Epsilon delta definition of limit gives seemingly nonsense result

Question

I want to calculate $$\lim_{x\to 0}\frac{\sin\frac1x}{\sin\frac1x}$$ This obviously equals one, but the epsilon delta definition would then be:

For all $\epsilon$>0, there exists a $\delta$>0, such that $$0<|x|<\delta \Rightarrow \Bigg|\frac{\sin\frac1x}{\sin\frac1x}-1\Bigg|<\epsilon$$ But $0<|x|<\delta$ also implies that there exists an x where $\big|\frac1x\big|$ is a multiple of pi, and $\frac{\sin\frac1x}{\sin\frac1x}=\frac00$, which is undefined. Assuming the limit equals one leads to a contradiction. Am I doing something wrong?

Answer 1

Notice that the (natural) domain of the function

$$f(x)=\frac{\sin(1/x)}{\sin(1/x)}$$

is $\operatorname{dom}(f) = \{x\in\mathbb{R} : x \neq 0 \text{ and } \sin(1/x) \neq 0\}$. So, the correct $\varepsilon$-$\delta$ definition of that limit should be as follows:

For any $\varepsilon > 0$, there exists $\delta > 0$ such that $$ x \in \operatorname{dom}(f) \text{ and }0 < |x - 0| < \delta \quad \implies \quad |f(x) - 1| < \varepsilon. $$

Of course, this condition is obviously true since $f$ is identically $1$ on its domain.