Limit square roots of polynomials

Question

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?

Answer 1

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.

Answer 2

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 $${\sqrt{n^3+1}-\sqrt{n^3} \over \sqrt{n^2+1}-n}=\dfrac{1}{n^{1/2}}\left({\sqrt{1+\dfrac{1}{n^2 }}+1 \over \sqrt{1+\dfrac{1}{n^3}}+1}\right)\to 0.$$