Limit of trig identity So I've come across bit of a stump for myself, and I'm hoping I can get some help.
So I know from the squeeze theorem, that limit of $x\sin (\frac1x)$ is equal to zero as $x \to 0$, and that makes a lot of sense to me. 
But can someone take the time to help me prove using the epsilon-delta definition why $x \sin(\frac1x) $ doesn't approach $1$, please?
For those curious, I got the one from limit of $\displaystyle\frac{\sin(x)}{x} \to 1$ as $x \to 0$. I know I didn't use it right, but I just want a nice epsilon-delta proof that I'm not completely insane.
Thank you
 A: Informally, as $x\to 0$, $\frac{1}{x}$ takes on integral multiples of $2\pi$ infinitely often, so the limit cannot be $1$. More formally, choose $\varepsilon>0$; we may assume $\varepsilon<1$. Then for any $\delta>0$, choose a positive integer $n$ such that $\frac{1}{2\pi n} < \delta$. Then
$$ \left|\frac{1}{1/2\pi n}\sin\frac{1}{1/2\pi n} - 1\right|
     = \left|2\pi n\sin 2\pi n - 1\right| = 1 > \varepsilon.$$
A: Given $\epsilon>0$, let $\delta=\epsilon$. Then if $|x-0|<\delta$ you have $\left|x\sin\frac1x\right|<|x|<\delta=\epsilon$. Essentially, the squeeze theorem lets you use the $\delta$ needed to prove that $|x|\to 0$ as $x\to 0$.
Given $$\lim_{x\to 0}\frac{\sin x}{x}= 1$$  and setting $y=\frac 1 x$, you can conclude that $$\lim_{y\to\infty} y\sin \frac{1}{y} = 1$$ But you are seeking $\lim_{y\to0}$ not $\lim_{y\to\infty}$.
A: Given $\varepsilon>0$, let $\delta=\varepsilon$.
If $|x-0|<\delta$ then
$$
|f(x) - 0| = \left|x\sin\frac1x - 0\right| = |x|\left|\sin\frac1x\right|\le|x|\cdot 1 < \delta.
$$
Therefore $f(x)\to0$ as $x\to0$.
That's one way to see that the limit is not $1$.
A question arises: Is there some reason to think the limit would be $1$??
A: There is no reason for this limit to approach 1 (don't simplify $x\sin\frac1x$ as 1) !
As $\sin x$ is bounded, $x\sin\frac1x$ tends to $0$, even using $\delta$s.
A: They Taylor expansion of $\sin y$ is:
$$
\sin y = y - {y^3\over 3!} + {y^5 \over 5!} + \cdots
$$
As $y \to 0$, the higher order terms become very small, so only the initial term matters, and $\sin y \to y$ as $y \to 0$.
However, $\sin {1\over x}$ as $x\to 0$, the initial terms becomes irrelevant and the infinite series of higher-order terms are all that matter.  Thus, $\sin {1\over x}$ does not approach ${1\over x}$ as $x\to 0$.
