Continuous from the left/right

Question

Here's a related problem from: Piecewise Function

$ 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?

I want to know what intervals I should take when discussing the continuity from the left of $x=4$. Should it be $(-\infty, 4)$ or $(1,4)$? As the first interval would yield a discontinuity unlike in the second interval.

Answer 1

The short answer: you can just look at $(1,4)$.

More formally, recall from the definition of continuity that $f$ will be continuous at $x=4$ if:

• $f(4)$ exists;
• the limit $L = \lim_{x\to 4}f(x)$ exists; and
• $f(4) = L$

The limit here doesn't care whether there are other discontinuities; the behaviour at $x=1$ is irrelevant. You are free to choose as small an interval around $x=4$ as you like, so for simplicity it is fine to assume that $x>1$.