"$x^p \sin(\frac{1}{x^q})$" absolutely continuous / of bounded variation

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$f(x)\ =\ x^p \sin(\frac{1}{x^q})\ (f(0)=0)$

(1) For $x$ on $[0,1]$, Find the condition of $p,q>0$ that the function f is absolutely continuous.

(2) For $x$ on $[0,1]$, Find the condition of $p, q>0$ that the function f is of bounded variation.

I've got no idea to solve this problem. Maybe I would start with dividing in two steps, "$p>q$, $p\leq q$" To solve this problem, maybe some intervals "good for showing F is AC or BV" should be given. However, I have no idea to make some good family of intervals. Could someone give me an idea?