Prove a function is not of Class $C^1$ Relating to the problem here: Show that the derivative of a function is not continuous, i.e. we have a function: 
$$g(x)=\begin{cases}
x+2x^2\sin\left(\frac{1}{x}\right)&\text{ if }x\neq0\\\
0&\text{ if }x=0
\end{cases}$$
Check that the function g is not of Class $C^1$ in any open interval around $x=0$, the Jacobian matrix of g at $x=0$ is nonsingular/invertible, but g is not injective in any open interval around $x=0$.
My understanding is that if g is of Class $C^1$, then partial derivatives exists and are continuous, and thus g is differentiable. But I'm not sure how to check the above properties of g, any help is appreciated.
 A: You are correct in your definition of $C^1$. Just realize that since $g$ is a function of one variable, in this particular case $C'$ simply means $g'$ exists and is continuous, and the Jacobian matrix will be the $1 \times 1$ matrix with entry $g'$. So the Jacobian matrix being nonsingular at $x = 0$ just means that $g'(0) \neq 0$.
Here's a sketch of what you need to do:


*

*Compute $g'(x) = 1 - 2\cos(1/x) + 4x\sin(1/x)$ at $x \neq 0$ using elementary differentiation rules. Show that $g'(0) = 1$ using the difference quotient definition. This shows that the Jacobian matrix of $g$ at $x = 0$ is nonsingular. To show that $g'$ is not continuous, show that $g'$ does not tend to $1$ as $x \to 0$. You could use an $\epsilon, \delta$ proof, but I think a more efficient method is to recall that if $g'$ is continuous at $x = 0$, then for every sequence $\{x_n\} \to 0$, the sequence $\{g'(x_n)\} \to g'(0) = 1$. Hint: when is $\cos(1/x) = 1?$.

*For noninjectivity note that $g$ is continuous. So to show that $g$ is not injective on any interval $(-c,c)$, you need only show that $g'$ takes both negative and positive values on any such interval. Hint: any such interval contains zeroes of both $\cos(1/x)$ and $\sin(1/x)$.
