By Fréchet derivative prove that $DL(x)=L(x)$ where $L:\Bbb R^m\to \Bbb R^n$ is a lineal application.
My try: We have, $$\begin{align*}\dfrac{L(a+h)-L(a)-DL(a)h}{||h||}&=\dfrac{L(a)+L(h)-L(a)-DL(a)h}{||h||}\\ &= \dfrac{L(h)-DL(a)h}{||h||}\end{align*}$$ How do I make it store to $ L(h) $ when $ h \to 0 $, I'm stuck on that.
Hint: The derivative of any function, at any point where it exists, is unique. So for $f: \mathbb{R}^m\to \mathbb{R}^n$ an arbitrary map and $\Lambda: \mathbb{R}^m\to \mathbb{R}^n$ is a linear map such that
$$\lim_{h\to0}\dfrac{|f(x+h)-f(x)-\Lambda (h)|}{|h|}=0,$$
then ($f$ is differentiable at $x$ and) $\Lambda=Df(x)$.
Given the hint, in your situation you need to show that
$$\lim_{h\to0}\dfrac{|L(x+h)-L(x)-L(h)|}{|h|}=0.$$
Alternatively one can think of linear approximations:
$$f(x+h)=f(x)+\Lambda(h)+\text{o}_{_{h\to 0}}(|h|),$$
where $\text{o}_{_{h\to 0}}(|h|)$ is some function decaying faster than $|h|$ does as $h\to0$, e.g. the function that is constantly $0$.