Is $f(x)=x\sin(x^{-2})$ uniformly continuous on (0,1)? Is $f(x)=\sin(x^{-2})$?
Potentially useful equations: For all $\epsilon\gt 0$ $$|f(x)-f(y)|\le\delta$$ $$|x-y|\le\epsilon$$ If the function is uniformly continuous.
This is the only way I know of to prove uniform continuity, but I'm not entirely sure how to apply it in this case. Generally, I would also apply Triangle's inequality in these problems, but I feel that it is unhelpful in this case
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Just in case, $f(x)=\sin(x^{-2})$ is not uniformly continuous on $(0,1)$. Proof: $f(x)$ reaches minima of $-1$ at $x^{-2}=2n\pi+\pi/2$ and maxima of $+1$ at $x^{-2}=2n\pi+3\pi/2$, for any positive integer $n$. That is to say, the minima are reached for $x=1/(2n\pi+\pi/2)^{1/2}$ and the maxima at $x=1/(2n\pi+3\pi/2)^{1/2}$. The distance between a minimum $x_{1n}$ and the next maximum $x_{2n}$ (in decreasing $x$) is $$ {1\over (2n\pi+\pi/2)^{1/2}}-{1\over (2n\pi+3\pi/2)^{1/2}}$$ This distance is less than the first element of the sum, and so may be made arbitrarily small by making $n$ large. Furthermore we have $|f(x_{1n})-f(x_{2n})| = 2$. If $f$ were uniformly continuous, for some $\delta>0$ we'd have $|f(x)-f(y)|\leq 1$ for $|x-y|\leq \delta$. If we make $n$ so large that $|x_{1n}-x_{2n}|\leq \delta$, we get a contradiction.
Note that $x \sin(x^{-2})$ satisfies $\lim_{x \to 0^+} x \sin(x^{-2}) = 0$, so that the function $$f(x) = \left\{ \begin{array}{ll} x \sin(x^{-2}) & x > 0 \\ 0 & x = 0 \end{array} \right.$$ is continuous on $[0,\infty)$. It is thus uniformly continuous on any subset of any compact subset of $[0,\infty)$.