Is the Determinant of the Jacobian a continuous function? i.e. $$f:\mathbb{R}^n \rightarrow \mathbb{R}^n $$ $$ \forall \varepsilon >0 \quad \exists \delta >0 : |x-x_0 |<\delta \Longrightarrow |det(Jf(x))-det(Jf(x_0))|<\varepsilon $$
How would I go about showing this? Does the $\varepsilon - \delta$ definition help in this case or would it be more appropriate to use the inverse function theorem?
Any suggestions appreciated
The standard example $f(t):=t^2\sin{1\over t}$, $\>f(0):=0$, shows that a differentiable function need not be continuously differentiable. In this case $J_f=f'$ is not continuous either.
For given $n\geq2$ consider the vector-valued function $$g(x):=\bigl(f(x_1),x_2,x_3,\ldots, x_n\bigr)\ .$$ Then $$J_g(x)=f'(x_1)$$ is not continuous at the origin.
If, however, you have a function $g:\>{\Bbb R}^n\to {\Bbb R}^n$ which is $C^1$ to begin with then its Jacobian determinant $J_g$ is continuous: $J_g$ is an $n$-th degree polynomial in terms of the partial derivatives $x\mapsto g_{i.k}(x)$. Since the latter are assumed to be cotinuous $J_g$ is continuous as well.