercice is just there for you to remember that some limits are better calculated when seen as derivatives at some point. One typical example is $\lim \limits_{x \to 0} \frac{\sin(x)}{x} = 1$, though depending on how you define $\sin$ in the first place, this one might be trivial.