There are some necessary definitions for the theorem:
There is the theorem:If
$(a)$$V$ is open in $R^{k}$.
$(b)$$T : V \to R^{k}$ is continuous, and
$(c)$$T$ is differentiable at some point $x \in V$, then $$ \lim_{r \to 0} \frac{m(T(B(x,r)))}{m(B(x,r))} = \Delta(T'(x)). $$
There is the proof: Assume, without loss of generality, that $x = 0$ and $T(x) = 0$. Put $A = T'(0)$.
The following elementary fact about linear operators one finite-dimensional vector spaces will be used: A linear operator $A$ on $R^{k}$ is one-to-one if and only if the range of $A$ is all of $R^{k}$. In that case, the inverse $A^{-1}$ of $A$ is also linear.
Accordingly, we split the proof into two cases.
Case I $A$ is one-to-one. Define $$ F(x) = A^{-1}T(x) \ \ ( x \in V).$$
Then $F'(0) = A^{-1}T'(0) = A^{-1}A = I$, the identity operator. We shall prove that $$ \lim_{r \to 0} \frac{m(F(B(0,r)))}{m(B(0,r))} = 1. $$
Since $T(x) = AF(x)$, we have $$ m(T(B)) = m(A(F(B))) = \Delta(A)m(F(B))$$
for every ball $B$, by $7.22(3)$. ( Check it in the photo of definition)
Hence $\lim_{r \to 0} \frac{m(F(B(0,r)))}{m(B(0,r))} = 1$ will give the desired result.
Choose $\epsilon \gt 0 $. Since $F(0) = 0$ and $F'(0) = I$, there exists a $\delta \gt 0$ such that $0 \lt |x| \lt \delta$ implies $$ |F(x) - x| \lt \epsilon |x|. $$
I don't understand How does it follow from $F(0) = 0$ and $F'(0) = I$, that there exists a $\delta \gt 0$ such that $0 \lt |x| \lt \delta$ implies $|F(x) - x| \lt \epsilon |x|.$
I also don't understand about the first assumption, when we assume without loss of generality, that $x = 0$ and $T(x) = 0$. What does this assumption on single point gives us? How is it to be generalized on the entire $V$?
Any help Would be appreciated.