limit, quotient of roots How can I find
$$
\lim_{n \to \infty} \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt{n + 2} - \sqrt{n + 1}} \sqrt[6]{n - 3}?
$$
If I multiply by $\sqrt{n + 2} + \sqrt{n + 1}$ I could get no divisor, but I cannot get any result on this way. On the other hand I could also transform this expression:
$$
\frac{(n + 2)^{1 / 3} \left(1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 3}\right)}{(n + 2)^{1 / 2} \left(1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 2}\right)} \sqrt[6]{n - 3} = {\left(\frac{n - 3}{n + 2}\right)}^{1 / 6} \frac{1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 3}}{1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 2}}.
$$
However I also cannot see how continue to see that the limit is $\frac 2 3$. Any help would be appreciated.
 A: Your idea of multiplying and dividing by $\sqrt{n+2}+\sqrt{n+1}$ (or by the conjugate) is a good one, and we can do the same thing for the conjugate of the numerator. Since the denominator had square roots, its conjugate came from the difference of squares formula
$$a-b = (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})$$
Similarly, since the numerator has cube roots, its conjugate must come from the difference of cubes formula
$$a-b = (\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a}^2+\sqrt[3]{a}\sqrt[3]{b}+\sqrt[3]{b}^2)$$
Applying these factors the expression, we obtain
$$\lim_{n\to\infty} \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt{n + 2} - \sqrt{n + 1}} \cdot \frac{\sqrt{n+2}+\sqrt{n+1}}{\sqrt{n+2}+\sqrt{n+1}} \cdot \frac{\sqrt[3]{n+2}^2+\sqrt[3]{n+2}\sqrt[3]{n+1}+\sqrt[3]{n+1}^2}{\sqrt[3]{n+2}^2+\sqrt[3]{n+2}\sqrt[3]{n+1}+\sqrt[3]{n+1}^2} \cdot\sqrt[6]{n - 3} $$
$$= \lim_{n\to\infty} \frac{\sqrt{n+2}+\sqrt{n+1}}{\sqrt[3]{n+2}^2+\sqrt[3]{n+2}\sqrt[3]{n+1}+\sqrt[3]{n+1}^2}\cdot\sqrt[6]{n - 3}$$
Then pull factors of $n$ from their respective roots
$$\lim_{n\to\infty} \frac{\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}}{\sqrt[3]{1+\frac{2}{n}}^2+\sqrt[3]{1+\frac{2}{n}}\sqrt[3]{1+\frac{1}{n}}+\sqrt[3]{1+\frac{1}{n}}^2}\cdot\sqrt[6]{1 - \frac{3}{n}} \cdot \frac{n^{\frac{1}{2}}n^{\frac{1}{6}}}{n^{\frac{2}{3}}} = \frac{1+1}{1+1+1} = \frac{2}{3}$$
A: Put $ m=n+1$.
The limit becomes after factorisation and simplification
$$\lim_{m\to\infty}\frac{(1+\frac 1m)^{\frac 13}-1}{(1+\frac 1m)^{\frac 12}-1}(1-\frac 4m)^{\frac 16}$$
Use the result
$$\lim_{N\to \infty}N\Bigl((1+\frac 1N)^\alpha-1\Bigr)=\alpha$$
