what does $(n+\sin n)/\sqrt{n^2+1}$ converge to? 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

This is what was done to the square root: $$\sqrt{n^2+1}=\sqrt{n^2 (1+1/n^2)}=\sqrt{n^2}\sqrt{1+1/n^2}.$$
For positive $n$ we have $\sqrt{n^2}=n$, and this $n$ then cancels the $n$ in the numerator.
For the other estimate, clearly $n^2+1 > n^2$, and since the square root function is increasing this gives $\sqrt{n^2+1} > \sqrt{n^2} = n$. The numbers are positive, so taking reciprocals reverses the inequality:
$$\frac{1}{\sqrt{n^2+1}} < \frac{1}{n}.$$
Therefore
$$0 \le \left| \frac{\sin n}{\sqrt{n^2+1}} \right| < \frac{1}{n},$$
and because of squeezing the limit of $c_n$ must be zero.
