Limit of $\frac{\sin\frac 1 x}{\sin\frac 1 x}$ as $x$ goes to zero

Question

I was taught that this limit does not exist, as when we get close to zero, the function is not defined an arbitrarily large number of times, and consequently there does not exist a neighbourhood of zero where it is defined throughout.

Is this reasoning correct? If yes, how do I formalise it?

Answer 1

This depends on the definition you're using. Some calculus books require the function to be defined in a punctured neighbourhood of the point $a$ in order for the limit as $x \to a$ to (possibly) exist, but I think this is mainly to simplify things and not confuse beginners. My impression is that a more common definition in advanced books is to only require that every punctured neighbourhood of the point $a$ contains points where the function is defined. With this definition (which is the one I use myself), your function has the limit 1 as $x \to 0$.