1. State the problem. From the graph we observe a vertical asymptote at $x=0$ with the left branch going to $+\infty$ as $x\to0^-$ and the right branch going to $-\infty$ as $x\to0^+$. The left branch levels off near $y\approx1$ as $x\to-\infty$ and the right branch levels off near $y\approx3$ as $x\to+\infty$. There is a local minimum on the right branch near $x\approx2$ but it does not affect the end behavior. 2. Infinite limits at the vertical asymptote. From the graph we read the one-sided limits. $$\lim_{x\to0^-} f(x)=+\infty$$ $$\lim_{x\to0^+} f(x)=-\infty$$ This confirms a vertical asymptote at $x=0$. 3. Limits at infinity. The left-end behavior gives the limit as $x\to-\infty$. $$\lim_{x\to-\infty} f(x)=1$$ The right-end behavior gives the limit as $x\to+\infty$. $$\lim_{x\to+\infty} f(x)=3$$ These are horizontal asymptotes for the corresponding ends. 4. Asymptotes summary. Vertical asymptote: $x=0$. Horizontal asymptotes: $y=1$ as $x\to-\infty$ and $y=3$ as $x\to+\infty$. There is no oblique asymptote because both ends approach horizontal lines. 5. Final answer. Infinite limits: $$\lim_{x\to0^-} f(x)=+\infty$$ $$\lim_{x\to0^+} f(x)=-\infty$$ Limits at infinity: $$\lim_{x\to-\infty} f(x)=1$$ $$\lim_{x\to+\infty} f(x)=3$$ Asymptotes: vertical $x=0$, horizontal $y=1$ and $y=3$.