Answer
$\lim\limits_{\theta \to \frac{\pi}{2}^+}\frac{1}{3}\tan \theta =\lim\limits_{\theta \to \frac{\pi}{2}^+}\frac{\sin \theta }{3\cos \theta }=-\infty \\
as\,\theta\,\,approach\,\,\frac{\pi}{2}\,from\,right\,(second\,quadrant)\\\sin \theta\,is\,positive\,and\,approach\,1\\
\cos \theta\,is\,negative\,and\,approach\,0 \\
$
Work Step by Step
$\lim\limits_{\theta \to \frac{\pi}{2}^+}\frac{1}{3}\tan \theta =\lim\limits_{\theta \to \frac{\pi}{2}^+}\frac{\sin \theta }{3\cos \theta }=-\infty \\
as\,\theta\,\,approach\,\,\frac{\pi}{2}\,from\,right\,(second\,quadrant)\\\sin \theta\,is\,positive\,and\,approach\,1\\
\cos \theta\,is\,negative\,and\,approach\,0 \\
$
