Answer
$$\lim_{\theta\to0^+}\frac{1-2\cot^2\theta}{5\cot^2\theta-7\cot\theta-8}=-\frac{2}{5}$$
Work Step by Step
$$A=\lim_{\theta\to0^+}\frac{1-2\cot^2\theta}{5\cot^2\theta-7\cot\theta-8}$$
Divide both numerator and denominator by $\cot^2\theta$: $$A=\lim_{\theta\to0^+}\frac{\frac{1}{\cot^2\theta}-2}{5-\frac{7}{\cot\theta}-\frac{8}{\cot^2\theta}}$$
As $\theta\to0^+$, we have $\sin\theta\to0$ while $\cos\theta\to1$, so $\cot\theta=\frac{\cos\theta}{\sin\theta}\to\infty$.
Therefore, as $\theta\to0^+$, $\frac{1}{\cot^2\theta}\to0$ and $\frac{7}{\cot\theta}\to0$ and $\frac{8}{\cot^2\theta}\to0$.
$$A=\frac{0-2}{5-0-0}=-\frac{2}{5}$$
