Finding set of limit points as $ x \rightarrow 0 $ If the function $ f $ is defined as
$$
f(x) 
= 
\begin{cases}
|x|^{\phi - 1} \sin(1/x), &amp; x \neq 0 \\
0, &amp; x = 0 
\end{cases}
$$ 
where $ \phi &gt; 0 $, how does one find the set of limit points of $ f(x) $ as $ x \rightarrow 0 $, for each $ \phi &gt; 0 $?
HINT

We need to consider the following cases


$\phi&gt;1$


$$0\le||x|^{\phi - 1} \sin(1/x)|\le |x|^{\phi - 1}\to 0$$


$\phi=1$


$$|x|^{\phi - 1} \sin(1/x)=\sin(1/x)$$


$0&lt;\phi&lt;1$


$$|x|^{\phi - 1} \sin(1/x)=\frac{\sin(1/x)}{|x|^{1-\phi}}$$
So, $\lim_{x \to 0+} x^{\phi - 1}\sin(\frac{1}{x}) = \lim_{x \to 0+} x^{\phi-2} x\sin(\frac{1}{x})$. As $x\sin(\frac{1}{x}) \to 1$ as $x \to 0$ that part has no importance, limit exists where limit $x^{\phi - 2}$ exists and has same value as that limit, and the latter limit is zero when $\phi &gt; 2$, 1 when $\phi = 2$ and doesn't exist when $\phi &lt; 2$