Value of the limit $\lim_{x \to 0} \frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}$

Question

What is the value of the limit: $$\lim_{x \to 0} \frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}$$ I think the answer should be $1$, but one I overheard one of my teachers saying that it is actually undefined.

Answer 1

The function $\sin(1/x)$ is defined for $x \in D=\mathbb R \setminus(\{0\} \cup \{\frac{1}{k \pi}: k \in \mathbb Z , k \ne 0\})$.

$0$ is an accumulation point of $D$, hence $\lim_{x \to 0, x \in D} \frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}$ makes sense.

Since $\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}=1$ for all $x \in D$, the limit $=1$.