(Using Heine's definition and using it without actually using it) Proving the limit of $x\sin(\frac{1}{x-1})$ as $x\rightarrow 1$ does not exist.

So far I am nowhere near a proof. The strategy I'm planning to use is the standard one. Fixed an $\epsilon$, pick $x_1$ and $x_2$ that satisfies the conditions of being within $\delta$ of $1$; that is $0<\vert x-1\vert <\delta$, then show $\vert f(x_1)-f(x_2)\vert <\epsilon$ cannot be true for all $\epsilon$. The given function is $x\sin(\frac{1}{x-1})$ and I am trying to compute for $x\sin(\frac{1}{x-1})=1,-1$ then set those as my $x_1$ and $x_2$. But, I am having difficulty solving it. So far I have trimmed it down to $1=(x-1)\sin(\frac{1}{x})$. But it just gets more complicated from there.

EDIT: I think have successfully proved this with Heine's definition and the hint given by gimusi, but would like hints on how to do it without Heine's definition. Thanks!

Proof(by contradiction, using Heine's definition and gimusi's hint.): Suppose that $\forall\epsilon >0\exists N\in\Bbb{N}\forall n\geq N(\vert x\sin(\frac{1}{x-1})-L\vert<\epsilon)$. We let $x=x_n$ such that $x_n=1+\frac1{n\frac{\pi}2}$ since $x\to 1$ as $n\to\infty$. Consider that , $x_n\sin(\frac{1}{x_n-1})=(1+\frac{1}{\frac{n\pi}{2}})\sin(n\frac{\pi}{2})$ so if $n$ is even, $\vert x_n\sin(\frac{1}{x_n-1})-L\vert=\vert 1-L\vert$, and if $n$ is odd, $\vert x_n\sin(\frac{1}{x_n-1})-L\vert=\vert -(1+L)\vert=\vert 1+L\vert$. Given $\epsilon =1$ with our assumption, it must be that $\exists$ an even $n_i$ and an odd $n_j$ such that $n_i,n_j>N$ and $\vert x_{n_i}\sin(\frac{1}{x_{n_i}-1})-L\vert<1$ and $\vert x_{n_j}\sin(\frac{1}{x_{n_j}-1})-L\vert<1$. Therefore, by the triangle inequality, \begin{align} 2 &= 1+1>\vert x_{n_i}\sin(\frac{1}{x_{n_i}-1})-L\vert+\vert x_{n_j}\sin(\frac{1}{x_{n_j}-1})-L\vert\\ &\geq \vert (x_{n_i}\sin(\frac{1}{x_{n_i}-1})-L)+(L-x_{n_j}\sin(\frac{1}{x_{n_j}-1}))\\&=\vert x_{n_i}\sin(\frac{1}{x_{n_i}-1}-x_{n_j}\sin(\frac{1}{x_{n_j}-1})\vert\\ &=\vert 1-(-1)\vert=2\end{align} Therefore $2>2$ yields the contradiction we want.

• Do you know about Heine definition for limits? And if yes, can you use the fact it is equivalent to limit? And what about arithmetic of limits?
– Holo
Commented Jun 13, 2018 at 7:42
• I don't know what that definition is and I don't think I could use that since the book has yet to cover it at this point. Commented Jun 13, 2018 at 7:43
• If you choose a sequence $x_n\to 1$ with $\sin\frac1{x_n-1}=1$ and another sequence $y_n\to 1$ with $\sin\frac{1}{y_n-1}=-1$, then $x_n \sin\frac1{x_n-1}>\frac12$ and $y_n \sin\frac{1}{y_n-1}<-\frac12$ for $n$ large. Commented Jun 13, 2018 at 7:43

HINT

Let consider

• $x_n=1+\frac1{n\frac{\pi}2}\to 1$

then

$$x_n\sin\left(\frac{1}{x_n-1}\right)=\left(1+\frac1{n\frac{\pi}2}\right)\sin \left(n\frac{\pi}2\right)$$

and then consider the limits for the subsequences for $n$ odd and $n$ even.

• Is this $x_n=1+\frac1{n\frac{\pi}2}\to 1$ a sequence as $n\rightarrow\infty$? Commented Jun 13, 2018 at 7:52
• @TheLastCipher yes of course $x_n\to 1$ as $n\to \infty$, showing that we have subsequences with different limits we can prove that the limit doesn't exist.
– user
Commented Jun 13, 2018 at 8:01
• So from $x_n=1+\frac1{n\frac{\pi}2}\to 1$, I continue with "as $n\rightarrow\infty$, we satisfy $0<\vert x_n-1\vert<\delta$ for all $x$ and $\delta>0$", correct? Commented Jun 13, 2018 at 8:05
• @TheLastCipher , this is Heine, to avoid using sequences prove that for any deleted neighborhood there is an element of that form, and hence $\epsilon=1/2$ will never work(think about another form like that with limit to $1$ but different $f(x)$)
– Holo
Commented Jun 13, 2018 at 8:34
• this seems tougher without Heine's definition. But what form are you referring to? Did you mean $\sin(\frac{1}{x-1})$? how do I find a similar form without changing the problem itself? thanks Commented Jun 13, 2018 at 8:39

First, let me explain what is Heine definition of limits:

A function has a limit at a point $c$ if for every sequences $x_n$ that is defined on some deleted neighborhood $c$, $\lim_{n\to\infty}x_n=c\implies\lim_{n\to\infty}f(x_n)=L$.

It is possible to show that this definition is equivalent to the definition of Cauchy, I recommend you to try.

Using this we can see(Like @gimusi said) that $x_n=1+\frac1{n\frac{\pi}2}$ satisfy the condition but $f(x_n)=x_n\sin\left(\frac{1}{x_n-1}\right)=\left(1+\frac1{n\frac{\pi}2}\right)\sin \left(n\frac{\pi}2\right),$ which is not converge, hence there is no limit.

You can still use this method even without Heine:

Because of the Archimedean property in every deleted neighborhood there is values in the form of $1+\frac1{2n\pi}$ and $1+\frac1{2n\pi+\frac\pi2}$, therefore for every epsilon there is at least one element in the form of $f(1+\frac1{2n\pi})=0,f(1+\frac1{2n\pi+\frac\pi2})=1+\frac1{2n\pi+\frac\pi2}$, so for $\epsilon=0.5$ no $\delta$ will work. This is how we "use" Heine without actually using it

• I added a proof attempt while trying to use Heine's definition. I hope I used it correctly. Also I will go through your answer. thanks!! Commented Jun 13, 2018 at 9:29
• @TheLastCipher it seems like it is correct, notice that in the second part of my answer we don't assume Heine, see if you understand it
– Holo
Commented Jun 13, 2018 at 9:37
• You mentioned you used the arichimedean property to derive $1+\frac{1}{2n\pi}$ and $1+\frac{1}{2n\pi+\frac{\pi}{2}}$, but I don't know on where and how to use this? I need more context. Thank you! Commented Jun 13, 2018 at 9:40
• @TheLastCipher How do I know that in every set: $(1-\delta,1+\delta)\setminus\{0\}$(all the $x$ such that $0<|x-1|<\delta$) there is an element from those forms? From the archimedean property(otherwise you may find $\delta$ small enough such that there won't be such elements)
– Holo
Commented Jun 13, 2018 at 9:43
• @TheLastCipher you had a mistake in the algebra: $f(x)=x\sin(\frac{1}{x-1})\implies f(1+\frac1{2n\pi})=(1+\frac1{2n\pi})\sin\left(\dfrac{1}{(1+\frac1{2n\pi})-1}\right)=(1+\frac1{2n\pi})\sin\left(\dfrac{1}{\frac1{2n\pi}}\right)=0$, now with $\epsilon=0.5$ you can find for any $\delta>0$ $2$ elements, $x_1,x_2$, such that $|f(x_1)-f(x_2)|>1\ge0.5=\epsilon$
– Holo
Commented Jun 13, 2018 at 10:00

You can prove this with sequences.

Take two sequences ${x_n}$, $y_n$ that converge to 1 s.t. the sequences $f(x_n), f(y_n)$ have diferent limits. Then by a Theorem the Limit does not exist.