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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).

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    $\begingroup$ for negative integers, you can simplify the problem by recognizing that sin(x)/x goes to 1 as x goes to 0--then look at what's left $\endgroup$ Nov 22 '17 at 17:20
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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 $g(x)=x^a\sin x\sin(1/x)$.

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  • $\begingroup$ can you please explain why $x^asin(x)sin(1/x)=x^{a+1}sin(x)xsin(1/x)$? $\endgroup$ Nov 22 '17 at 17:25
  • $\begingroup$ @BeginningMath Doesn't $x^{a+1}/x=x^a$? $\endgroup$
    – egreg
    Nov 22 '17 at 17:26
  • $\begingroup$ thank you, understood $\endgroup$ Nov 22 '17 at 17:30
  • $\begingroup$ I don't get why you say that "the limit exists if and only if the limit of $$x^{\alpha+1} \mathrm{sin} \left( \frac{1}{x} \right)$$ does". I mean you're not allowed to say that if a part of a function diverges when x goes to $0$, then the whole function diverges. Basically, if I did understand your answer correctly you used the product rule for limits, namely you said that finding the limit of our function is the same as finding the limit of $sin(x)/x$ times $x^a\ sin(1/x)$, which is correct of both limits exists, but which is incorrect when they do not $\endgroup$
    – Skyris
    Nov 22 '17 at 19:47
  • $\begingroup$ @Skyris It's just applying the limit product rule. See the addition. $\endgroup$
    – egreg
    Nov 22 '17 at 22:06
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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.

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  • $\begingroup$ Thank you though, I finally get it. $\endgroup$
    – Skyris
    Nov 22 '17 at 20:03
  • $\begingroup$ @Skyris yes I see it, I removed the remark that was only true for real numbers. $\endgroup$
    – Shashi
    Nov 22 '17 at 20:07
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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

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  • $\begingroup$ You are using the squeeze theorem to show that the limit exists. However, when it comes to showing that a limit doesn't exist, I'm not sure if we can use the squeeze theorem. $\endgroup$
    – Skyris
    Nov 22 '17 at 19:54

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