Limit with cube root conjugate

Question

Calculate the limit as $n -> \infty$ of $$n(\sqrt[3]{(1+1/n)} -1)$$

Summary of my work:

1. Put n into the denominator in the form of $1/n$
2. Multiplied by $n/n$
3. Multiplied by conjugate of numerator: $[\sqrt[3]{(n^3 + n^2)^2} + 2\sqrt[3]{n^3 +n^2} + n^2]/[\sqrt[3]{(n^3 + n^2)^2} + 2\sqrt[3]{n^3 +n^2} + n^2]$
4. Multiplied by $((1/n^2)/(1/n^2))$

Ended up with a final answer of 1/2

Answer 1

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

which tends to $\frac{1}{3}$ as $n\to\infty$.