# To show following function is discontinous

Given $f(x) = [x + 1] (\sin(1/x))$, where[.] denotes greatest integer function ;

when $x\in (-1,0) \cup (0,1)$

$$f(x) = 0 , \text{ otherwise}$$

Question is to show f has discontinuity of second kind at $x=0$ and discontinuity of first kind at $x=1$

Attempt :At x=0 , LHL is zero , While RHL does not exist . I have seen this plugging values like 0.000000000001 and -0.00000000000001 .RHL does not exist because of lim as x goes to 0 sin(1/x) doesnt exist .But i also did this by putting $x = 0 + h$ and $x= 0 -h$ resp. Am i correct in drawing conclusions ? Thanks

HINT: $f(x)=\sin{1/x}$ for $x\in(0,1)$ (if $[x]$ means $\lfloor x \rfloor$).
• @SophieClad I suppose yes, if non-existence of limit in $0^+$ is obtained from properties of $\sin(1/x)$. (Your attempt gives no details). – Przemysław Scherwentke Nov 23 '14 at 8:11