## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to\pi^-}\csc x=\infty$$
$$A=\lim_{x\to\pi^-}\csc x=\lim_{x\to\pi^-}\frac{1}{\sin x}$$ As $x\to\pi^-$, $\sin x$ approaches $0$ from the left, where the values of $\sin x$ are positive. Therefore, as $x\to\pi^-$, $1/\sin x$ will approach $\infty$. $$A=\infty$$