<s>[INST] Limit square roots of polynomials I am trying to find $\lim \limits_{n \to \infty} {\sqrt{n^3+1}-n\sqrt{n} \over \sqrt{n^2+1}-n}$. I rewrite the fraction as
$${(\sqrt{n^3+1}-n\sqrt{n})(\sqrt{n^3+1}+n\sqrt{n}) \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})} = {1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}$$
I notice $$\sqrt{n^2+1}-n > 0$$
so the denominator is growing to infinity while the whole limit is $0$. In the material I'm covering, the properties of polynomials haven't been discussed (but the properties of the square root have been). Is there a more basic way to conclude about the denominator going to infinity?
 [/INST] One may proceed the following way, as $n \to \infty$,
$$
\begin{align}
{1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}&={\sqrt{n^2+1}+n \over ((n^2+1)-n^2)\cdot(\sqrt{n^3+1}+n\sqrt{n})}
\\\\&={\sqrt{n^2+1}+n \over 1\cdot(\sqrt{n^3+1}+n\sqrt{n})}
\\\\&=\frac1{n^{1/2}}{\sqrt{1+1/n^2}+1 \over (\sqrt{1+1/n^3}+1)}
\\\\&\sim \frac1{n^{1/2}}
\end{align}
$$ then one may conclude easily.
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