# 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 $\sin n/\sqrt{n^2+1}$?

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Does $n\to\infty$? –  Rasmus Jan 9 '11 at 14:44
I think it's more likely that, for $b_n$, they divided by $n$ instead of $n^2$; notice that $n=\sqrt{n^2}$, so the effect of dividing by $n$ on the bit inside the square root is that it gets divided by $n^2$. –  Clive Newstead Mar 27 '13 at 14:33

Hint: The square root $\sqrt{n^2+1}$ behaves just like $n$ for large $n$. Since $\lvert\sin(n)\rvert$ is bounded this summand does not contribute to the limit. To see that $\frac{n}{\sqrt{n^2+1}}$ converges to 1 divide numerator and denominator by $n$.

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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.

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EDIT: I just now see that this is an old question...

This is different that the approach that you take in your question. But maybe it is helpful nonetheless. You want to find $$\lim_{x\to \infty}\frac{x + \sin(x)}{\sqrt{x^2 + 1}}$$ Divide through by $x$ on the top an bottom: $$\lim_{x\to \infty}\frac{1 + \frac{\sin(x)}{x}}{\sqrt{1 + x^{-2}}}$$ You see that
$$\lim_{x\to \infty} \frac{\sin(x)}{x} = 0$$ (because $\sin(x)$ is bounded) and so you get $$\lim_{x\to \infty}\frac{1 + \frac{\sin(x)}{x}}{\sqrt{1 + x^{-2}}} = \frac{1+0}{\sqrt{1+0}} = 1.$$

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