1. **Stating the problem:** We need to determine the limit $$\lim_{x \to 2^-} \frac{f(x)}{x-2}$$ where the function $f(x)$ has vertical asymptotes at $x=0$ and $x=1$, and the behavior near $x=2$ is indicated by the graph. 2. **Understanding the limit:** The expression involves the denominator $x-2$, which approaches $0$ as $x$ approaches $2$ from the left. To evaluate the limit, we need to understand the behavior of $f(x)$ near $x=2$. 3. **From the graph description:** At $x=2$, the graph shows an arrow pointing upward from the point $(2,0)$ on the x-axis, suggesting $f(2)$ is close to $0$ and $f(x)$ is positive just to the right of $2$. Since we are approaching $2$ from the left, we consider values slightly less than $2$. 4. **Analyzing the numerator and denominator:** As $x \to 2^-$, $x-2$ approaches $0$ from the negative side (since values are less than $2$). If $f(x)$ approaches a positive number close to $0$ from the left, then $f(x)$ is positive and small. 5. **Evaluating the limit:** The fraction $$\frac{f(x)}{x-2}$$ will have a positive numerator close to $0$ and a denominator approaching $0$ from the negative side, so the fraction tends to $-\infty$. 6. **Conclusion:** Therefore, $$\lim_{x \to 2^-} \frac{f(x)}{x-2} = -\infty.$$