Limits paradox in case of sine function

The limit does not exist, but not because "both numerator and denominator are in [-1,1], so we can not determine what it exactly is". The denominator is $0$ whenever $\frac{1}{x}$ is a multiple of $π$, so in any open interval containing $0$ there is a point at which the expression is not defined, and hence the limit does not exist. This has nothing to do with the oscillatory behaviour of the numerator and denominator.

In your other example, the limit is indeed $1$ because it is indeed defined and equal to $1$ in any open interval of $0$ less $0$ itself.

When considering $\lim_{x\to 0} f(x)$, it does not matter whether or not $f(0)$ is defined at all. However, we still should have a look at the domain where your function $f(x)=\frac{\sin(1/x)}{\sin(1/x)}$ is defined. The problem is that we may sometimes divide by $0$, namely first of all $\frac1x$ is not defined for $x=0$, but also $\sin(1/x)=0$ whenever $\frac1x$ is an integer multiple of $\pi$. Hence the maximal domain of $f$ is $$D=\mathbb R\setminus\bigl(\{0\}\cup\{\,\tfrac1{k\pi}\mid k\in\mathbb Z\setminus\{0\}\,\}\bigr).$$ We note that for arbitrary $\epsilon>0$, the set $D\cap(-\epsilon,\epsilon)$ has more gaps than just at $x=0$ and I can only imagine that this is what your book objects to. On the other hand, it makes sense to define $\lim_{x\to a} f(x)$ for all $a\in\overline D$ and that would include tha case $a=0$. Only an "unlucky" wording of the definition of limit may prevent this, so you should check back what the exact definition of $\lim_{x\to a}f(x)$ is in your book (including, what conditions are imposed about where $f$ is defined).

I don't know if it should be a comment or an answer:

In most books, the definition of limit is: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $0<|x| < \delta$. Using this definition, there is no limit, as explained by many other answers.

However, some books use: given $ \epsilon > 0$, there exists $\delta > 0$ such that $|f(x) - l| < \epsilon$ for all $x$ in the domain of $f$ satisfying $0<|x| < \delta$. With this second definition, then your function is constant on its domain of definition, and the limit exists.

I guess your book is using the first definition.