Show if the function $f(x) = x^2 sin(1/x)$ is Lipchitz on the interval $(0,\infty)$. If it is not, show if it is uniformly continuous or not.
My idea: I started by calculating the derivate which is $f'(x) = 2xsin(1/x)-cos(1/x)$. By showing that the derivate is bounded in the interval $(0, \infty)$, I proved that the function is Lipchitz. However, I found out that $\lim_{x \rightarrow 0} [2xsin(1/x)-cos(1/x)]=indeterminate$. Now I am stuck and I do not know how to show if the function is Lipchitz or not.
$f$ is Lipchitz and uniformly continuous. Actually, if $f$ is lipchitz continuous then it is uniformly continuous too. it is to show that $f$ has a continuous derivative on $\Bbb R$. Now I will show that $f$ has a bounded derivative on $\Bbb R$. For a good choice of $M$, on the interval $[M,\infty)$,we would show that $f^{\prime}$ is bounded by $5$.
$$ |f^{\prime}| \leq |2x\sin(\frac{1}{x})|+|\cos(\frac{1}{x})| \leq |2x\sin(\frac{1}{x})| + 1 .$$ Now let $M>1$ such that for every $x>M$ we have $|x\sin(\frac{1}{x})|\leq 2$. This is possible since $\lim_{x\to \infty} \frac{\sin(\frac{1}{x})}{\frac{1}{x}} = 1 $. And we get the desired inequality. Now on the interval $(0,M]$ the derivative of $f$ is bounded by $2M+1$, because :
$$ |f^{\prime}| \leq |2x\sin(\frac{1}{x})|+|\cos(\frac{1}{x})| \leq |2x\sin(\frac{1}{x})| + 1 \leq 2 |x||\sin(\frac{1}{x})| + 1 \leq 2 M +1.$$
So the derivative of $f$ is bounded on $(0,\infty)$, hence $f$ is lipchitz continuous.