Is this simple calculus proof formal enough and correct? There is function $f$ differentiable at $x=0$ and  $f'(0) = m > 0, f(0) = 0.$
I need to prove that there is $K > 0$ and $\delta > 0$ that for every $0<x<\delta$ : $f(x) > Kx$.
So I look at the right differential at $0$:
\begin{align}
f'(0^+) &= \lim_{x\to 0^+} {f(x) - f(0)\over x - 0}
\\
&= \lim_{x\to 0^+} {f(x)\over x} = m.
\end{align}
And from that limit we can understand that there is  $0<x<\delta$: 
$\left|{f(x)\over x} - m\right|<\epsilon$ so ${f(x)\over x} - m> -\epsilon$ and so
$${f(x)\over x} > m  - \epsilon$$
and if we choose $0 < K < m/2$ then ${f(x)\over x} > m - K > K$. This allows us to conclude ${f(x)\over x} > K$ and so $f(x) > Kx$ which is what we sought to prove.
Any objections?
 A: Your solution is good, but there is room for improvement.
Main comments:


*

*When presenting your problem to others, give the domain of $f$.
Is zero on the boundary of the domain or inside it?
I would consider it more natural to work with the usual derivative instead of a right sided one.
(I assume that your function $f$ is defined in some neighborhood of the origin. If it is only defined on, say, $[0,1)$, then using the right derivative is the way to go.)

*Choice of words: The thing you are looking at is not the right differential but the right derivative.

*Using definition of limits: What are the roles of $\epsilon$ and $\delta$?
They are not very clear.
There are not quantifiers, so I don't know if that holds for all $\epsilon$ and $\delta$ or if one should be chosen before the other.
How are $K$ and $\epsilon$ related?
This confusion is the biggest problem; otherwise your solution is very good.
(Elaboration on this last point below.)


Suggestion:
Use the definition of a limit explicitly.
(This is often a good idea and makes the argument clearer.)
That is, you know that
$$
\lim_{x\to0+}\frac{f(x)}{x}=m.
$$
This means that for any $\epsilon>0$ there is $\delta>0$ such that $|f(x)/x-m|<\epsilon$ for all $x\in(0,\delta)$.
Choose $\epsilon=m/2$.
Then you have $|f(x)/x-m|<m/2$ and hence $f(x)/x>m/2$ for all $x\in(0,\delta)$; here of course $\delta$ is the one corresponding to this choice of $\epsilon$.
If you set $K=m/2$, this is the estimate you wanted.
There is, of course, freedom in making the choices.
You can choose $\epsilon<m$ first as I did above and then set $K=m-\epsilon$, or you can first choose $K<m$ and then set $\epsilon=m-K$, or do something yet different.
The way you handle $\epsilon$ and $\delta$ and relate them to $K$ is a bit unclear; using explicit quantifiers and doing so in the correct order is strongly recommended.
