How do $\lim_{n\to\infty} \frac{\tan(n)}{2^n} =0 $? How do one show $\lim_{n\to \infty} \frac{\tan(n)}{2^n}=0 $
But, I didn't compute $\lim_{n\to \infty} \frac{\tan(n)}{2^n} $, as $\lim_{n\to\infty} \tan(n) $ doesn't exist even in extended real line. So, I am unable to use L'Hospital's rule.
What should I do?
My other question is what is the limit of $\lim_{n\to\infty} |\tan(n)| $, is it $\infty$ or doesn't exist infinitely?
Sorry to everyone for edit, i am now curious about this limit. That's why I edit it.
I knew, I asked a different question before in this post, but please don't do vote for closing it, I didn't know much about this $\lim_{n\to \infty} \frac{\tan(n)}{2^n}$ because of the tan(n) term, so , please help me how should I deal with it?
For b), if the sequence diverged, then $\forall L&gt;0 \ \ \exists n_0$ s.t.$ \tan(n)&gt;L \ \forall n&gt;n_0$. Yet $\tan (\pi k)=0 \ \forall k=2n, n \in \mathbb{N}$. Hence the sequence doesn't diverge. Use the same logic to show sequence doesn't converge either. Hence the limit doesn't exist.
For b), if the sequence diverged, then $\forall L&gt;0 \ \ \exists n_0$ s.t.$ \tan(n)&gt;L \ \forall n&gt;n_0$. Yet $\tan (\pi k)=0 \ \forall k=2n, n \in \mathbb{N}$. Hence the sequence doesn't diverge. Use the same logic to show sequence doesn't converge either. Hence the limit doesn't exist.