One-sided infinite limit near a vertical asymptote

As $x \to 4$, the numerator $5-2x \to 5-8 = -3$, a nonzero constant. The denominator $x-4 \to 0$, so the overall fraction grows unboundedly in magnitude. To find the sign, we need to determine whether $x-4$ approaches $0$ from the positive or negative side as $x \to 4^-$.

Since $x \to 4^-$ means $x$ is slightly less than $4$, we have $x-4$ is a small negative number. So the fraction is $\frac{\text{(negative, near }-3)}{\text{(small negative number)}}$, which is a positive number growing without bound. Therefore $\lim_{x \to 4^-} \frac{5-2x}{x-4} = +\infty$, indicating a vertical asymptote at $x=4$.