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?

Observe that $${\sqrt{n^3+1}-\sqrt{n^3} \over \sqrt{n^2+1}-n}=\dfrac{n^{3/2}}{n}\left({\sqrt{1+\dfrac{1}{n^3
}}-1 \over \sqrt{1+\dfrac{1}{n^2}}-1}\right)$$ and by multiplying conjugates of both top and bottom yields $${\sqrt{1+\dfrac{1}{n^3
}}-1 \over \sqrt{1+\dfrac{1}{n^2}}-1}=\dfrac{\dfrac1{n^3}}{\dfrac1{n^2}}\left({\sqrt{1+\dfrac{1}{n^2
}}+1 \over \sqrt{1+\dfrac{1}{n^3}}+1}\right).$$ Combinig these results we have $${\sqrt{n^3+1}-\sqrt{n^3} \over \sqrt{n^2+1}-n}=\dfrac{1}{n^{1/2}}\left({\sqrt{1+\dfrac{1}{n^2
}}+1 \over \sqrt{1+\dfrac{1}{n^3}}+1}\right)\to 0.$$
