Limit Question (Calculus 2) Simply put, is there a way to show that $$\lim \limits_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$$
where $$a_n = \frac {(n^2-n)^{1+1/n}}{(1+\frac {1}{n})^\sqrt{(n^2-n)}}$$
This is a problem I may incorporate into a project for AP Calculus BC/Calculus 2, so using methods with that skill level in mind is preferable (if not possible, I will still appreciate a proof nonetheless).
The most progress I've made was through taking the natural logarithm and using those properties to arrive at $$\lim \limits_{n \to \infty} ln\left(\frac{a_{n+1}}{a_n}\right) = \lim \limits_{n \to \infty}\left[\left(1 + \frac{1}{n+1}\right)\ln\left(n^2+n\right)+\ln\left(1+\frac{1}{n}\right)\sqrt{n^2-n}-\left(1+\frac{1}{n}\right)\ln(n^2-n)-\ln\left(1+\frac{1}{n+1}\right)\sqrt{n^2+n}\right]$$
and although I can deal with the first and third terms using L'Hôpital's Rule, I was unable to finish off the second and fourth terms.
(Also, apologies if anything seemed confusing in the equations. I'm new to MathJax and the website in general).
Thank you kindly!
 A: If you already know and use Taylor series, there are some simple possible approches.
Considering $$a_n = \frac {(n^2-n)^{1+1/n}}{(1+\frac {1}{n})^\sqrt{n^2-n}}$$ as you did, $$\log(a_n)=\left(1+\frac 1n\right)\log(n^2-n)-\sqrt{n^2-n}\log\left(1+\frac 1n\right)$$ $$\log(a_n)=\left(1+\frac 1n\right)\left(2\log(n)+\log(1-\frac 1n)\right)-n\sqrt{1-\frac 1n}\log\left(1+\frac 1n\right)$$ Now, using the classical Taylor series, you could get $$\log(a_n)=\left(2 \log \left(n\right)-1\right)+\frac{2 \log
   \left(n\right)}{n}+O\left(\frac{1}{n^2}\right)$$ So, limiting to the first order $$\log(a_{n+1})-\log(a_n)=2\log\left(\frac{n+1}n\right)+O\left(\frac{1}{n}\right)=2\log\left(1+\frac{1}n\right)+O\left(\frac{1}{n}\right)= \frac 2 n+O\left(\frac{1}{n^2}\right)$$ Now, using $x=e^{\log(x)}$ and Taylor again $$\frac{a_{n+1}}{a_n}=1+\frac 2 n+O\left(\frac{1}{n^2}\right)$$ which shows the limit and also how it is approached.
Playing with my old pocket calculator $a_{10}\approx 57.1454$, $a_{11}\approx 67.7086$ which makes a ratio $\approx 1.18485$ while the Taylor expansion gives $1.2$. Similarly, $a_{100}\approx 4032.93$, $a_{101}\approx 4111.06$ which makes a ratio $\approx 1.01937$ while the Taylor expansion gives $1.02$ 
A: Hint:
L'Hospital's rule is not the alpha and omega of limits computation!
You can show  the log of the denominator tends to $1$, hence the denominator is equivalent to its limit $\mathrm e$. Here are some details:
\begin{align*}
\ln\biggl(\Bigl(1+\frac1n\Bigr)^{\sqrt{n^2-n}}\biggr)&=\sqrt{n^2-n}\ln\Bigr(1+\frac1n\Bigr)\\&=\sqrt{1-\frac1n}\,n\,\ln\Bigr(1+\frac1n\Bigr)\\&=\sqrt{1-u}\,\frac{\ln(1+u)}{u},\quad\text{setting}\enspace u=\frac1n.
\end{align*}
As to the numerator, split it as
$$(n^2-n)^{1+1/n}=(n^2-n)\,n^{2/n}\Bigl(1-\frac1n\Bigr)^{1/n}.$$
Now $n^2-n\sim_\infty n^2$ and the last two factors are equivalent to their limit $1$, hence
$$(n^2-n)^{1+1/n}\sim_\infty n^2,\enspace\text{so}\quad a_n\sim_\infty \frac{n^2}{\mathrm e}.$$
A: Noting $\sqrt {n^2-n} = n\sqrt {1-1/n},$ we have
$$ \frac{a_n}{n^2/e} = \frac{n^2-n}{n^2}\cdot \frac{e}{[(1+1/n)^n]^{\sqrt {1-1/n}} }\cdot(n^2-n)^{1/n}.$$
The first fraction on the right $\to 1.$ In the second fraction, the term in brackets in the denominator $\to e,$ and the power $\sqrt {1-1/n}\to 1.$ So the full denominator $\to e^1 = e.$ Hence the second fraction $\to 1.$ For the last term we have $1\le n^2-n \le n^2$ for $n>1.$ Hence $1\le (n^2-1)^{1/n}\le (n^2)^{1/n}$ $ = (n^{1/n})^2.$ I take as knows that $n^{1/n} \to 1.$ Hence the last term $\to 1.$
Conclusion: $a_n/(n^2/e) \to 1.$ Therefore
$$\frac{a_{n+1}}{a_n} = \frac{a_{n+1}/[(n+1)^2/e]}{a_n/[n^2/e]}\frac{(n+1)^2/e}{n^2/e} \to \frac{1}{1} \cdot 1 = 1.$$
A: You can use this identity - If $\lim_{n \to \infty }\frac{a_{n+1}}{a_n}=l$ then $\lim_{n\to\infty} (a_n)^{1/n}=l.$ So you just havr to find the limit 
$$L=\lim_{n\to\infty} (a_n)^{1/n}=\lim_{n\to\infty}[{ \frac {(n^2-n)^{1+1/n}}{(1+\frac {1}{n})^\sqrt{(n^2-n)}}}]^{\frac {1}{n}}$$
So now just take ln both the sides and put $n=\frac {1}{t}$, so limit will change to $$lnL=\lim_{t\to\ 0}t[(1+t)ln(\frac{1-t}{t^2})-\frac{\sqrt{1-t}}{t}ln(1+t)]$$
Now, you can solve this limit very easily which is equal to zero. So
$$lnL=0$$ So $$L=1$$
Hence Proved
