<s>[INST] Limit problems and quandaries: finding $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. Find  $\lim_\limits{n\to \infty } {({n^2-n\over n^2+1})^{n+10} }$. What I did is: 
$\lim_\limits{n\to \infty }{({n^2-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty } {({n^2+1-1-n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty }{(1-{1+n\over n^2+1})^{n+10}}=\lim_\limits{n\to \infty}[{(1-{1+n\over n^2+1})^{ n^2+1\over 1+n}}]^{(n+10)(1+n)\over (n^2+1)}$. Denoting: ${ n^2+1\over 1+n}=t$ ($t\to \infty$ as $n\to \infty$) we get: $\lim_\limits{n\to \infty}[{(1+{(-1)\over t})^{ t}}]^{(n+10)(1+n)\over (n^2+1)}=\lim_\limits{n\to \infty}e^{-{(n+10)(1+n)\over (n^2+1)}}=e^{\lim_\limits{n\to \infty}-{(n+10)(1+n)\over (n^2+1)}}=e^{-1}$. 
My question is: how can I know it is defined and lawful? I know I can use the continuity of $e$, but I don't know if it is okay that I separated the limits that way(Because maybe the first limit is not even $e$ considering other things)
.I Would appreciate your help...
 [/INST] It is much simpler to take logs. We need to use the following theorem for that purpose.
If $a_{n} > 0$ then $\lim_{n \to \infty}a_{n}$ exists if and only if $\lim_{n \to \infty}\log a_{n}$ exists or is $-\infty$. If $\lim_{n \to \infty}\log a_{n} = A$ then $\lim_{n \to \infty}a_{n} = e^{A}$ and if $\log a_{n} \to -\infty$ then $a_{n} \to 0$.
Supposing then that the desired limit is $L$, we have $$\begin{aligned}\log L &= \log\left\{\lim_{n \to \infty}\left(\frac{n^{2} - n}{n^{2} + 1}\right)^{n + 10}\right\}\\
&= \lim_{n \to \infty}\log\left(\frac{n^{2} - n}{n^{2} + 1}\right)^{n + 10}\text{ (by continuity of log)}\\
&= \lim_{n \to \infty}(n + 10)\log\left(\frac{n^{2} - n}{n^{2} + 1}\right)\\
&= \lim_{n \to \infty}n\log\left(\frac{n^{2} - n}{n^{2} + 1}\right) + 10\cdot\log\left(\frac{n^{2} - n}{n^{2} + 1}\right)\\
&= \lim_{n \to \infty}n\log\left(\frac{n^{2} - n}{n^{2} + 1}\right) + 10\cdot\log 1\\
&= \lim_{n \to \infty}n\log\left(1 - \frac{n + 1}{n^{2} + 1}\right)\\
&= \lim_{n \to \infty}n\left(-\dfrac{n + 1}{n^{2} + 1}\right)\dfrac{\log\left(1 - \dfrac{n + 1}{n^{2} + 1}\right)}{\left(-\dfrac{n + 1}{n^{2} + 1}\right)}\\
&= \lim_{n \to \infty}n\left(-\dfrac{n + 1}{n^{2} + 1}\right)\cdot\lim_{x \to 0}\frac{\log(1 + x)}{x}\text{ (by putting }x = -(n + 1)/(n^{2} + 1))\\
&= -\lim_{n \to \infty}\dfrac{n^{2} + n}{n^{2} + 1}\\
&= -\lim_{n \to \infty}\dfrac{1 + \dfrac{1}{n}}{1 + \dfrac{1}{n^{2}}}\\
&= -1\end{aligned}$$ Hence $L = e^{-1} = 1/e$.
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