Is $x\sin(\frac{1}{x^2})$ uniformly continuous on $(0,1]$? I am really unsure how to start this one off.
I have done checks for "similar" functions like:
$f=\sin(\frac{1}{x}), x\in(0,1]$, by using $|f(x_1-f(x_2)|=|\sin(\frac{1}{x_1})$-$\sin(\frac{1}{x_2})|=2$ when $\frac{1}{x_1}=\frac{\pi}{2}+2\pi n$ and $\frac{1}{x_2}=\frac{3\pi}{2}+2\pi n$ and showing that for these values, $|x_1-x_2|<\delta$ holds for any $\delta$ by choosing the appropriate value of $n$, hence $|x_1-x_2|<\delta\rightarrow|f(x_1-f(x_2)|≥2$ and by setting $\epsilon=2$ we see it is not uniformly continuous.
$f=x\cos x, x\in(-\infty,\infty)$by showing that for $x_1=\frac{\pi}{2}+2\pi n$ and $x_2=x_1+h$ where $0<h<\min(\pi,\delta)$, $|f(x_1)-f(x_2)|≥1$ by choosing $n≥\frac{1}{2\pi \sinh}$, and thus for $\epsilon=1, |x_1-x_2|=h<\delta\rightarrow|f(x_1-f(x_2)|≥\epsilon$ and thus it is not uniformly continuous.
I get the feeling I am supposed to use a similar method here, but I cannot for the love of me find out where to start with this particular problem. Am I getting too hung up in what I have already done to see a new way of approaching it?