Let $U\in \mathbb R^n$ an open set and $f:U\to \mathbb R^m$ an differentiable function. Thus we can write:

$$f(a+v)-f(a)=f'(a)\cdot v+r(v)$$

Where, $\lim_{v\to 0}\frac{r(v)}{|v|}=0$.

My question is really simple, my doubt is can I say that $r(0)=0$? because when we replace $v=0$ we get $0=f'(a)\cdot0+r(v)$, since $f'(a)$ is a linear transformation $f'(a)\cdot 0=0$ and $r(v)=0$.

  • $\begingroup$ Yeap, that's right (if we were trying to approximate $f(x)$ near $x$, and we don't get jt right even at that point, we're doing something wrong). $\endgroup$ – YoTengoUnLCD Jan 6 '17 at 5:24
  • $\begingroup$ It must be, of you set $v=0$ in the first equation you get $f(a)-f(a)=f'(a)\cdot 0 + r(0)$ and so $r(0) = 0$. $\endgroup$ – copper.hat Jan 6 '17 at 5:30

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