Discontinuity of $\sin\ (1/x) $ at $x=0$? Why is $\sin\ (1/x)$ discontinous at $x=0$ ? Isn't the value of sine always between $1$ and $0$? Why doesnot $\sin\ (1/x)$ exist?
 A: The function $f(x)=\sin(1/x)$ isn't discontinuous at $x=0$, it is undefined. That's a different thing.
However, if you remedy that by defining it (say we set the value to $f(0)=0$), then it will necessarily be discontinuous. This follows quite immediately from (the negation of) any reasonable definition of continuity, for instance the $\epsilon$-$\delta$ definition using $\epsilon=\frac13$.
A: In every interval centered at $x=0$ there are points where the function is $1$ and points where the function is $-1$. Then it has no limit at $x=0$.
A: Your function is undefined at $x=0$. But furthermore, you can't fix the situation because a limit for $x\rightarrow 0$ does not exist. Graphically, for $x\rightarrow 0$ your function oscillates rapidly between minus one and one, without settling on any value.
A simple way to prove a limit of a function ${\rm lim}_{x\rightarrow x_{0}}f\left(x\right)$ does not exist is to find two sequences $a_{n}$ and $b_{n}$ with limit $x_{0}$ such that $f\left(a_{n}\right)$ and $f\left(b_{n}\right)$ are different for $n\rightarrow\infty$. Can you think of such sequences in the case of your function? Hint: $\sin\left(\pi n\right)=0$ and $\sin\left(\dfrac{\pi}{2}\left(2n+1\right)\right)=1$.
A: When x = 0, for sin(1/x) you have the fraction 1/0. Division by 0 is undefined, therefore there is a discontinuity at x= 0 + pik
