Answer
$-\frac{2}{5}$
Work Step by Step
$\lim_{\theta\to0^+}\frac{1-2cot^2\theta}{5cot^2\theta-7cot\theta-8}=\lim_{\theta\to0^+}\frac{sin^2\theta-2cos^2\theta}{5cos^2\theta-7sin\theta cos\theta-8sin^2\theta}=\lim_{\theta\to0^+}\frac{0-2(1)}{5(1)-7(0) (1)-8(0)}=-\frac{2}{5}$
