How to prove it using $\epsilon$-$\delta$ defination of limit? The question is :
Show that,
$$\lim_{x \to 0} \frac {\sin \frac {1} {x}} {\sin \frac {1} {x}}$$ does not exist.
How to solve it by the defination of limit? Can it be solved using sequential criterion? Please help me. Thank you in advance.
 A: You wanted an $\varepsilon-\delta$ proof, but the claim is wrong.  So I will provide an $\varepsilon-\delta$ that the claim is wrong.  More specifically, I will show that 
$$\lim_{x \to 0} \frac {sin \frac {1} {x}} {sin \frac {1} {x}}=1$$
Given $\varepsilon>0$, put $\delta=\varepsilon$. Then whenever $0<|x-0|<\delta$ (and $1\neq x\pi k$ for some $k\in\mathbb{Z}$), we have 
$$\left| \frac {sin \frac {1} {x}} {sin \frac {1} {x}} -1\right|=0<\varepsilon.$$
A: Yeah for all $x\in\mathbb R$, and $x\ne0$, $$\lim_{x \to 0} \frac {\sin \frac {1} {x}} {\sin \frac {1} {x}} =1$$ since it became a constant function!
A: Let $f(x)$ be an abbreviation for $(\sin 1/x)/\sin 1/x .$ The statement $$\forall e>0\;\exists d>0\;\forall x \;(|x|<d\implies |f(x)-1|<e)$$ is FALSE. Because for any $d>0$ there exists $x\in (-d,d)$ at which $f(x)$ does not exist :
Given $d>0$ : For any large enough $n\in \mathbb Z^+$ we have $0<1/\pi n <d $, so if $x=1/\pi n$ then $x\in (-d,d) ,$ and $\sin 1/x=0$ so the formal expression $(\sin 1/x)/\sin 1/x$ is $0/0.$ 
