Prove that f(x) is uniformly continuous Show that $f(x)=x^2cos(\frac{1}{x^2})$ on (0,1).
Let $\epsilon > 0$ and $x,y \epsilon (0,1)$. $| f(y) - f(x)| = |y^2cos(\frac{1}{y^2}) - x^2cos(\frac{1}{x^2})|$.
After this I do not know how to proceed.
 A: A common way to show a function is uniformly continuous is to show it is Lipschitz. That is, to show that $|f(x)-f(y)|\leq L |x-y|$ on a set $A$, where $L$ is some constant independent of $x,y$. A common way to show a function is Lipschitz is the use the mean value theorem: $|f(x)-f(y)|=|f'(\xi)||x-y|$. For $1>\eta>0$, show that your function $f(x)$ is Lipschitz on $[\eta, 1)$.
Show also that for $x, y \in (0, \eta]$ that $|f(x)-f(y)|\leq 2 \eta^2$.
Combine both of these facts to show that $f$ is uniformly continuous. (That is, given $\epsilon>0$, pick $\eta>0$ and then pick $\delta > 0$ so that $|f(x)-f(y)|$ is small. You will have to find a way to "bridge the gap" between the intervals $(0, \eta]$ and $(\eta, 1)$. That is, you have to deal with the case $x\in (0, \eta]$ and $y \in (\eta, 1)$. Add and subtract and use the triangle inequality.)
[P.S. If this is homework, please remember to cite all help you receive.]
A: If you define $f(0) = 0$, you can prove that $f$ is continuous on $[0,1]$.
