Strange behavior of $\lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}$ Alright, scratch everything below the line. Let me present one cohesive question not marred by repeated edits.
The limit $\lim_{x\to a}f(x)=L$ exists iff for every $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-L|<\epsilon$ when $0<|x-a|<\delta$.
Thus, $\lim_{x\to0}\sin\left(\frac1x\right)$ does not exist because, being that it oscillates infinitely near $0$, there is no $\epsilon,\delta$.
On the other hand, with the limit$$\lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}\\\lim_{x\to0}x\sin\left(\frac1x\right)=0\\y=x\sin\left(\frac1x\right)\\\lim_{y\to0}\frac{\sin y}{y}=1\\\lim_{x\to0}\frac{\sin\left(x\sin\left(\frac1x\right)\right)}{x\sin\left(\frac1x\right)}=1$$
this proof can be shown. However, since $\sin\left(\frac1x\right)$ oscillates infinitely, by the same definition of limit we used to show the above, the limit does not exist. How do I resolve this discrepancy?
 A: A useful thing to know about limits, is that if you are evaluating limit of the form $\lim_{x \to a} f(g(x))$, and you know that $\lim_{x \to a} g(x) = b$ then $\lim_{x \to a} f(g(x)) = \lim_{y \to b} f(y)$ (assuming everything is nice and continuous). So, because you have already worked out the limit $\lim_{x\to 0} x \sin\frac{1}{x} = 0$, you get:
$$ \lim_{x \to 0} \frac{\sin(x \sin\frac{1}{x})}{x \sin\frac{1}{x}} = \lim_{y\to 0}\frac{\sin y}{y} = 1$$
Edit 
Let me be a little more precise. Suppose you know that $\lim_{y \to b} f(y) = L$ (so for $\varepsilon$ you have $\delta_f(\varepsilon)$ such that if $|y -b| < \delta_f(\varepsilon)$ you have $|f(y) - L| < \varepsilon$) and that $\lim_{x \to a} g(x) = b$ (so for $\varepsilon$ you have $\delta_g(\varepsilon)$ such that if $|x -a| < \delta_g(\varepsilon)$ you have $|g(x) - b| < \varepsilon$). I claim that then, with no extra assumption  $\lim_{x \to a} f(g(x)) = L$. With $g(x) = x \sin\frac{1}{x}$ and $f(y) = \frac{1}{y}\sin y$, this solves your problem.
The reasoning is as follows: Take $\varepsilon > 0$, and define $\delta := \delta_g(\delta_f)$. I claim that if $|x-a| < \delta$ then $|f(g(x))-L| < \varepsilon$. First, because $|x-a| < \delta_g(\delta_f(\varepsilon))$, you have $|g(x)-b| < \delta_f(\varepsilon)$. Next, because $|g(x)-b| < \delta_f(\varepsilon)$, you have $|f(g(x)) - L| < \varepsilon$, as promised.
A: The limit is undefined: no matter how small a $\delta>0$  you choose, there will exist positive $n\in\mathbb{Z}$ with $0<\frac{1}{n\pi}<\delta$. Let's call your function $f(x)$. Then the denominator of $f(x)$ is zero, making $f(x)$ itself undefined at such values. Therefore it would be untrue to say "$|f(x)-l|<\epsilon$ for all $0<|x|<\delta$", since $f(x)$ is undefined for $x=\frac{1}{n\pi}$.
A: As we know that $\lim_{x \to 0} \frac{\sin x}{x} = 1$. Point to be note here is that $lim_{x\to 0} x = 0$. I hope you got your answer with this hint.
A: (Hint) What is $\displaystyle \lim_{x \to 0} \frac{\sin x}{x}$?
A: You have something of the form $\frac{\sin(u)}u$, if $u=x\sin(1/x)$. Can you transform this into a limit of $u$? If $x\to 0$, $u\to?$
A: I'll prove this result: if $\displaystyle\lim_{x\to a} g(x)=b$ then $\displaystyle\lim_{x\to a} f(g(x))=\ell=\lim_{y\to b}f(y)$.
Let $\epsilon>0$, there's $\delta>0$: if $|x-a|<\delta|$ then $|f(g(x))-\ell|<\epsilon$
For the founded $\delta$ there's $\delta'>0$ and if $|x-a|<\delta'$ then $|g(x)-b|<\delta$
hence take $y=g(x)$ and if $|x-a|<\min(\delta,\delta')$ then $|y-b|<\delta$ and $|f(y)-\ell|<\epsilon$.
From this I think you can apply the result to your question.
