Find the values of $\alpha$ and $\beta$ for which the function $f$ is of bounded variation on $[0,1]$? The question is :

Let $f : [0,1] \longrightarrow \mathbb {R}$ be defined by 

$f(x) = x^{\alpha} \sin {\frac {1} {x^{\beta}}}$ , whenever $x \in (0,1]$ and $=0$ , whenever $x = 0$.
Then find the values of $\alpha$ and $\beta$ for which $f$ becomes a function of bounded variation on $[0,1]$.
Please give me a right way to proceed.Thank you in advance.
I have just obtained the result which is 
$\alpha > 2$ and $\beta < \alpha - 2$.Is it correct at all?Please verify it.
 A: Hint: If $f$ is continuously differentiable on $[a,b],$ then the total variation of $f$ on $[a,b]$ is $\int_a^b|f'(x)|\, dx.$ 
A: Claim. Let $\alpha>0$, $\beta > 0$ and let $f$ be as given. Then $f \in \operatorname{BV}(\left[0,1\right])$ if and only if $\alpha > \beta$.
Proof. Suppose that $\alpha>\beta$. Let $h:\left[0,1\right]\to\mathbf{R}$ be defined by
$$
h(x)=
\begin{cases}
\displaystyle{x^{\alpha/(\alpha-\beta)}\sin\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)}&,\text{ if } 0<x\le 1\\
0&,\text{ if } x = 0.
\end{cases}
$$
Then $h$ is differentiable at every $x\in \left[0,1\right]$. In fact, we have
$$
h^{\prime}(x)=
\begin{cases}
\displaystyle{\frac{\alpha}{\alpha-\beta}x^{\beta/(\alpha-\beta)}\sin\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)-\frac{\beta}{\alpha-\beta}\cos\left(\frac{1}{x^{\beta/(\alpha-\beta)}}\right)}&,\text{ if }0<x\le1\\
0&,\text{ if }x=0.
\end{cases}
$$
In particular, $h^{\prime}$ is bounded, hence $h$ is Lipschitzian of order $1$, i.e., there is some $C>0$ such that for all $x\in\left[0,1\right]$ and all $y\in\left[0,1\right]$, we have
$$ \left|h(x)-h(y)\right|\le C\left|x-y\right|. $$
Substituting $x\mapsto x^{\alpha-\beta}$, $y\mapsto y^{\alpha-\beta}$, we finally get
$$ \left|f(x)-f(y)\right|\le C\left|x^{\alpha-\beta}-y^{\alpha-\beta}\right|, $$
therefore the sums appearing in the definition of total variation are bounded above by $C(t_{n}^{\alpha-\beta} - t_{0}^{\alpha-\beta})=C(1-0)=C$, which implies $f\in\operatorname{BV}(\left[0,1\right])$.
Suppose that $\alpha\le\beta$ and consider the following partition
$$ \left\lbrace 0, \left(\frac{2}{\pi(2n+1)}\right)^{1/\beta}, \left(\frac{2}{\pi(2n)}\right)^{1/\beta},\left(\frac{2}{\pi(2n-1)}\right)^{1/\beta},\cdots,\left(\frac{2}{\pi}\right)^{1/\beta},1 \right\rbrace. $$
The following holds:
$$ \begin{aligned}
\sum_{k=1}^{2n+2}\left|f(t_{k})-f(t_{k-1})\right|&\ge\sum_{k=2}^{2n+1}\left|f(t_{k})-f(t_{k-1})\right|\\
&=\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=2}^{2n+1}\left|\frac{1}{(2n+2-k)^{\alpha/\beta}}\sin\left(\frac{k\pi}{2}\right)-\frac{1}{(2n+2-(k-1))^{\alpha/\beta}}\sin\left(\frac{(k-1)\pi}{2}\right)\right|
\\
&\ge\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=1}^{n}\frac{1}{\left(2(n-k)+1\right)^{\alpha/\beta}}\\
&=\left(\frac{2}{\pi}\right)^{\alpha/\beta}\sum_{k=0}^{n-1}\frac{1}{\left(2k+1\right)^{\alpha/\beta}}.
\end{aligned} $$
Because $0<\alpha\le\beta$, the last sum can be made arbitrarily large. Consequently, $f\notin \operatorname{BV}(\left[0,1\right])$ and the claim is proved.
