Think Weierstrass

The above function $f(x) = \frac{1}{x^3 - x^2 - 2x} = \frac{1}{x(x-2)(x+1)}$ has asymptotes at $x = -1, 0 , 2.$ . Taking the given choices for $a$ towards the closed interval $[a, a+1]$ yields:

$a = -2 \Rightarrow [-2,-1]$ which the contains asymptote $x = -1$ ;

$a = 0 \Rightarrow [0,1]$ which the contains asymptote $x = 0$ ;

$a = 1 \Rightarrow [1,2]$ which the contains asymptote $x = 2$ ;

$a = 2 \Rightarrow [2,3]$ which the contains asymptote $x = 2.$

The Weierstrass Theorem cannot be applied across any of these intervals, hence choice C is the answer.