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What I am trying is finding $A_n < X_n < B_n$, proving that $A_n$ and $B_n$ converge, and then $X_n$ converges. I first found that $A_n = \frac{n}{n^2 + 1}$ converges because the limit is zero. I still need to find $B_n$.

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Not very clearly put, but you are choosing, I think, $A_n=\frac{n^2}{n^2+n}$ and $B_n=\frac{n^2}{n^2}$. Then Squeezing gives limit $1$. – André Nicolas Jun 30 '12 at 23:39
up vote 4 down vote accepted

Take the upper bound$B_n=n \frac{n}{n^2+1}$ and the lower bound $A_n= n \frac{n}{n^2+n}$ Hence $ \lim_{ n \rightarrow \infty} X_n =1$

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Thanks for the help! – metroxylon Jun 30 '12 at 23:34
I am glad I could help:) – clark Jun 30 '12 at 23:37
I just got one more question, if $n >= 0$ then $A_{{n}}$ would be undefined at $n = 0$? I could just consider the sequence for $n >= 1$, right? – metroxylon Jun 30 '12 at 23:44
As $X_n$ is defined it implies that is it defined for $ n \geq 1$. But even if it had a problem at $n=0$ it would not change the strenght of the argument because for a limit you just need a $k_0$ such that the above inequalities hold $ \forall n \geq k_0$ – clark Jun 30 '12 at 23:50

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