How to find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$? I know that it exists and can't find it with Stolz theorem. How do I find $\lim_{n\to+\infty} \frac{n^2}{7^{\sqrt{n}}}$?
I thought about Stolz theorem so I got:
$$ \frac{(n+1)^2-n^2}{7^{\sqrt{n+1}}-7^{\sqrt{n}}}, $$
but it doesn't seem to lead anywhere (at least not through $e$ and $\ln$).
And I know for fact that there is some limit because:
$$\lim_{n \to +\infty} \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)^2}{7^{\sqrt{n+1}}}}{\frac{n^2}{7^{\sqrt{n}}}} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n+1}}*n^2} = \frac{(n+1)^2*7^{\sqrt{n}}}{7^{\sqrt{n}\sqrt{\frac{1}{n}+1}}*n^2} \implies \frac{(n+1)^2}{n^2}\implies 1$$
 A: Somewhere along the timeline you learned that exponential growth wipes out polynomial growth. So if we let $n=x^2,$ we are looking at
$$\lim_{x\to \infty}\frac{x^4}{7^x}.$$
This is polynomial growth divided by exponential growth, and the limit is $0.$
A: Hint: it suffices to consider the subsequence $n_k = k^2$.
A: This is $\lim\limits_{n\to\infty}\exp\left(2\ln n-\sqrt n\ln 7\right)=0$, because $\lim\limits_{x\to\infty}2\ln x-\sqrt x\ln 7=-\infty$.
For the record, you should be warned that Riemann rearrangement theorem implies that for all $0\le x\le y\le\infty$ there is a sequence $a_n>0$ such that $$\begin{cases}\liminf\limits_{n\to\infty}a_n=x\\ \limsup\limits_{n\to\infty}a_n=y\\ \lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1\end{cases}$$
Therefore $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=1$ is inconsequential. Also, while the value of this limit is certainly $1$, I think you made some mistake either of algebra or in manipulation of limits; I won't inspect it any further, because I am not a tutor.
