Answer
$$\lim_{\theta\to(\pi/2)^-}\frac{4\tan^2\theta+\tan\theta+1}{\tan^2\theta+5}=4$$
Work Step by Step
$$A=\lim_{\theta\to(\pi/2)^-}\frac{4\tan^2\theta+\tan\theta+1}{\tan^2\theta+5}$$
Divide both numerator and denominator by $\tan^2\theta$: $$A=\lim_{\theta\to(\pi/2)^-}\frac{4+\frac{1}{\tan\theta}+\frac{1}{\tan^2\theta}}{1+\frac{5}{\tan^2\theta}}$$
As $\theta\to(\pi/2)^-$, we have $\sin\theta\to1$ while $\cos\theta\to0$, so $\tan\theta=\frac{\sin\theta}{\cos\theta}\to\infty$.
Therefore, as $\theta\to(\pi/2)^-$, $\frac{1}{\tan\theta}\to0$ and $\frac{1}{\tan^2\theta}\to0$ and $\frac{5}{\tan^2\theta}\to0$.
$$A=\frac{4+0+0}{1+0}=4$$
