图形直觉

Link to image above

$$\lim_{x\rightarrow -4^-}f(x) = 2$$ $$\lim_{x\rightarrow -4^+}f(x) = 2$$ $$\lim_{x\rightarrow -4}f(x) = 2$$

$f(-4)$ DNE (Does Not Exist). The open circle at $x = -4$ means that $f$ is not defined at this point.

$$\lim_{x\rightarrow 1^-}f(x) = 4$$ $$\lim_{x\rightarrow 1^+}f(x) = -2$$ $$\lim_{x\rightarrow 1}f(x) = DNE$$

$\lim_{x\rightarrow 1}f(x) =$ DNE because $\lim_{x\rightarrow 1^-}f(x) \ne \lim_{x\rightarrow 1^+}f(x)$. $f(1) = 4$ because there is a closed circle at $x = 1$

$$\lim_{x\rightarrow 6^-}f(x) = 5$$ $$\lim_{x\rightarrow 6^+}f(x) = 5$$ $$\lim_{x\rightarrow 6}f(x) = 5$$

$f(6) = 2$ because there is a closed circle at the point $(6, 2)$

Test Yourself!

Link to image above