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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

  • 0
    Does $n\to\infty$?2011-01-09
  • 0
    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 $n^2$.2013-03-27

3 Answers 3