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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.

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    Are you sure the exponent is $n\epsilon$, not $-n\epsilon$? – 2017-02-14
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    (i.e., how do you define $a^x$ when $a<0$ and $x\in\mathbb{R}$, otherwise?) – 2017-02-14
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    Hi, yes, that is how my term appears as when I was calculating some error bounds. – 2017-02-14
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    That must be a mistake, then. This is not a well-defined quantity. – 2017-02-14
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    Fully agree with @ClementC. I wonder a bit as there are answers given for a wrong problem. The expression is not a real number if $\epsilon>0$ and $n$ is odd. – 2017-07-26

3 Answers 3

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

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    You seem to assume that $-1/e^{n\epsilon}\to - \infty$ but it tends to $0$. -1 – 2017-07-26
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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.

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$\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 $-\infty$.

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    Slight problem if $n$ is odd as the expression is no longer real. – 2017-07-26
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    True. The stated problem is probably not what OP intended. – 2017-07-26