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?
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1You can rewrite this as $(1-z^n)^n$ where $0<z<1$. – Thomas Andrews Feb 18 '17 at 02:45
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Hint: Using Bernoulli's inequality
$$1 \geqslant \left(1-e^{-nk}\right)^n \geqslant 1 - n (e^{-k})^n$$
RRL
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METHODOLOGY $1$: Apply L'Hospital's Rule
If one wishes to use L'Hospital's Rule, then one can proceed by writing
$$\begin{align} \lim_{n\to \infty}\left(1-e^{-kn}\right)^n&=\lim_{n\to \infty}e^{n\log\left(1-e^{-kn}\right)}\tag 1\\\\ &=e^{\lim_{n\to \infty}n\log\left(1-e^{-kn}\right)}\\\\ &=e^{\lim_{n\to \infty}\frac{\log\left(1-e^{-kn}\right)}{1/n}}\\\\ &=e^{\lim_{n\to \infty}\frac{-kn^2}{e^{kn}-1}}\\\\ &=e^{0}\\\\ &=1 \end{align}$$
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.
Mark Viola
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Can I ask how you went from $\frac{\log\left(1-e^{-kn}\right)}{1/n}$ to $\frac{kn^2}{e^{kn}-1}$? – user321627 Feb 18 '17 at 02:43
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Applying L'Hospital's Rule, we get $$\lim \frac{ke^{-kn}/(1-e^{-kn})}{(-1/n^2)}$$Then, just simplify. ;-)) – Mark Viola Feb 18 '17 at 02:45