# Testing for Continuity

Given the function $f(n) = \begin{cases} \sin (1/x), & x \neq 0 \\ 0, & x=0 \end{cases}$

describe the interval on which the function is continuous.

I know that the function is continuous at all real values except $x=0$, because $y=1/x$ is continuous at all points except for $x=0$, and $\sin (1/x)$ is a composite function.

But what about the bottom half of the function? It states that there exists a point at $(0,0)$, so why wouldn't it be continuous there as well?

• Because $\nexists \lim\limits_{x \to 0} \sin(1/x))$, so $\lim\limits_{x \to 0} f(x) \neq f(0)$ – Botond Jun 10 '18 at 19:52
• Recall the definition of continuity requires the limit at the point be equal to the value at that point. You need to check if $$\lim_{x\to 0} \sin \frac{1}{x} = 0$$ – Osama Ghani Jun 10 '18 at 19:52
• @Aniket Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – user Aug 4 '18 at 20:57

Note that $\lim_{x\to 0} \sin \frac{1}{x}$ doesn't exist, indeed for
• $x_n=\frac1{2\pi n}\to 0\implies \sin\frac1{x_n}=\sin 2\pi n=0$
• $x_n=\frac2{\pi (4n-3)}\to 0\implies \sin\frac1{x_n}=\sin \frac{\pi (4n-3)}2 =1$