I just cannot wrap my head around this.
I encounter this issue when trying to understand the quadratic approximation and I don't seem to find a good explanation of where the 1/2 coefficient comes from by using the definition of derivatives and I think that may be related to the below?
Let $f(x) = ax^2+bx+c$ so the $f'(x)=2ax+b$ and if we take the derivative of $f'(x)$ then $f''(x)=2a$
However, if we use the definition of derivatives $$f'(x)=\lim \limits_{x \to k} \frac{f(x)-f(k)}{x-k}$$ $$f'(x) = \lim \limits_{x \to k} \frac{ax^2+bx+c-ak^2-bk-c}{x-k}$$ $$f'(x) = \lim \limits_{x \to k} \frac{a(x^2-k^2)+b(x-k)}{x-k}$$ $$f'(x) = \lim \limits_{x \to k} \frac{a(x+k)(x-k)+b(x-k)}{x-k}$$ $$f'(x) = \lim \limits_{x \to k} a(x+k)+b$$ if we stop here and don't make x+k =2x but apply the definition of derivative on this again
we will get: $$f''(x) = \lim \limits_{x \to k} \lim \limits_{x \to k} \frac{a(x+k)+b-a(k+k)-b}{x-k}$$ $$f''(x) = \lim \limits_{x \to k} \lim \limits_{x \to k} \frac{a(x+k-k-k)}{x-k}$$ $$f''(x) = \lim \limits_{x \to k} \lim \limits_{x \to k} \frac{a(x-k)}{x-k}$$ $$f''(x) = a$$
Where do I get it wrong?