<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] Here is an alternative approach which only uses "elementary" calculations (i.e. no MacLaurin or any other kind of expansion, no continuity, no $\log$, no $\exp$):
Note that 
$$\bigg(\frac{n^2-n}{n^2+1}\bigg)^{n+10}=\frac{\bigg(1-\frac{1}{n}\bigg)^n}{\bigg(1-\frac{1}{n^2}\bigg)^n}\cdot\frac{\bigg(1-\frac{1}{n}\bigg)^{10}}{\bigg(1-\frac{1}{n^2}\bigg)^{10}}$$
The last factor on the right hand side clearly converges to $1$. So, we only have to consider the term
$$\frac{\bigg(1-\frac{1}{n}\bigg)^n}{\bigg(1-\frac{1}{n^2}\bigg)^n}\tag{$*$}$$
Let us first compute the limit of the numerator of $(*)$:
$$\begin{align} 
\bigg(1-\frac{1}{n}\bigg)^n & = \bigg(\frac{n}{n-1}\bigg)^{-n} \\
                            & = \bigg(1+\frac{1}{n-1}\bigg)^{-n} \\
                            & = \bigg(1+\frac{1}{n-1}\bigg)^{-1}\bigg(1+\frac{1}{n-1}\bigg)^{-(n-1)} 
\end{align}$$
and this shows that the numerator tends to $e^{-1}$
Let us now compute the limit of the denominator of ($*$):
$$\begin{align}
\bigg(1-\frac{1}{n^2}\bigg)^n & = \bigg(\frac{n^2-1}{n^2}\bigg)^n \\
                              & = \bigg(\frac{n-1}{n}\cdot\frac{n+1}{n}\bigg)^n \\
                              & = \bigg(1-\frac{1}{n}\bigg)^n\bigg(1+\frac{1}{n}\bigg)^n \\
\end{align}$$
and therefore, the denominator tends to $1$.
This shows that the limit you are searching for is $e^{-1}$.
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