The limit of $f(x)=\sin \frac{1}{x}$ at $x=0$ On page 96 of Spivak's Calculus, 4th Edition, he writes:

... For this function it is false that $f$ approaches $0$ near $0$. This amounts to saying that it is not true for every number $\epsilon > 0$ that we can get $|f(x)-0| \lt \epsilon$ by choosing $x$ sufficiently small, and $\neq 0$. To show this we simply have to find one $\epsilon > 0$ for which the condition $|f(x)-0| \lt \epsilon$ cannot be guaranteed, no matter how small we require $|x|$ to be. In fact, $\epsilon = \frac{1}{2}$ will do. It is impossible to ensure that $|f(x)| < \frac{1}{2}$ no matter how small we require $|x|$ to be; for if $A$ is any interval containing $0$, there is some number $x=\frac{1}{(\frac{1}{2}\pi+2n\pi)}$ which is in this interval and for this $x$ we have $f(x)=1$.

My questions are:


*

*How did he find $x=\frac{1}{(\frac{1}{2}\pi+2n\pi)}$ and what does the $n$ stand for?

*How can we show that $x=\frac{1}{(\frac{1}{2}\pi+2n\pi)}$ is contained in any interval that contains $0$?


Thank you in advance for any help provided.
 A: Recall this result 

Sequential characterization of the limit: $\lim_{x\to a} f(x)=\ell\iff$ for every sequence $(x_n)$ convergent to $a$, the sequence $(f(x_n))$ is convergent to $\ell$

Now the sequence $x_n=\displaystyle\left(\frac{1}{\frac{1}{2}\pi+2n\pi}\right)_n$ is convergent to $0$ and $f(x_n)=\sin(\frac{1}{2}\pi+2n\pi)=1$ is convergent to $1$ so if $f$ has a limit at $0$ it can only be $1$ but by taking $y_n=\displaystyle\left(\frac{1}{\pi+2n\pi}\right)_n$ which's also convergent to $0$ we have $f(y_n)=\sin(\pi+2n\pi)=0\to 0\neq 1$ so by the sequential characterization the function $f$ hasn't a limit at $0$.
A: *

*solve $\sin(\frac{1}{x})=c$ where $c\gt\frac{1}{2}$.
Here $c=1.$ $n$ stands for any integer because $\sin x$ is a periodic function.


To find $x$ given $c=1$, you have to know $sin y=1$ when $y=\frac{1}{2}\pi+2n\pi$.


*

*For any $\epsilon$, you can find an $n$ such that $x=\frac{1}{(\frac{1}{2}\pi+2n\pi)}\lt\epsilon$. That means $x$ is contained in any interval that contains 0.


To find $n$, solve $\frac{1}{(\frac{1}{2}\pi+2n\pi)}\lt\epsilon$, it is $\frac{1}{2}\pi+2n\pi \gt \frac{1}{\epsilon}$.
So, $n\gt\frac{1}{2\pi}\cdot{}(\frac{1}{\epsilon}-\frac{1}{2}\pi)$. 
A: It is being said that for every interval $A$ that contains $0$, you can always find a number within the interval that is of the form $\dfrac{1}{(\frac{1}{2}\pi+2n\pi)} = x$ where $n$ is some suitably chosen integer. It's not hard to see that if you choose $n$ to be sufficiently large, you can always make $x$ small enough to be within $A$, no matter how small the interval $A$ is.
Now note that $\sin \frac1x = \sin(\frac{1}{2}\pi+2n\pi) = \sin(\frac\pi2) = 1$. Thus no matter how small an interval you choose around $A$, $\sin \frac1x$ will always take on the value of $1$ somewhere within it.
I suggest you look at the graph of $\sin \frac1x$. The function oscillates infinitely often as you get close to $0$.
