Proof of there is no limit at $x=0$ for $f(x)=\sin(\frac{1}{x})$ I've seen a few questions posted before about mine, but this is a bit different.
The original form of the question can be found here: http://librarun.org/book/10452/159.
It says that prove by contradiction that $f(x)=\sin\left(\frac{1}{x}\right)$ has no limit value as $x$ $\rightarrow$ $0$.
I attempted to take $$x=\frac{1}{n\pi}$$ in which case $f(x)=0$ for every integer $n$. After, I write $x$ as follows $$x=\frac{1}{\frac{\pi}{2}+n\pi}$$
in which case we have $f(x)=1$ or $-1$ for every integer $n$. How do I proceed from here? Note that until this question writer did not mention about inifinite limits and limits of composite functions and sequences.
 A: If it is not true that "for every $\epsilon > 0$ there is some $\delta > 0$ such that if $0 < |x-a| < \delta$ then $|f(x) - L| < \epsilon$", then
$$\color{blue}{\text{ there is some } \epsilon > 0 \text{ such that for every }\delta > 0 \text{ there is some } x \text{ which satisfies } \\
0 < |x - a| < \delta \text{ but not } |f(x) - L| < \epsilon}$$
So for your limit with $f(x) = \sin(1/x)$, suppose we want to show it does not approach $0$ near $0$, consider $\epsilon = 1/2$ and note that for every $\delta > 0$ there is some $x$ with $0 < |x - 0| < \delta$ but not $|\sin(1/x) - 0| < 1/2$. Namely, $x = 1/(\pi/2 + 2\pi n )$ which satisfies $0 < |x-0| < \delta$ for some sufficiently large $n$.
More generally, any limit $L$ must lie in the interval $[-1,1]$ as that is the range of the function $\sin$. For any such $L \in [-1,1]$, there is some $x$ with $0 < |x - 0| < \delta$ but not $|\sin(1/x) - L| < 1/2$, namely 
$$x = \frac{1}{\arcsin(L) \pm \pi/2 + 2\pi n}$$
for some sufficiently large $n$ and the appropriate choice of sign. (The easiest way to see this might be to look at the graphs of $y = \left|\sin(x) - \sin(x \pm \pi/2)\right|$.)

Added: Explicitly


*

*For $L \in [0,1]$ choose the $-$ sign, i.e.,
$$x = \frac{1}{\arcsin(L) - \pi/2 + 2\pi n}$$
Then $$|f(x) - L| = |-\cos(\arcsin(L))-L| = \sqrt{1 - L^2} + L$$ which has minimum value of $1$ on the domain $[0,1]$. 

*For $L \in [-1,0)$, chose the $+$ sign. Then $$|f(x) - L| = |\cos(\arcsin(L)) - L| = \sqrt{1-L^2} - L$$ which likewise has minimum of $1$ on $[-1,0)$.
A: Hint: for $f:D\rightarrow \mathbb R$ with $(a,x_0)\subset D, (b,x_0)\subset D$ (assuming $a<x_0<b$) the following are equivalent:


*

*$\lim\limits_{x\to x_0} f(x)=L$

*For every sequence $(x_n)_{n\in\mathbb N}$ in $D\setminus\{x_0\}$ with $x_n\rightarrow x_0$ it follows that $f(x_n)\rightarrow L$.

A: If you set $a_n=\frac{1}{2\pi n}$ and $b_n=\frac{1}{\frac{\pi}{2}+2\pi n}$ you'll get that $a_n\to 0$, $b_n\to 0$ but $$0=\lim_{n\to\infty }f(a_n)\neq \lim_{n\to\infty }f(b_n)=1,$$
what prove the claim.
