Evaluating a one-sided infinite limit near a vertical asymptote

As $x \to 1^-$, the numerator $\sqrt{x} \to \sqrt{1} = 1$, a positive constant. The denominator $x - 1$ approaches $0$, but since $x$ is slightly less than $1$, we have $x - 1 < 0$, so the denominator approaches $0$ through negative values.

A positive numerator near $1$ divided by a denominator approaching $0^-$ produces values that are large in magnitude and negative. For example, at $x = 0.99$, $f(x) \approx \dfrac{0.995}{-0.01} \approx -99.5$, and this magnitude increases without bound as $x \to 1^-$. Therefore $\lim_{x\to 1^-} \dfrac{\sqrt{x}}{x-1} = -\infty$, indicating a vertical asymptote at $x=1$ approached from below toward $-\infty$.