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.

@ClementC. "Does Not Exist" – MasterOfBinary Oct 12 '15 at 20:49 — MasterOfBinary, Oct 12 '15 at 20:49

score 7 · Answer 1 · answered Oct 12 '15 at 20:50

7

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

answered Oct 12 '15 at 20:50

Clement C.

67,323

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

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