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:

*

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

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

*$$a_n\sim\frac{9}{7}n^{-5/14},\quad n\to+\infty$$
 A: 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.
A: Let $b_n=\sqrt[7]{n+5}$ and let $c_n=\sqrt[7]{n-4}$. Then\begin{align}9&=b_n^{\,7}-c_n^{\,7}\\&=(b_n-c_n)\left(b_n^{\,6}+b_n^{\,5}c_n+b_n^{\,4}c_n^{\,2}+\cdots+c_n^{\,6}\right)\\&\geqslant(b_n-c_n)7\sqrt[7]{n-4}^6,\end{align}and therefore$$a_n=\sqrt n(b_n-c_n)\leqslant\frac{9\sqrt n}{7\sqrt[7]{n-4}^6}.$$Can you take it from here?
A: One more, multistorey, without series, way
$$\sqrt{n}(\sqrt[7]{n+5}-\sqrt[7]{n-4})=$$
$$=\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-\sqrt[7]{1-\frac{4}{n}}\right)
=\\
=\sqrt{n}\sqrt[7]{n}\left(\sqrt[7]{1+\frac{5}{n}}-1-\sqrt[7]{1-\frac{4}{n}}+1 \right) =\\
=\sqrt{n}\sqrt[7]{n}\left(\frac{\sqrt[7]{1+\frac{5}{n}}-1}{\frac{5}{n}}\cdot \frac{5}{n}-\frac{\sqrt[7]{1-\frac{4}{n}}-1}{\frac{4}{n}}\cdot \frac{4}{n}\right)
=\\
=\frac{\sqrt{n}\sqrt[7]{n}}{n}\left(\frac{\sqrt[7]{1+\frac{5}{n}}-1}{\frac{5}{n}}\cdot 5+\frac{\sqrt[7]{1-\frac{4}{n}}-1}{-\frac{4}{n}}\cdot 4\right) \sim \frac{\sqrt{n}\sqrt[7]{n}}{n}\left(\frac{5}{7}+\frac{4}{7}\right) = \\
=\frac{9}{7n^{\frac{5}{14}}}$$
where it was used $\frac{(1+x)^\alpha-1}{x}\to \alpha, x\to 0$.
