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

$$\lim_{\theta\to0^-}(1+\csc\theta)=-\infty$$
$$A=\lim_{\theta\to0^-}(1+\csc\theta)=1+\lim_{\theta\to0^-}\csc\theta=1+\lim_{\theta\to0^-}\frac{1}{\sin\theta}$$ As $\theta\to0^-$, $\sin x$ approaches $\sin0=0$ from the left, where $\sin\theta\lt0$. ($\theta\to0^-$ are values of $\theta$ like $-\pi/6, -\pi/4$, etc. and all these values give $\sin \theta$ negative values) Therefore, $1/\sin \theta$ will approach $-\infty$. In other words, $$A=1+(-\infty)=-\infty$$