I am reading a textbook about the derivatives, and it says..
Let $x, y, f(x)\in \mathbb{R}$ then we can define $f'(x)$ as the following:
$$\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=f'(x)$$
However if $x$ is a vector, then equivalent to the above is the condition
$$f(x+h) = f(x)+f'(x)h+R(h) \Rightarrow \lim_{h\rightarrow 0}\frac{R(h)}{|h|}=0$$
I am quite confused about why is this equivalent and what is $R(h)$ here? Please advise, thanks!
Think of the existence of the derivative as saying the function is “approximately linear” at that point. For the one variable case say we want to approximate the value of $f$ at a point near $a$, say $x$, using the value of $f(a)$. We can write $f(x)= f(a) + f’(a) h + R(h)$ where we took $x=a+h$. Do you see how this says $f$ is locally like a line given that the remainder $R$ gets sufficiently small? That is, $R(h)/h \to 0$ as $h \to 0$. The same thing happens in higher dimensions, except we can’t express it as a quotient because we’re dealing with vectors. Instead we express that the remainder (which is now a vector) has a magnitude that gets sufficiently small so the function $f$ is “locally linear”. Or $\|R(h)\| / |h| \to 0$
In your formula, it is missing that the limit of $\frac{R(h)}{|h|}$ should be 0.
R(h) stands for the (order one) Taylor reminder, which is defined implicitly by the first equation i.e.:
$$ R(h)= f(x+h) - [f(x)+ f^\prime(x) h] $$
Geometrically, it measures the error in the approximation to the graph of f by the tangent line at x.
