(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.

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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.