How do I compute the limit of a radical outside of a fraction?

Question

How do I compute this?

$$\lim_{x\to\infty}\left(\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}\right)$$

I tried multiplying it by the conjugate exactly as if it were in a fraction, but it didn’t quite work out, because you end up with a way bigger expression.

Arriving at the solution likely uses the sandwich theorem, but with two separate expressions. I find it to be quite the challenge.

Answer 1

$$\lim_{x\to\infty}\left(\frac{\sqrt[3]{n² +5} \ - \ \sqrt[3]{n² + 3}}{1}\right)$$

Multiply numerator and denominator by $$\left(\sqrt[3]{n² +5}\right)^2+\left(\sqrt[3]{n² +5}\right)\left(\sqrt[3]{n² +3}\right)+\left(\sqrt[3]{n² +3}\right)^2$$ the fraction becomes $$\frac{\left(\sqrt[3]{n² +5}-\sqrt[3]{n² + 3}\right)\left[\left(\sqrt[3]{n² +5}\right)^2+\left(\sqrt[3]{n² +5}\right)\left(\sqrt[3]{n² +3}\right)+\left(\sqrt[3]{n² +3}\right)^2\right]}{\left(\sqrt[3]{n² +5}\right)^2+\left(\sqrt[3]{n² +5}\right)\left(\sqrt[3]{n² +3}\right)+\left(\sqrt[3]{n² +3}\right)^2}$$ Recalling the identity $(a-b)(a^2+ab+b^2)=a^3+b^3$ we get $$\lim_{x\to\infty}\frac{(n^2+5)-(n^2+3)}{\sqrt[3]{(n² +5)^2}+\sqrt[3]{(n² +5)(n^2+3)}+\sqrt[3]{(n² +3)^2}}=$$ $$=\lim_{x\to\infty}\frac{2}{\sqrt[3]{(n² +5)^2}+\sqrt[3]{(n² +5)(n^2+3)}+\sqrt[3]{(n² +3)^2}}=0$$ Hope this helps

Answer 2

HINT: use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$

Answer 3

Use Taylor's formula at order $1$ near $0$: $$(1+u)^{\tfrac13}=1+\frac u3+o(u).$$ Your expression rewrites as \begin{align} \sqrt[3]{n² +5} \ - \ \sqrt[3]{n² + 3}&=n^{\tfrac23}\biggl[\biggl(1+\frac 5{n^2}\biggr)^{\tfrac13}-\biggl(1+\frac 3{n^2}\biggr)^{\tfrac13}\biggr]\\ &=n^{\tfrac23}\biggl[\biggl(1+\frac 5{3n^2}\biggr)-\biggl(1+\frac 3{3n^2}\biggr)+o\biggl(\frac1{n^2}\biggr)\biggr]\\ &=\frac 2{3n^{\tfrac43}}+o\Biggl(\frac1{n^{\tfrac43}}\Biggr)\to 0\quad\text{as}\enspace n\to \infty. \end{align}