Let $\Omega \subset R^n$ is a open set of $R^n$ , $f:\Omega\rightarrow R^n $ is $C^2$ function.The Jacobi of $f$ is : $$ J_f(x)=\frac{D(f_1,...,f_n)}{D(x_1,...,x_n)}=|\frac{\partial f_i}{\partial x_j}| $$
Then, I guess that :
$$
\forall x\in \Omega,~~~J_f(x)=\lim\limits_{r\rightarrow 0}\frac{V(f(B(x,r)))}{V(B(x,r))}
$$
$B(x,r)$ is the ball of centered at $x$ of radius $r$.
$V(B(x,r))$ is volume of $B(x,r)$.
I feel it's right ,but don't know how to prove it .And ,if it's not right ,are there any right result after suitable deform ?
Up to a sign, the answer is "you are right". Let's have a look at the transformation formula for integrals, we have \begin{align*} \def\v{\operatorname{vol}}\v f[B_r(x)] &= \int_{f[B_r(x)]}\, dy\\ &= \int_{B_r(x)} |\det Df(\xi)| \, d\xi\\ &= |\det Df(\xi_{r,x})| \cdot \v B_r(x) \end{align*} where $\xi_{r,x} \in B_r(x)$ exists as the integrated function is continuous. Now $\lim_{r\to 0} \xi_{r,x} = x$, and hence $$ |J_f(x)| = |\det Df(x)| = \lim_{r\to 0} \frac{\v f[B_r(x)]}{\v B_r(x)} $$