Can I use l'Hopital for $\lim_{n\rightarrow\infty}(a_n) = 0$ where $a_n = \tan(n) (\frac{1}{e})^{n}$

Wolfram Alpha tells me that $\lim_{n\to\infty} \tan(n) = -\infty$ to $\infty$. Does it mean that the limit doesn't exist? Furthermore can I use l'Hopital for the following limit: $\lim_{n\to \infty}(a_n) = 0$ where $a_n = \tan(n) (\frac{1}{e})^{n}$ ?

• The limit of $\tan n$ does not exist. – André Nicolas Aug 19 '15 at 16:26
• The limit is $0$, as you can see in your previous question's answer, and using l'Hopital does not make any sense here. – Crostul Aug 19 '15 at 16:29
• @Crostul the question is whether this is possible. – Symeof Aug 19 '15 at 16:35
• @Symeof The answer is no. – Crostul Aug 19 '15 at 16:41