Limit with cube root conjugate Calculate the limit as $n -> \infty$ of $$n(\sqrt[3]{(1+1/n)} -1)$$
Summary of my work:


*

*Put n into the denominator in the form of $1/n$

*Multiplied by $n/n$

*Multiplied by conjugate of numerator: $[\sqrt[3]{(n^3 + n^2)^2} + 2\sqrt[3]{n^3 +n^2} + n^2]/[\sqrt[3]{(n^3 + n^2)^2} + 2\sqrt[3]{n^3 +n^2} + n^2]$

*Multiplied by $((1/n^2)/(1/n^2))$


Ended up with a final answer of 1/2

$$n\left(\sqrt[3]{1+\frac{1}{n}}-1\right)=n\left(\frac{\frac{1}{n}}{\sqrt[3]{(1+\frac{1}{n})^{2}}+\sqrt[3]{1+\frac{1}{n}}+1}\right)$$
which tends to $\frac{1}{3}$ as $n\to\infty$.
