Evaluate $\lim_{n\to{\infty}}\frac{e(1-\frac{1}{n})^n-1}{n^{\alpha}}=c$ If $$\lim_{n\to{\infty}}\frac{e(1-\frac{1}{n})^n-1}{n^{\alpha}}=c$$
  
  where $c$ is a non zero constant. Then evaluate $12(c-\alpha)$.


My Attempt:

I know that $$\lim_{n\to{\infty}}\bigg(1-\frac{1}{n}\bigg)^n=\frac{1}{e}$$

i.e. as $$\lim_{n\to{\infty}}{e\bigg(1-\frac{1}{n}\bigg)^n-1}=0$$

And as the question says limit is non zero, we can conclude that $\frac{e(1-\frac{1}{n})^n-1}{n^{\alpha}}$ is an $\frac{0}{0}$ form.

i.e. $$\lim_{n\to{\infty}}n^{\alpha}=0$$

From this I can conclude that $\alpha\lt{0}$.

But I am not able to solve this further. I know that the exact value of $\alpha$ can be found using the fact that the limit exists and is a finite non zero constant. But I don't know how to approach this? I tried using L-Hopital but it wasn't of much help. How do I proceed?

Thank you.
$\begin{array}\\
(1-1/n)^n 
&amp;=\exp(n\ln(1-1/n))\\
&amp;=\exp(-n(1/n+1/(2n^2)+O(1/n^3)))\\
&amp;=\exp(-1-1/(2n)+O(1/n^2))\\
&amp;=\exp(-1/(2n)+O(1/n^2))/e\\
&amp;=(1-1/(2n)+O(1/n^2))/e\\
\end{array}
$

so the numerator is
$-1/(2n)+O(1/n^2)$.

For the limit to exist, we must have
$a=-1$.
The limit is then
$c=-1/2$
so
$12(c-a)=12(-1/2-(-1))=12(1/2)=6$.
Hint:

By the Binomial theorem,
$$\left(1-\frac1n\right)^n=1-\frac nn+\frac{n(n-1)}{2n^2}-\frac{n(n-1)(n-2)}{3!n^3}+\cdots$$

and this should be made close to 

$$\frac{1+cn^\alpha}e.$$

Develop the RHS in terms of $1$ and $n^{-1}$.