#### Answer

a) $\lim\limits_{x \to 2-}$ f(x) = 3
b) $\lim\limits_{x \to 2+}$ f(x) = 1
c) $\lim\limits_{x \to 2}$ f(x) doesn't exist.
d) f(2) = 3
e) $\lim\limits_{x \to 4}$ = 4
f) f(4) doesn't exist.

#### Work Step by Step

a) As x approaches 2 from the left, the value of $\lim\limits_{x \to 2-}$ = 3
b) As x approaches 2 from the right, the value of $\lim\limits_{x \to 2+}$ = 2
c) Because the value of $\lim\limits_{x \to 2-}$ $\ne$ the value of $\lim\limits_{x \to 2+}$, the $\lim\limits_{x \to 2}$ doesn't exist.
d) At x = 2, f(2) = 2 ( The dark point)
e) When x approaches 4, the left and right limits are the same ( = 4), so the $\lim\limits_{x \to 4}$ = 4
f) When x = 4, f(4) is undefined ( the circle, white point).