I was trying to answer the following Question:
Let $f$ be a function satisfying $$f(x+y)=f(x)+f(y)+xy,\,\quad\forall x,y\in\mathbb{R}.$$ Suppose that $f'(a)=1+a$ for some $a\in\mathbb{R}.$ Then show that $f'(x)=1+x\,\forall x\in\mathbb{R}.$
One can easily notice that $f(0)=0.$ Thus, i considered two cases: $a=0$ and $a\neq0.$ For if $a=0,$ then, using the given condition for $f,$ we can easily deduce that $f'(a)=f'(0)=1$ and then it is straightforward to see that $$f'(x):=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=f'(0)+1=x+1$$ for any $x\in\mathbb{R}-\{0\}.$
If $a\neq0,$ then using $f'(a)=1+a$, we arrive at $$1+a=\lim_{h\to0}\left(\frac{f(h)}{h}+a\right)$$
My question is that whether one can safely deduce from this that $\lim_{h\to0}\frac{f(h)}{h}$ exists and $=1$ (i.e. f'(0)=1)? Do we need an additional argument on this?
[Please, forgive me if it sounds like a naive question!]