$ f(x)= \begin{cases} 4^x&\text{if}\, x\leq 1\\ \frac{9-x^2}{3-x}&\text{if}\, 1<x\leq 4\\ \sqrt x&\text{otherwise} \end{cases} $
(b) at which of these numbers is f continuous from the right, the left, or neither?
(c) sketch the graph of $f$
My question is what does the "otherwise" mean? How do I use it to find the limit on the right or left? and when sketching do I include it?
I already got the number in which $f$ is continuous at $4^x$ and $\frac{9-x^2}{3-x}$. I'm just confused on the $\sqrt x$ one.
Any help is much appreciated.