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

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.

• Multiply by $\frac {(\sqrt[3]{n^2+5})^2 + (\sqrt[3]{n^2+5})(\sqrt[3]{n^2+3})+(\sqrt[3]{n^2+3})^2}{(\sqrt[3]{n^2+5})^2 + (\sqrt[3]{n^2+5})(\sqrt[3]{n^2+3})+(\sqrt[3]{n^2+3})^2}$ – Doug M Aug 23 '17 at 18:42

$$\lim_{x\to\infty}\left(\frac{\sqrt[3]{n² +5} \ - \ \sqrt[3]{n² + 3}}{1}\right)$$
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
HINT: use that $$a^3-b^3=(a-b)(a^2+ab+b^2)$$
• I don't quite see how that's going to work with $$a^{1/3} + b^{1/3}$$ – Leroy Aug 23 '17 at 18:50
• @leroy let $a = \sqrt[3]{n^2+5}, b = \sqrt[3]{n^2+3}$ and $(a-b) = \frac {2}{a^2 + ab + b^2}$ – Doug M Aug 23 '17 at 18:53
• it is $$(a^{1/3}+b^{1/3})(a^{2/3}-(ab)^{1/3}+b^{2/3})$$ – Dr. Sonnhard Graubner Aug 23 '17 at 18:53
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}