Do I need $|u|=1$ to prove $D_uf(a)=\langle \nabla f(a),u\rangle$

Question

I'm reading this book and I would like to know where the authors are using the fact of $|u|=1$ in the proof.

Answer 1

It is used in the first equation, which stands alone, namely $$\frac{f(a+tu)-f(a)}{t}-\left<\nabla f(a),u\right>=\frac{|t|}{t}\epsilon(a+tu)\|u\|=\frac{|t|}{t}\epsilon(a+tu),$$ but you don't need to do this there.

Answer 2

Notice $\|tu\|=|t|$ if and only if $\|u\|=1$ (for $t\neq 0$, but we are taking limit as $t\to 0$ so it doesn't matter).

When you go from $f(a+tu)=f(a)+\langle\nabla f(a),u \rangle +\epsilon(a+tu)\|tu\|$ to next line, you need precisely the fact that $\|tu\|=|t|$