How to compute the limit of $\lim_{n\to \infty}\left(1-e^{-nk}\right)^n$ for $k>0$ a constant?

Question

I am wondering how I can compute the limit of $\lim_{n\to \infty}\left(1-e^{-nk}\right)^n$ for $k>0$ a constant. I want to think L'Hospital's rule would work here but am unable to get it. Is there an easy trick here?

Answer 1

Hint: Using Bernoulli's inequality

$$1 \geqslant \left(1-e^{-nk}\right)^n \geqslant 1 - n (e^{-k})^n$$

Answer 2

METHODOLOGY $1$: Apply L'Hospital's Rule

If one wishes to use L'Hospital's Rule, then one can proceed by writing

$$\begin{align} \lim_{n\to \infty}\left(1-e^{-kn}\right)^n&=\lim_{n\to \infty}e^{n\log\left(1-e^{-kn}\right)}\tag 1\\\\ &=e^{\lim_{n\to \infty}n\log\left(1-e^{-kn}\right)}\\\\ &=e^{\lim_{n\to \infty}\frac{\log\left(1-e^{-kn}\right)}{1/n}}\\\\ &=e^{\lim_{n\to \infty}\frac{-kn^2}{e^{kn}-1}}\\\\ &=e^{0}\\\\ &=1 \end{align}$$

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METHODOLOGY $2$: Apply Elementary Inequalities and the Squeeze Theorem

Alternatively, we can use the fact that the logarithm function satisfies the inequalities SEE THIS ANSWER

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le\log(x)\le x-1 }\tag 2$$

Using $(2)$ in $(1)$ we have

$$e^{\frac{-ne^{-kn}}{1-e^{-kn}}}\le e^{n\log\left(1-e^{-kn}\right)}\le e^{-ne^{-kn}}$$

whence application of the squeeze theorem yields the coveted limit.