n-th root in limits I'm having trouble with this kind of series: $$a_n = \sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})$$
I tried to make something like perfect square to simplify those $7$th root junk which is obviously impossible, and as $n$ tends to infinity L'Hopital's rule also can't solve the problem in this form.
The multiple choice:

*

*$$\lim_{n\to\infty}a_n=\frac{9}{7}$$

*$$\lim_{n\to\infty}a_n=+\infty$$

*$$a_n\sim\frac{9}{7}n^{-5/14},\quad n\to+\infty$$

Rewrite it $$\sqrt{n}\left(\sqrt[7]{n+5} - \sqrt[7]{n-4}\right) = \\ n^{\frac{9}{14}}\left((1+\frac{5}{n})^{\frac{1}{7}}-(1-\frac{4}{n})^{\frac{1}{7}}\right) \\ \approx n^{\frac{9}{14}}\left(1+\dfrac{1}{7}\cdot\dfrac{5}{n} - 1 +\dfrac{1}{7}\cdot\dfrac{4}{n} + o(\dfrac{1}{n})\right) \\ \approx \dfrac{9}{7n^{\frac{5}{14}}}$$. Hence the answer is choice $4$-th down the list.
