Limit proof involving "e" How would you prove the following?
$$\lim_{n \rightarrow \infty} (1+c/n^2 - 2/n)^n = e^{-2}$$
where $c\geq 1$ is a constant.
I can find proofs online of $\lim_{n \rightarrow \infty} (1-2/n)^n = e^{-2}$ but how do you make the above rigorous?
 A: Note that
$$\frac 1 {n^2}-\frac 2 n +1=\left(\frac 1 n -1\right)^2$$
ADD
Let 
$$x_n= \left(1- \frac 2 n+\frac c {n^2}\right)^n $$
so that $$y_n=\log x_n=n \log \left(1- \frac 2 n+\frac c {n^2}\right)$$
Then
$$n{\log \left(1- \frac 2 n+\frac c {n^2}\right)}=$$
$$n\left(\frac c {n^2}- \frac 2 n\right)\frac{\log \left(1+\left(\frac c {n^2}- \frac 2 n\right)\right)}{\left(\frac c {n^2}- \frac 2 n\right)}=$$
$$\left(\frac c {n}- 2\right)\frac{\log \left(1+a_n\right)}{a_n}=$$
Now $a_n\to 0$; so that
$$\left(\frac c {n}- 2\right)\frac{\log \left(1+a_n\right)}{a_n}\to (0-2)\cdot 1 =-2$$
This means $x_n\to e^{-2}$
A: $$\lim _{n\to\infty} \left(\left(1+\frac{c-2n}{n^2}\right)^{\frac{n^2}{c-2n}}\right)^{\frac{c-2n}{n}}$$
$$= \left(\lim _{n\to\infty}\left(1+\frac{c-2n}{n^2}\right)^{\frac{n^2}{c-2n}}\right)^{\lim _{n\to\infty}\left(\frac{c-2n}n\right)}$$
$$= \left(\lim _{m\to\infty}\left(1+\frac1m\right)^m\right)^{\lim _{n\to\infty}\left(\frac{c-2n}n\right)}$$ where $m=\frac{n^2}{c-2n},n\to \infty\implies m\to\infty$
$$=e^{-2}$$ for any finite value of $c$
