For $\epsilon >0$, how to show that $\lim_{n \to \infty} (1-e^{n\epsilon})^{\frac{1}{1-\frac{1}{n}}} = 0$?

Question

For $\epsilon >0$, I would like to show that:

$$ \lim_{n \to \infty} \left(1-e^{n\epsilon}\right)^{\frac{n}{n-1}} = \lim_{n \to \infty} (1-e^{n\epsilon})^{\frac{1}{1-\frac{1}{n}}} = 0 $$

I can see it intuitively that $\frac{1}{1-\frac{1}{n}} \to 1$ so no matter what is under the exponent, it will be placed to the power of $1$. However, I can not see a formal way to do this. Does anyone have any ideas? Thanks.

Answer 1

This post was accepted as the answer but it was not correct as Paramanand Singh states.

Answer 2

You can use L'Hospital's rule for the exponent, but more importantly, I think you made an error in your question. The exponent over e goes to infinity without a minus sign, and so your entire expression would go to negative infinity. With the minus sign the term involving e goes to one and then you have 1-1=0, done.

Answer 3

$\lim_{n \to \infty} \left(1-e^{n\epsilon}\right)^{\frac{n}{n-1}} $

For any $\epsilon>0$, $e^{n\epsilon} \to \infty$ so the limit is $-\infty$.