Maybe this question is dumb but I dont know how to approach it. In a book it is stated that, for $f\in C^1(\Bbb R,\Bbb R)$ two times differentiable at $c$ then it holds that
$$f''(c)=\lim_{h\to 0^+}\frac{f(c+2h)-2f(c+h)+f(c)}{h^2}\tag1$$
But I'm unable to justify the above formula from the natural one
$$f''(c)=\lim_{h\to 0}\lim_{s\to 0}\frac{f(c+h+s)-2f(c+s)+f(c)}{hs}\tag2$$
Can someone show me (or give a reference) about how to prove $(1)$? I'm also very intrigued by the reasons about why it is stated the limit in $(1)$ as just a lateral limit. Thank you.
Hint Do you know derivative of a function at a point can be approximated by
$$f(a+h)-f(a)-f'(a)h=h\epsilon\tag 1$$ where $\epsilon$ is a error term
Write the similar expression for second derivative,
$$f'(a+h)-f'(a)-f''(a)h=h\epsilon'$$
Some more work,
Now $$hf'(a+h)-hf'(a)-f''(a+h)h^2=h^2\epsilon$$ Replace $$hf'(a+h)=f(a+2h)-f(a+h)-h\epsilon$$ we get $$f(a+2h)-f(a+h)-h\epsilon -f(a+h)+f(a)+h\epsilon-f''(a)h^2=h^2\epsilon'$$ $$f(a+2h)-2f(a+h)+f(a)-f''(a)h^2=h^2\epsilon'$$
Dividing by $h^2$, you get the desired expression.