I'm self-studying the analysis from Zorich and the next definition of differentiability is given: $f:E\to \mathbb{R}^n$ is differentiable at the point $x$, which is a limit point of $E\subset \mathbb{R}^m$, if $f(x+h)-f(x)=L(x)\cdot h+\alpha(x;h)$, so that $L(x):\mathbb{R}^m\to \mathbb{R}^n$ is linear map wrt $h$ and $\dfrac{\alpha(x;h)}h\to 0$ while $h\to 0$ for $x+h \in E$.
Why do we need the linearity of $L$ wrt $h$? What do we lose if $L$ is not linear? Why $x$ is the argument of $L$, isn't it fixed? Shouldn't it be $L(h):\mathbb{R}^m\to \mathbb{R}^n$?
Why do we need the linearity of $L$ wrt $h$? What do we lose if $L$ is not linear?
The derivation at a point shall be the best linear approximation of the function at this point. More precisely $L$ is the best linear approximation of the difference function $g(h):=f(x+h)-f(x)$ for a fixed point $x$.
Why $x$ is the argument of $L$, isn't it fixed? Shouldn't it be $L(h):\mathbb{R}^m\to \mathbb{R}^n$?
For each point $x$ you will get a different linear map $L(x):\mathbb{R}^m\to \mathbb{R}^n$. Note that $L(x)$ is a matrix / linear map not a vector. This linear map, which is the best linear approximation of the function at this point, depends only on the function $f$ and the point $x$.
Because $L(x)$ is a map, $L(x)\cdot h = L(x)(h)$ is the approximation of the difference $f(x+h)-f(x)$. -> $L(x)(h)$ is a vector.