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...
 A: 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 $e^{-1}$.
A: Easier: write the original limit as $\bigg(\frac{1-\frac{1}{n}}{1+\frac{1}{n^2}} \bigg)^{n+10}$ and keep in mind $(1+\frac{1}{n^2})^n = e^{\frac{\log(1+\frac{1}{n^2}}{\frac{1}{n}}} \sim e^{\frac{1}{n}} = 1$. The last approximation uses Maclaurin expansion. 
Also keep in mind $(1+\frac{1}{n})^a \to_n 1$ if a is a constant. 
A: I think it's simpler to rewrite it as an exponential and use the MacLaurin expansion of the logarithm:
$$\begin{align}
\lim_{n\to \infty }{\left({n^2-n\over n^2+1}\right)^{n+10}} &= \lim_{n\to \infty }{\exp\left((n + 10)\ln\left(1 - {\frac{n + 1}{n^2+1}}\right)\right)} =\\
&= \lim_{n \to \infty}\exp\left(-\frac{(n + 10)(n + 1)}{n^2 + 1} + o(1)\right) = e^{-1}
\end{align}$$
So that you don't have to worry about moving limits inside and outside. However, if you really want to do it algebraically, you are right to say that $e^x$ must be continuous for that transformation to make sense.
As for the step
$$\lim_{n \to \infty}\left(1 + \frac1n\right)^{n\cdot a_n} = \lim_{n \to +\infty} \exp(a_n)$$
observe that it is valid only because we know a priori that
$$\lim_{n \to \infty}\left(1 + \frac1n\right)^n = e \in \mathbb R.$$
A: $$\frac{n^2-n}{n^2+1}=1-\frac{n+1}{n^2+1}=1-\frac1{\frac{n^2+1}{n+1}}=1-\frac1{n-\frac{n-1}{n+1}}\implies$$
$$\left(\frac{n^2-n}{n^2+1}\right)^{n+10}=\left(\frac{n^2-n}{n^2+1}\right)^{n-\frac{n-1}{n+1}}\cdot\left(\frac{n^2-n}{n^2+1}\right)^{10}\cdot\left(\frac{n^2-n}{n^2+1}\right)^{\frac{n-1}{n+1}}=$$
$$=\left(1-\frac1{n-\frac{n-1}{n+1}}\right)^{n-\frac{n-1}{n+1}}\left(\frac{n^2-n}{n^2+1}\right)^{10}\left(\frac{n^2-n}{n^2+1}\right)\left(\frac{n^2-n}{n^2+1}\right)^{-\frac2{n+1}}\xrightarrow[n\to\infty]{}e^{-1}\cdot1\cdot1\cdot1=e^{-1}$$
A: 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 $L = e^{-1} = 1/e$.
