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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.

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@Davide: Very good point, I am sorry I am not on the ball today. I don't know what I was thinking. –  Eric Naslund Sep 19 '11 at 11:22
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up vote 3 down vote accepted

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 $0$.

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