1. **State the problem:** We need to graph the function $f(x) = \csc x$ and discuss its behavior near $x=0$. 2. **Recall the definition:** The cosecant function is defined as $\csc x = \frac{1}{\sin x}$. 3. **Important rules:** - $\sin x$ is zero at $x = k\pi$ for all integers $k$. - Since $\csc x$ is the reciprocal of $\sin x$, it is undefined where $\sin x = 0$. 4. **Behavior near $x=0$:** - At $x=0$, $\sin 0 = 0$, so $\csc 0$ is undefined. - As $x \to 0^+$, $\sin x \approx x$, so $\csc x = \frac{1}{\sin x} \approx \frac{1}{x}$, which tends to $+\infty$. - As $x \to 0^-$, similarly $\csc x \approx \frac{1}{x}$, which tends to $-\infty$. 5. **Graph features:** - Vertical asymptote at $x=0$. - The function has vertical asymptotes at every integer multiple of $\pi$. - Between asymptotes, $\csc x$ has branches going to $+\infty$ and $-\infty$. 6. **Summary:** Near $x=0$, $f(x) = \csc x$ is not defined and has a vertical asymptote. The function tends to $+\infty$ from the right and $-\infty$ from the left. **Final answer:** The graph of $f(x) = \csc x$ has vertical asymptotes at $x = k\pi$, including $x=0$, and near $0$, $f(x)$ tends to $+\infty$ as $x \to 0^+$ and $-\infty$ as $x \to 0^-$.