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${b_n} = n -\sqrt{n^2+2n}$. Taking $(\frac{1}{n})\rightarrow 0 $ as given, using Algebraic Limit Theorem, show $\lim(b_n)$ exists and find value

The question also says to use the fact that if $(x_n)\rightarrow x$ then $\sqrt{x_n}\rightarrow \sqrt{x}$. I've simplified the ${b_n}$ down to $n(1-\sqrt{1+\frac{2}{n}})$, which appears to be going to zero as n gets large using the algebraic limit theorem. However, I know that $\lim{b_n} = -1$. I'm stuck on how to prove this, and I haven't found any help elsewhere. Any advice would be helpful!

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    $\begingroup$ $$b_n=\frac{-2n}{n+\sqrt{n^2+2n}}=\frac{-2}{1+\sqrt{1+2/n}}$$ $\endgroup$ – Did Sep 20 '16 at 5:25
  • $\begingroup$ Some similar questions: math.stackexchange.com/questions/789772/… and math.stackexchange.com/questions/1766582/… (Notice that the latter is limit at $-\infty$, so after a simple substitution it becomes this limit.) $\endgroup$ – Martin Sleziak Sep 20 '16 at 6:47
  • $\begingroup$ A very naive intution: $\sqrt{n^2+2n}$ is rather close to $\sqrt{n^2+2n+1}=n+1$. So the whole expression is close to $(n+1)-1=1$. (But this is far from rigorous proof and similar arguments can lead to incorrect results if the change we make in the expression are not negligible. But in this case $\sqrt{n^2+2n+1}-\sqrt{n^2+2n}$ tends to zero for $n\to\infty$, so here this naive approach works. $\endgroup$ – Martin Sleziak Sep 20 '16 at 6:50

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