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?
Answer
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.
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