Show the limit of a function doesn't exist So, we are asked to say for which  $\alpha \in \mathbb Z$, the following limit exists: $$\lim_{x\to\ 0}\ \left(x^{\alpha}\ \mathrm{sin}(x)\ \mathrm{sin}\left( \frac{1}{x}\right)\right)$$
So far, I've been able to prove that for $\alpha\geq 0$, the limit exists. But when it comes to negative integers, I'm stuck. I checked with a program and it turns out that this limit does not exist if $\alpha$ is negative. 
Thus, I've tried to find two sequences, $a_n$ and $b_n$ that both converge to $0$, but where $$\lim_{n\to+\infty} f(a_n) \neq \lim_{n\to+\infty} f(b_n)$$
where $f$ is the function above.
I've tried taking $a_n = \frac{1}{n}$, which gives me the following (knowing that alpha is negative) : $$\lim_{n\to+\infty} \left( n^{|\alpha |}\ \mathrm{sin}\left(\frac{1}{n} \right)\ \mathrm{sin}(n)\ \right) $$ which I don't know what it equals to, because $n^{|\alpha |}$ goes to infinity and $\mathrm{sin}\left(\frac{1}{n} \right)$ goes to $0$ (so it's an undetermined form).
I've also tried with $b_n=\frac{1}{\pi n}$, giving me: $$\lim_{n\to+\infty} \left( \pi^{|\alpha |}\ n^{|\alpha |}\ \mathrm{sin}\left(\frac{1}{\pi n} \right)\ \mathrm{sin}(n\pi)\ \right)=0$$ since $\mathrm{sin}(n\pi)$ is equal to $0$ for all n.
But apart from this, I really don't know what to do (and clearly, right now I haven't shown that for $\alpha<0$ the limit of the function as $x$ goes to $0$ doesn't exist).
 A: Note that, for $x\ne0$,
$$
x^a\sin(x)\sin\Bigl(\frac{1}{x}\Bigr)=
x^{a+1}\frac{\sin(x)}{x}\sin\Bigl(\frac{1}{x}\Bigr)
$$
so the limit you're looking for exists if and only if the limit of
$$
x^{a+1}\sin\Bigl(\frac{1}{x}\Bigr)
$$
does, because of the known limit $\lim_{x\to0}\frac{\sin(x)}{x}=1$. If the limits exist, then they are equal.
If $a+1>0$, the limit is zero, because
$$
-|x|^{a+1}\le x^{a+1}\sin\Bigl(\frac{1}{x}\Bigr)\le |x|^{a+1}
$$
If $a+1\le0$, the limit does not exist, because…

Note added after comments.

Suppose $\lim_{x\to a}f(x)=1$; if $\lim_{x\to a}g(x)$ exists (finite or infinite), then also $\lim_{x\to a}f(x)g(x)$ exists and is the same.

Now apply this with $f(x)=\frac{\sin x}{x}$ and $g(x)=x^{a+1}\sin(1/x)$ and with $f(x)=\frac{x}{\sin x}$ and $g(x)=x^a\sin x\sin(1/x)$.
A: So we have this function $f$:
\begin{align}
f(x)=x^\alpha \sin(x)\sin\left( \frac 1 x\right)
\end{align}
Okay, let's show that the limit of $x\to 0$ does not exist  for $\alpha\leq -1$ using your thoughts with finding sequences. Let us use $a_n= \frac{2}{\pi n}$ (only one sequence is sometimes enough!). Now we have:
\begin{align}
f(a_n)= \left(\frac{\sin\left(\frac{2}{\pi n}\right)}{\frac{2}{\pi n}} \right)\left(\frac{2}{\pi n}\right)^{\alpha+1} \sin\left(\frac{\pi n}{2}\right)
\end{align}
So we can write:
\begin{align}
f(a_n) = b_n \left(\frac{2}{\pi n}\right)^{\alpha+1} \sin\left(\frac{\pi n}{2}\right)
\end{align}
Where $b_n \to 1$ (why?). Now look at the following:
\begin{align}
\limsup_{n\to\infty}f(a_n) = 
\begin{cases}
1 & \text{ if } &\alpha=-1\\
\infty & \text{ if } & \alpha<-1
\end{cases}
\end{align}
Similarly we have:
\begin{align}
\liminf_{n\to\infty}f(a_n) = 
\begin{cases}
-1 & \text{ if } &\alpha=-1\\
-\infty & \text{ if } & \alpha<-1
\end{cases}
\end{align}
You see that the limit does not even exist for one sequence. Therefore the limit cannot exist.
A: since we have $$\left|x^{\alpha}\sin(x)\sin\left(\frac{1}{x}\right)\right|\le |x^{\alpha}|$$
and this tends to Zero if $$\alpha\geq 0$$
for $\alpha<0$ the Limit doesn't exist
