prove rate of growth of $(\frac{3 + \sqrt{17}}{2})^n$ is $o(n^{\sqrt{n}})$, $\Theta(n^{\sqrt{n}})$, or $\Omega(n^{\sqrt{n}})$

Question

Prove the rate of growth of $(\frac{3 + \sqrt{17}}{2})^n$ is $o(n^{\sqrt{n}})$, $\Theta(n^{\sqrt{n}})$, or $\Omega(n^{\sqrt{n}})$

I try to take the limit: $$\lim \limits_{n \to \infty}\frac{(\frac{3 + \sqrt{17}}{2})^n}{n^{\sqrt{n}}}$$ But when I try to apply L'Hopital Rule, the function becomes more complicated. How should I do?

Answer 1

Hint: For the $o(n^\sqrt{n})$, instead of using L'Hôpital, why not write the limit as: $$ \lim_{n\rightarrow \infty} \left(\frac{\frac{3+\sqrt{17}}{2}}{n^{\frac{1}{\sqrt{n}}}}\right)^n.$$ Then compute $\lim_{n\rightarrow \infty} \left(\frac{\frac{3+\sqrt{17}}{2}}{n^{\frac{1}{\sqrt{n}}}}\right).$

Answer 2

For $k \in \mathbb{R^+}$ fixed, $$\frac{k^n}{n^\sqrt{n}} = \frac{k^n}{\left( n^\frac{1}{\sqrt{n}}\right)^n}$$ and $$\lim_{n \to \infty} n^\frac{1}{\sqrt{n}} = \exp \lim_{n \to \infty} \frac{\log n}{\sqrt{n}} \log n = \exp \lim_{n \to \infty} \frac{1/n}{1/(2\sqrt{n})} = \exp 0 = 1$$ so therefore $$\lim_{n \to \infty} \frac{k}{n^\frac{1}{\sqrt{n}}} = k$$ and \begin{align*}\lim_{n \to \infty} \frac{k^n}{n^\sqrt{n}} = \begin{cases} 0 & 0 < k \leq 1 \\ \infty & k > 1. \end{cases} \end{align*} (The $k = 1$ limit has to be deduced from examining the original form $\frac{k^n}{n^\sqrt{n}}$ directly, as the rewrite as $\frac{k^n}{\left( n^\frac{1}{\sqrt{n}}\right)^n}$ is an indeterminate $1^\infty$ form.)