[tex]\tan^{-1}x[/tex] is continuous over its domain, so
[tex]\displaystyle\lim_{x\to\infty}5\tan^{-1}(\ln3x)=5\tan^{-1}\left(\lim_{x\to\infty}\ln3x\right)[/tex]
and since [tex]\ln3x\to\infty[/tex] and [tex]\tan^{-1}x\to\dfrac\pi2[/tex] as [tex]x\to\infty[/tex], the value of the limit here would be [tex]\dfrac{5\pi}2[/tex]
