How do I find the limit of $8^{\log n}+1+\sin(n)$?

Question

How do I find the limit of $8^{\log n}+1+\sin(n)$ as $n$ goes to infinity? I know that $8^{\log_2n}$ is $n^3$ but $\sin n$ does not exist when you take limit to $\infty$. Can I use L'Hospital's somehow? I can't figure out how to make it a fraction.

Answer 1

Hint: $\sin n \geq -1$, so $8^{\log_2 n}+1+\sin n\geq 8^{\log_2 n} = n^3$.