One-sided and two-sided limits of a simplifiable rational function

Factor both numerator and denominator: $x^3 - 8 = (x-2)(x^2+2x+4)$ and $x^2 - 4 = (x-2)(x+2)$. For $x \neq 2$, the common factor $(x-2)$ cancels, giving $f(x) = \dfrac{x^2+2x+4}{x+2}$. This simplified form agrees with $f(x)$ everywhere except at $x=2$, which is exactly what matters for evaluating limits since limits ignore the point $x=a$ itself.

Substituting $x=2$ into the simplified expression gives $\dfrac{4+4+4}{4} = \dfrac{12}{4} = 3$. Since this simplified function is continuous at $x=2$, both the left-hand and right-hand limits equal 3, matching the table's trend (values approach 3 from both sides). Therefore $\lim_{x\to2^-}f(x) = 3$, $\lim_{x\to2^+}f(x) = 3$, and since these one-sided limits agree, $\lim_{x\to2}f(x) = 3$. However, $f(2)$ itself is undefined (0/0), illustrating that the limit exists even though the function has no value at that point.