A question regarding $\lim_{x\to0}\frac{\sin\frac{1}{x}}{\sin\frac{1}{x}}$ I can not solve this question 
$$ \lim_{x\to0}\frac{\sin\frac{1}{x}}{\sin\frac{1}{x}} $$
Because I am getting two different answers for different situation.


*

*Situation - 1 ) If $\sin(1/x) = 0$ then the answer will be indeterminate.

*Situation - 2 ) If $\sin(1/x) \neq 0$ then the answer will be "one".
Which one is correct and why?
 A: The limit is $1$. Any point $x$ for which $\sin \frac{1}{x}=0$ is not in the domain of the function $\frac{\sin \frac{1}{x}}{\sin \frac{1}{x}}$, so we do not consider those values when taking the limit. However, there are points $x$ in the domain that are arbitrarily close to $0$, and for each such point we have $\frac{\sin \frac{1}{x}}{\sin \frac{1}{x}}=1$. Hence the limit is $1$.
A: Note that $\frac{\sin\left(\frac1x\right)}{\sin\left(\frac1x\right)}$ is equal to $1$ if $x$ is not of the form $\frac1{k\pi}$, with $k\in\mathbb Z\setminus\{0\}$ and $x\neq0$; otherwise, that quotient is undefined. And therefore $\lim_{x\to0}\frac{\sin\left(\frac1x\right)}{\sin\left(\frac1x\right)}=1$.
A: On every point of the domain, 
$$\frac{\sin\dfrac1x}{\sin\dfrac1x}=1.$$
As the domain has an accumulation point at $x=0$, the limit exists.
A: As the other answers have pointed out, the limit is $1$. I'll attempt to address your two situations.
Situation $1$ seems to suggest that you think that it matters what $f(a)$ is when we're looking for the limit of $f(x)$ at $x=a$. This is untrue, since a limit is the behaviour of a function around the point, not at the point. Case in point, if we had two functions $f_1(x)$ and $f_2(x)$, which are equal everywhere (black curve) except at $0$, where $f_1(x)$ has the blue point and $f_2(x)$ has the red point. This makes no difference to the limit at $0$, which we see from the graph must be the green point. So it's essentially irrelevant what $\sin\left(\frac1x\right)$ is at $0$ when we're finding the limit. 

Furthermore, $\sin\left(\frac1x\right)$ is equal to $0$ at infinitely many places in the neighborhood around $x=0$, but this does not make the limit undefined. This is because these points (multiples of $\pi$) are not included in the domain of $\frac1{\sin\left(\frac1x\right)}$ due to the fact of them being undefined. Hence, $\frac{\sin(1/x)}{\sin(1/x)}$ is indeed equal to $1$ at every point at which it's defined. If instead, the undefined points were replaced by $1$'s, you'd be correct in your intuition that the limit would be undefined. 
So then situation $2$ basically tells you the answer. Since the limit is the behaviour of the function around the point (wherever it is well defined) and the function is equal to $1$ everywhere around the point, then the limit must be $1$.
