Vertical asymptote behavior

Combine the terms over a common denominator: $k(x) = \frac{1}{x^2} - \frac{1}{x} = \frac{1 - x}{x^2}.$ As $x \to 0$ from either side, the numerator $1-x \to 1 > 0$, while the denominator $x^2 \to 0^+$ (since $x^2$ is always positive for $x \neq 0$).

Since the denominator approaches $0$ through positive values from both sides, and the numerator stays near $1$, the whole fraction grows without bound in the positive direction from both sides. Thus $\lim_{x\to 0^-} k(x) = +\infty, \qquad \lim_{x\to 0^+} k(x) = +\infty.$

Write $k(x) = x^{-2} - x^{-1}$, so $k'(x) = -2x^{-3} + x^{-2} = \frac{-2}{x^3} + \frac{1}{x^2} = \frac{-2 + x}{x^3} = \frac{x-2}{x^3}.$ Setting the numerator equal to zero gives $x = 2$ as the only critical number (note $x=0$ is excluded from the domain).

Testing signs of $k'(x) = \frac{x-2}{x^3}$: for $0 < x < 2$, the numerator is negative and denominator positive, so $k'(x) < 0$ (decreasing); for $x > 2$, both numerator and denominator are positive, so $k'(x) > 0$ (increasing). This sign change from negative to positive means $x = 2$ gives a local minimum, with $k(2) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4}$.