I have the answer of $1$. But for the answer they split it up into $b_n = n/\sqrt{n^2+1}$ and $c_n =\sin n/\sqrt{n^2+1}$ which is fine.
Then for $b_n$ they divide evrything through by $n^2$ including everything inside the square root to give $1/\sqrt{1+1/n^2}$ and I thought you couldn't do this and checked with real numbers and they are not equivalent.
For $c_n$ they said $\sin n/\sqrt{n^2+1}\leq 1/\sqrt{n^2+1}$. Fine. But then they say this is $< 1/n$ by continuity of the square root? and this tends to $0$ so $\sin n/\sqrt{n^2+1}$ tends to $0$. I thought for this to be true $1/n$ would have to tend to $0$ slower then $\sin n/\sqrt{n^2+1}$?
Could you help please