From :
$$-1=\lim _{x\to 0^-}\:\frac{\lfloor x \rfloor \sin x}{x}\ne\lim _{x\to 0^+}\:\frac{\lfloor x \rfloor \sin x}{x}=0$$
I know that the limit does not exist but I need to prove it with $(\epsilon , \delta)$
Any help how to start with that.
Thanks a lot.
Let there be $L$ a possible limit. We need to find an $\varepsilon$ (the distance of the function from the limit) such that $$\forall \delta>0 \exists x\neq 0: -\delta <x<\delta \land |f(x)-L|\geq \varepsilon $$ If $L\neq 0$ we'll choose $\varepsilon = \frac{|L|}{2}$ because then for every delta, if we take $x=min\{ \frac{|L|}{10},\delta ,\frac{1}{2}\}$ we'll see that $-\delta<x<\delta , x\neq 0$ and $f(x)=0 $ meaning $|f(x)-L|=|-L|=|L|\geq \varepsilon$.
So now we agree that if there is a limit, it must be $0$. But it can't be, because we know that $$\forall \varepsilon \exists \delta \forall x\neq 0 : -\delta<x<\delta \land |\frac{\sin x}{x} - 1| < \varepsilon$$
Now just do some manipulations and it'll get the job done :) .
