I am relatively new to calculus and I'm trying to understand it rigorously. For this question, assume that I only consider the functions to have purely real domains and ranges.
I have seen that $\lim\limits_{x\to0}{\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})}}=1$ is considered true. This is what my calculator and an online limit solver gave, so unless they are wrong, there must be a flaw in my reasoning. I do not understand it or know how to prove it using epsilon and delta. Here's what I've thought so far:
For any function f, $\frac{f(x)} {f(x)}=1$ if f is nonzero and defined at x. If these conditions are met when $|x-c|\in(0, \delta)$, the quotient is one, so any epsilon satisfies the condition. This means that to prove $\lim\limits_{x\to c}{\frac{f(x)}{f(x)}}=1$, you just need to show that f is nonzero and defined when $|x-c|\in(0, \delta)$. I don't believe this can be shown for $f(x)=\sin(\frac{1}{x})$, since it has infinitely many x-intercepts within any $\delta$.
Have I made a mistake anywhere or failed to consider something? How would this be normally demonstrated?
The "trick" is that you only consider evaluations of the function in its domain. The point where you want to compute the limit ($x=0$) needs to be an accumulation point, and it is the case here. As for all values in the domain the ratio is $1$, this is indeed the value of the limit.