18. $f(x)=\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}$\nleft: $lim_{x\rightarrow1^{-}}f(x)=-infty$\nright…

Explanation:

Step1: Analyze left - hand limit

When $x\to1^{-}$, the numerator $(x - 2)(x - 4)=(1^{-}-2)(1^{-}-4)=(-1)(-3) = 3$ (a positive number), and the denominator $(x - 1)(x + 2)=(1^{-}-1)(1^{-}+2)=0^{-}\times3 = 0^{-}$ (a negative - infinitesimal). So, $\lim_{x\to1^{-}}\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}=-\infty$.

Step2: Analyze right - hand limit

When $x\to1^{+}$, the numerator $(x - 2)(x - 4)=(1^{+}-2)(1^{+}-4)=(-1)(-3)=3$ (a positive number), and the denominator $(x - 1)(x + 2)=(1^{+}-1)(1^{+}+2)=0^{+}\times3 = 0^{+}$ (a positive - infinitesimal). So, $\lim_{x\to1^{+}}\frac{(x - 2)(x - 4)}{(x - 1)(x + 2)}=\infty$.

Answer:

The left - hand limit as $x\to1$ is $-\infty$ and the right - hand limit as $x\to1$ is $\infty$.