Let $f$ be a real function defined as follows:
$f(x)=x, -1< x< 1, $
$f(x+2)=x,\forall x\notin (-1,1)$
Show that $f$ is discontinuous at every odd integer.
My work:
The function is continuous in $(-1,1)$. I have proved discontinuity at $x=-1, 1$.
Now, for any integer $x=2n+1, n\in I$, the function is coming out to be continuous as per the following reasoning.
At $x=2n+1$
LHL = $\lim \limits_{h \to 0}f(2n+1-h)=\lim \limits_{h \to 0}f[(2n+1-h-2)+2]=\lim \limits_{h \to 0}2n+1-h-2=2n-1$
RHL = $\lim \limits_{h \to 0}f(2n+1+h)=\lim \limits_{h \to 0}f[(2n+1+h-2)+2]=\lim \limits_{h \to 0}2n+1+h-2=2n-1$
Also, $f(2n+1)=f[(2n+1-2)+2]=2n-1$
So, according to me, LHL=RHL=$f(2n+1)$.
How is it discontinuous at odd integers?
There is a problem with the definition of $f(x)$.
Lets analyse it at $x=0.8$ (this would work for any $|x|<1$):
By the first definition, $f(0.8)=0.8$.
By the second definition, $f(0.8)=f(-1.2+2)=-1.2$ since $-1.2<-1$
Then, as for one value of $x$ we have two different values of $f(x)$, either this is not a function or it is not well defined.
(However, for odd integers greater than $3$, it should be continuous since for those sufficiently large values $f(x)=x-2$)