Find the limit $\sqrt[3]{n^3+n^2} - \sqrt[3]{n^3+1}$

Question

how to find the limit of

$$\sqrt[3]{n^3+n^2} - \sqrt[3]{n^3+1}$$

I tried to multiply it by

$$\frac{\sqrt[3]{n^3+n^2} + \sqrt[3]{n^3+1}}{\sqrt[3]{n^3+n^2} + \sqrt[3]{n^3+1}}$$

once or twice, also tried to multiply by :

$$\frac{\sqrt[3]{(n^3+n^2)^2} + \sqrt[3]{(n^3+1)^2}}{\sqrt[3]{(n^3+n^2)^2} + \sqrt[3]{(n^3+1)^2}}$$

but i couldnt get rid of the roots

Answer 1

Hint:

$$ a^3-b^3=(a-b)\left(a^2+ab+b^2\right)\ \Longrightarrow\ a-b=\dfrac{a^3-b^3}{a^2+ab+b^2}. $$

Now make a suitable choice for $a$ and $b$ that will simplify your limit.

Answer 2

Use

$$a-b = \frac{a^3-b^3}{a^2+a b+b^2} $$

where $a=\sqrt[3]{n^3+n^2}$ and $b=\sqrt[3]{n^3+1}$.

Answer 3

Using the powerful tool: Taylor expansion we get

$$\sqrt[3]{n^3+n^2} - \sqrt[3]{n^3+1}=n\left(1+\frac1n\right)^{1/3}-n\left(1+\frac1{n^3}\right)^{1/3}\sim_\infty n+\frac13-n-\frac1{3n^2}$$ hence we see that the desired limit is $\frac13$.