Limit of multiple third roots without L'Hospital

Question

I'm not able to solve the limit of a textbook question.

The limit:

$$\lim_{x\to \infty} (\sqrt[3]{x^2}(\sqrt[3]{x+1} - \sqrt[3]{x}))$$

I've been able to simplify the limit to: $$\lim_{x\to \infty} (\sqrt[3]{x^3+x^2} - x)$$

How do I solve this limit?

Note: no L'Hospital allowed.

Answer 1

Using $$a-b=\frac{a^3-b^3}{a^2+ab+b^2}$$ gives $$\sqrt[3]{x^3+x^2}-x=\frac{x^2}{(x^3+x^2)^{2/3}+x(x^3+x^2)^{1/3}+x^2} =\frac{1}{(1+x^{-1})^{2/3}+(1+x^{-1})^{1/3}+1}\to\frac13$$ as $x\to\infty$.

Answer 2

Pull out $x^{1/3}$ You get $\lim x((1+\frac 1 x)^{1/3}-1) =\lim x(1+\frac 1 {3x} +o(\frac 1x )-1)=\frac 1 3$.

Answer 3

Set $x=1/y^3$ to find

$$\lim_{y\to0}\dfrac1{y^2}\cdot\dfrac{\sqrt[3]{1+y^3}-1}y$$

Set $\sqrt[3]{1+y^3}-1=z\implies y^3=3z+3z^2+z^3$

So, the limit should be $$\dfrac13$$