Evaluate $\lim_{n\to{\infty}}\frac{e(1-\frac{1}{n})^n-1}{n^{\alpha}}=c$

Question

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.

Answer 1

$\begin{array}\\ (1-1/n)^n &=\exp(n\ln(1-1/n))\\ &=\exp(-n(1/n+1/(2n^2)+O(1/n^3)))\\ &=\exp(-1-1/(2n)+O(1/n^2))\\ &=\exp(-1/(2n)+O(1/n^2))/e\\ &=(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 $12(c-a)=12(-1/2-(-1))=12(1/2)=6$.