This was a practice exam and I got the question wrong but I do not understand why. The picture is the answer sheet that our professor gave us.
The left hand limit when $x=1$ is $1-x^{3}$, and the right is $sin(\pi x) $
Then after when $x=-1$ they switch and the left hand limit is $sin(\pi x) $ and the right is $1-x^{3}$
Why is that? Or is it error? 
Best Answer
The solution seems fine to me. Your confusion to me seems to be about why the left hand limit at $x=-1$ is the $\sin$ function while it is the right hand limit at $x=1$. This is because your function is defined to be $1-x^3$ for $|x|\leq1$, which is the same as saying that $-1\leq x\leq 1$.
The left hand limit is essentially the limit taking an arbitrarily small value close to $x$ which is less than $x$, which, here, puts in the region where it is defined as the $\sin$ function for $x=-1$ (because then we have $f(x_0)$ where $x_0<-1$, where this follows from definition) and as the polynomial for $x=1$. An analogous argument holds for the other sides at each point.