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.
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3The limit is 1, the actual function value is undefined (a hole). – Andrew Li May 08 '18 at 11:46
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1What is your definition of limit? For beginners the typical definition assumes that the function is defined in some deleted neighborhood of the point under consideration. This definition does not apply here. A more general definition of limit requires that the point under consideration be an accumulation point of domain. This is handled in one of the answers here. – Paramanand Singh May 08 '18 at 13:34
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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 $=1$.
Fred
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