Infinite gradient and continuity Would I be right in thinking that although the function $f(x)=x^2\sin({1\over x^2})$ has infinite gradient, it still uniformly continuous? Thanks.
 A: Since $x^2\sin\left(\frac{1}{x^2}\right)$ is not defined at zero, I assume that we let $f(0)=0$.  The function will be uniformly continuous on all of $\mathbb{R}$ since it is continuous and has a limit as we go to infinity.
The derivative takes arbitrary large and small values on any neighborhood of zero.  In particular the derivative is not continuous at $0$, and has a discontinuity of the second kind.
Remark: You can prove that if the derivative of a function is discontinuous, it must be a discontinuity of the second kind.
Lets prove some of the above:  For $x\neq 0$ we have that $$f^'(x)=2x\sin\left(\frac{1}{x^2}\right)-\frac{1}{x}\cos\left(\frac{1}{x^2}\right).$$  From this we see that the derivative does indeed take arbitrary large positive and negative values as we approach $0$ since $\frac{1}{x}$ becomes arbitrary large and since $\cos\left(\frac{1}{x^2}\right)$ oscillates.  
At $x=0$, the derivative is $$f^(0)=\lim_{h\rightarrow 0} \frac{h^2\sin\left(\frac{1}{h^2}\right)-0}{h}=\lim_{h\rightarrow 0} h\sin\left(\frac{1}{h^2}\right)=0$$ by the squeeze theorem.  Hence $f$ is differentiable everywhere, but $f^'$ is discontinuous at $0$.  
