For small $x>0$ we can write
$$
f(x) = x^{\alpha-1} \cdot x \sum_{n=1}^\infty \frac{\sin nx}{(nx)^\alpha},
$$
which is $x^{\alpha-1}$ times a Riemann sum for
$$
I_\alpha := \int_0^\infty \sin u \frac{du}{u^\alpha}.
$$
This suggests that $x^{1-\alpha} f(x) \rightarrow I_\alpha$
as $x \rightarrow 0^+$. We shall see that $I_\alpha > 0$, so
$x^{1-\alpha} f(x) \rightarrow I_\alpha$ means that
not only is $f$ discontinuous at $x=0$ but
$f(x) \rightarrow \pm \infty$ as $x \rightarrow 0^\pm$.

The fact that the Riemann sum approaches the integral
is not immediate, because the interval of integration is infinite;
but it can be shown by partial summation:
the partial sums of $x \sum_n \sin nx$ are bounded,
and $\{(nx)^{-\alpha}\}_{n=1}^\infty$ is a decreasing positive sequence,
so $x \sum_{n=n_0}^\infty \sin nx / (nx)^\alpha = O((n_0 n)^{-\alpha})$
which approaches zero as $n_0 n \rightarrow \infty$.

Positivity of $I_\alpha$ is obtained as usual by showing that
the integral is positive on each interval
$(0,2\pi)$, $(2\pi,4\pi)$, $(4\pi,6\pi)$, etc., which we do by writing each
$\int_{2\pi m}^{2\pi m + 2\pi} \sin u \, du/u^\alpha$ as
$$
= \int_{2\pi m}^{2\pi m + \pi} \sin u \frac{du}{u^\alpha}
+ \int_{2\pi m + \pi}^{2\pi m + 2\pi} \sin u \frac{du}{u^\alpha}
= \int_{2\pi m}^{2\pi m + \pi} \sin u
\, \left(\frac1{u^\alpha} - \frac1{(u+\pi)^\alpha}\right) \, du
$$
with the integrand positive on $2\pi m < u < 2\pi m + \pi$
(and small enough for the sum over $m$ to converge).

[In fact $I_\alpha$ can be evaluated in closed form $-$
it's $\Gamma(1+\alpha) \cos(\alpha\pi/2)$ $-$ but we do not need this.]