Denote $$f(x)=\frac{\sin \left(x\sin\dfrac{1}{x}\right)}{x\sin\dfrac{1}{x}}.$$We want to research the limit $$\lim_{x \to 0}f(x).$$ According to most of the textbooks for common calculus, especially in China which is named Advanced Mathematics not Mathematical Analysis, the existence of the functional limit at some point requires that the funciton has the definition over some deleted neighborhood of that point. Since the domain of $f(x)$ is $$D=\mathbb{R} \backslash M,M=\left\{x\bigg|x=0,~\text{or}~~ x=\pm\frac{1}{k\pi}(k\in \mathbb{N_+})\right\},$$ then there exists no deleted neighborhood of $x=0$ such that $f(x)$ has definition over it. That is because: $$\forall \delta>0, \exists x_0 \in M: 0<|x_0-0|<\delta.$$ Thus, the limit seems not to exist, for the reason that $f(x)$ can not satisfy the fundamental condition which is the premise for the definition of the functional limit.
But, according to Mathematical Analysis, the functional limit can be defined at the condensation point. Thus, we have to acknowledge that the limit really exists and may readily evaluate the limit $$\lim_{x \to 0}f(x)=1.$$
Now, the question comes out. Does the limit exist or not on earth? How to reconcile this contradiction especially in teaching and examination?
The assumption that the function exists on a punctured neighbourhood is the kiddie version of limits. You can even define the limit $\lim_{x \to c} f(x)$ over the rationals, for example. All you need (as you've said) is for $c$ to be an accumulation point. Then, you can define $\lim_{x \to c} f(x)$ over a domain $D$ to be the unique $L$ such that, for all $\varepsilon > 0$, there exists a $\delta > 0$ such that $$x \in (c - \delta, c + \delta) \cap D \setminus \lbrace c \rbrace \implies |f(x) - L| < \varepsilon.$$ In terms of what you are permitted to do in exams, this is really a question for your instructor.