My book says that $f(x) =\lfloor \sin x\rfloor$ is continuous at $3\pi/2$, but on drawing the graph of $\lfloor \sin x\rfloor$, there was a jump continuity at all values of $x$ where $y=1$ or $-1$.
The discontinuity is not removable, so how is the function continuous at $3\pi/2$?
Am I wrong somewhere?
(Editor's Note. I have replaced the original usage of "$[\;\;]$" to denote the "greatest integer function" with the unambiguous "$\lfloor\;\;\rfloor$". —@Blue)
On
The function “greatest integer” is not continuous at integers. So if you consider $g(x)=\lfloor f(x)\rfloor$, where $f$ is continuous, then $g$ may be not continuous where $f$ takes on an integer value.
However, the fact that $x\mapsto\lfloor x\rfloor$ is not continuous at an integer $n$ stems from the fact that $x\mapsto x$ can take on values greater and less than $n$ in every neighborhood of $n$.
In your case, there exists a neighborhood of $3\pi/2$ where $x\mapsto\sin x$ only takes on values $\ge\sin(3\pi/2)=-1$ (for the simple reason that $\sin x\ge-1$ for every $x$).
More generally, if $c$ is a local maximum or minimum point for the function $f$ and $f(c)$ is an integer, then $g(x)=\lfloor f(x)\rfloor$ is continuous at $c$.
As $[\sin(x)]=-1$ on $(\pi , 2\pi)$ so it is continuous at $\dfrac{3\pi}{2}$.